"This is a really fun forum game that i want to start it works like this, i start to tell a story, but i only include the first 3 words, then the next person copys what i wrote via copy and paste and adds anouther 3 words, next person does same, and so on "

also, please make the words you add in [i]italics[/i] so people can see what's new.

when a paragraph gets too long, it doesn't cost anything to make a new paragraph.

[i]Once upon a[/i]

Once upon a time. The End. [i]Once again a[/i]

Once upon a time. The End. Once again an [i]End is near[/i]

Once upon a time. The End. Once again an End is near [i]but also far[/i]

Once upon a time. The End. Once again an End is near but also far [i]from clucking chickens![/i]

Once upon a time. The End. Once again an End is near but also far from clucking chickens! [i]There lived a [/i]

Once upon a time. The End. Once again an End is near but also far from clucking chickens! There lived a [i]man who could[/i]

Once upon a time. The End. Once again an End is near but also far from clucking chickens! There lived a man who could [i]divide by zero![/i]

Once upon a time. The End. Once again an End is near but also far from clucking chickens! There lived a man who could divide by zero! [i]But only on[/i]

Once upon a time. The End. Once again an End is near but also far from clucking chickens! There lived a man who could divide by zero! But only on [i]Chuck Norris's calculator![/i]

Once upon a time. The End. Once again an End is near but also far from clucking chickens! There lived a man who could divide by zero! But only on Chuck Norris's calculator! [i]But it's not[/i]

Once upon a time. The End. Once again an End is near but also far from clucking chickens! There lived a man who could divide by zero! But only on Chuck Norris's calculator! But it's not [i] Norris's, it's Jesus's.[/i]

Once upon a time. The End. Once again an End is near but also far from clucking chickens! There lived a man who could divide by zero! But only on Chuck Norris's calculator! But it's not Norris', it's Jesus'.[i]So to get[/i]

Once upon a time. The End. Once again an End is near but also far from clucking chickens! There lived a man who could divide by zero! But only on Chuck Norris's calculator! But it's not Norris', it's Jesus'. So to get [i]those infidel Muslims[/i]

Once upon a time. The End. Once again an End is near but also far from clucking chickens! There lived a man who could divide by zero! But only on Chuck Norris's calculator! But it's not Norris', it's Jesus'. So to get those infidel Muslims, one must calculate the square root of a monkey, as accurately as Jesus will allow. Ekad! He's almost a constipated penguin. That will slow down the universe to the speed of a bad Bowling Ball, on a muddy field.

Since there was little speed, the bowling ball exploded, and little flowers sprouted and became various Pokemon. Then they exploded too into various Pokemon. A challenger appears! Named Chuck Norris/Jesus. He thinks 1=2. Which is true. The human race wants to lynch the man who proved that 1=2. On the way, Pokemon appeared again! Pikachu attacks! It explodes into little jackdillons that eat dirt. The end.

After the end, God said, "Let epsilon be positive." Gauss remarked that God also said: "People shall be defined as compact disks of love if for all people, there exists a positive delta within every twinkie. Even though twinkies are actually pokemon, who have twinkies which are pokemon and other creatures" The Grammar Police saw that there was no pizza up their butts with more Pokemon food. They started a Pokemon battle.

...he woke up from his long Pokemon battle nap and he saw that his pizza came alive and wanted a cookie. It craved the chocolate chips that killed the dinosaurs. It knew that it could run at the speed of cookies, but alas! It got eaten! It started anew in the bakery, when a new purple attacked him, and he died.

"Respawn point???" he screamed as he lost The Game. He is no good. He should end this blasphemy. He pushed a big slice of roast chicken up the hill, where Nurse Joy awaited with a Chansey. Nobody likes Chansey. Not even my extremely dead grandmother can endure the pain of perdition. Nor my extremely old and fat tabby cat had endured the pain of frolicking in the realm of pokemon and dragons.

Why is the cookie doe not disposed off properly? [/i]Because Kamikazee watermelons

[hide="Click to reveal the story which is exceeding many words! Also, please close your italics tags properly."]Once upon a time. The End. Once again an End is near but also far from clucking chickens! There lived a man who could divide by zero! But only on Chuck Norris's calculator! But it's not Norris', it's Jesus'. So to get those infidel Muslims, one must calculate the square root of a monkey, as accurately as Jesus will allow. Ekad! He's almost a constipated penguin. That will slow down the universe to the speed of a bad Bowling Ball, on a muddy field.

Since there was little speed, the bowling ball exploded, and little flowers sprouted and became various Pokemon. Then they exploded too into various Pokemon. A challenger appears! Named Chuck Norris/Jesus. He thinks 1=2. Which is true. The human race wants to lynch the man who proved that 1=2. On the way, Pokemon appeared again! Pikachu attacks! It explodes into little jackdillons that eat dirt. The end.

After the end, God said, "Let epsilon be positive." Gauss remarked that God also said: "People shall be defined as compact disks of love if for all people, there exists a positive delta within every twinkie. Even though twinkies are actually pokemon, who have twinkies which are pokemon and other creatures" The Grammar Police saw that there was no pizza up their butts with more Pokemon food. They started a Pokemon battle.

...he woke up from his long Pokemon battle nap and he saw that his pizza came alive and wanted a cookie. It craved the chocolate chips that killed the dinosaurs. It knew that it could run at the speed of cookies, but alas! It got eaten! It started anew in the bakery, when a new purple attacked him, and he died.

"Respawn point???" he screamed as he lost The Game. He is no good. He should end this blasphemy. He pushed a big slice of roast chicken up the hill, where Nurse Joy awaited with a Chansey. Nobody likes Chansey. Not even my extremely dead grandmother can endure the pain of perdition. Nor my extremely old and fat tabby cat had endured the pain of frolicking in the realm of pokemon and dragons.

Why is the cookie doe not disposed off properly? Because Kamikazee watermelons [i]decided to bomb[/i]

[/hide]

[hide="Click to reveal the story that is very long"]
Once upon a time. The End. Once again an End is near but also far from clucking chickens! There lived a man who could divide by zero! But only on Chuck Norris's calculator! But it's not Norris', it's Jesus'. So to get those infidel Muslims, one must calculate the square root of a monkey, as accurately as Jesus will allow. Ekad! He's almost a constipated penguin. That will slow down the universe to the speed of a bad Bowling Ball, on a muddy field.

Since there was little speed, the bowling ball exploded, and little flowers sprouted and became various Pokemon. Then they exploded too into various Pokemon. A challenger appears! Named Chuck Norris/Jesus. He thinks 1=2. Which is true. The human race wants to lynch the man who proved that 1=2. On the way, Pokemon appeared again! Pikachu attacks! It explodes into little jackdillons that eat dirt. The end.

After the end, God said, "Let epsilon be positive." Gauss remarked that God also said: "People shall be defined as compact disks of love if for all people, there exists a positive delta within every twinkie. Even though twinkies are actually pokemon, who have twinkies which are pokemon and other creatures" The Grammar Police saw that there was no pizza up their butts with more Pokemon food. They started a Pokemon battle.

...he woke up from his long Pokemon battle nap and he saw that his pizza came alive and wanted a cookie. It craved the chocolate chips that killed the dinosaurs. It knew that it could run at the speed of cookies, but alas! It got eaten! It started anew in the bakery, when a new purple attacked him, and he died.

"Respawn point???" he screamed as he lost The Game. He is no good. He should end this blasphemy. He pushed a big slice of roast chicken up the hill, where Nurse Joy awaited with a Chansey. Nobody likes Chansey. Not even my extremely dead grandmother can endure the pain of perdition. Nor my extremely old and fat tabby cat had endured the pain of frolicking in the realm of pokemon and dragons.

Why is the cookie doe not disposed off properly? Because Kamikazee watermelons decided to bomb [i]the cookie violently.[/i]
[/hide]

[hide]Once upon a time. The End. Once again an End is near but also far from clucking chickens! There lived a man who could divide by zero! But only on Chuck Norris's calculator! But it's not Norris', it's Jesus'. So to get those infidel Muslims, one must calculate the square root of a monkey, as accurately as Jesus will allow. Ekad! He's almost a constipated penguin. That will slow down the universe to the speed of a bad Bowling Ball, on a muddy field.

Since there was little speed, the bowling ball exploded, and little flowers sprouted and became various Pokemon. Then they exploded too into various Pokemon. A challenger appears! Named Chuck Norris/Jesus. He thinks 1=2. Which is true. The human race wants to lynch the man who proved that 1=2. On the way, Pokemon appeared again! Pikachu attacks! It explodes into little jackdillons that eat dirt. The end.

After the end, God said, "Let epsilon be positive." Gauss remarked that God also said: "People shall be defined as compact disks of love if for all people, there exists a positive delta within every twinkie. Even though twinkies are actually pokemon, who have twinkies which are pokemon and other creatures" The Grammar Police saw that there was no pizza up their butts with more Pokemon food. They started a Pokemon battle.

...he woke up from his long Pokemon battle nap and he saw that his pizza came alive and wanted a cookie. It craved the chocolate chips that killed the dinosaurs. It knew that it could run at the speed of cookies, but alas! It got eaten! It started anew in the bakery, when a new purple attacked him, and he died.

"Respawn point???" he screamed as he lost The Game. He is no good. He should end this blasphemy. He pushed a big slice of roast chicken up the hill, where Nurse Joy awaited with a Chansey. Nobody likes Chansey. Not even my extremely dead grandmother can endure the pain of perdition. Nor my extremely old and fat tabby cat had endured the pain of frolicking in the realm of pokemon and dragons.

Why is the cookie doe not disposed off properly? Because Kamikazee watermelons decided to bomb the cookie violently. [i]Pikachu and Agumon[/i][/hide]

[hide]Once upon a time. The End. Once again an End is near but also far from clucking chickens! There lived a man who could divide by zero! But only on Chuck Norris's calculator! But it's not Norris', it's Jesus'. So to get those infidel Muslims, one must calculate the square root of a monkey, as accurately as Jesus will allow. Ekad! He's almost a constipated penguin. That will slow down the universe to the speed of a bad Bowling Ball, on a muddy field.

Since there was little speed, the bowling ball exploded, and little flowers sprouted and became various Pokemon. Then they exploded too into various Pokemon. A challenger appears! Named Chuck Norris/Jesus. He thinks 1=2. Which is true. The human race wants to lynch the man who proved that 1=2. On the way, Pokemon appeared again! Pikachu attacks! It explodes into little jackdillons that eat dirt. The end.

After the end, God said, "Let epsilon be positive." Gauss remarked that God also said: "People shall be defined as compact disks of love if for all people, there exists a positive delta within every twinkie. Even though twinkies are actually pokemon, who have twinkies which are pokemon and other creatures" The Grammar Police saw that there was no pizza up their butts with more Pokemon food. They started a Pokemon battle.

...he woke up from his long Pokemon battle nap and he saw that his pizza came alive and wanted a cookie. It craved the chocolate chips that killed the dinosaurs. It knew that it could run at the speed of cookies, but alas! It got eaten! It started anew in the bakery, when a new purple attacked him, and he died.

"Respawn point???" he screamed as he lost The Game. He is no good. He should end this blasphemy. He pushed a big slice of roast chicken up the hill, where Nurse Joy awaited with a Chansey. Nobody likes Chansey. Not even my extremely dead grandmother can endure the pain of perdition. Nor my extremely old and fat tabby cat had endured the pain of frolicking in the realm of pokemon and dragons. [/hide]

Why is the cookie doe not disposed off properly? Because Kamikazee watermelons decided to bomb the cookie violently. Pikachu and Agumon [i]ate cookies like[/i]

[hide]Once upon a time. The End. Once again an End is near but also far from clucking chickens! There lived a man who could divide by zero! But only on Chuck Norris's calculator! But it's not Norris', it's Jesus'. So to get those infidel Muslims, one must calculate the square root of a monkey, as accurately as Jesus will allow. Ekad! He's almost a constipated penguin. That will slow down the universe to the speed of a bad Bowling Ball, on a muddy field.

Since there was little speed, the bowling ball exploded, and little flowers sprouted and became various Pokemon. Then they exploded too into various Pokemon. A challenger appears! Named Chuck Norris/Jesus. He thinks 1=2. Which is true. The human race wants to lynch the man who proved that 1=2. On the way, Pokemon appeared again! Pikachu attacks! It explodes into little jackdillons that eat dirt. The end.

After the end, God said, "Let epsilon be positive." Gauss remarked that God also said: "People shall be defined as compact disks of love if for all people, there exists a positive delta within every twinkie. Even though twinkies are actually pokemon, who have twinkies which are pokemon and other creatures" The Grammar Police saw that there was no pizza up their butts with more Pokemon food. They started a Pokemon battle.

...he woke up from his long Pokemon battle nap and he saw that his pizza came alive and wanted a cookie. It craved the chocolate chips that killed the dinosaurs. It knew that it could run at the speed of cookies, but alas! It got eaten! It started anew in the bakery, when a new purple attacked him, and he died.

"Respawn point???" he screamed as he lost The Game. He is no good. He should end this blasphemy. He pushed a big slice of roast chicken up the hill, where Nurse Joy awaited with a Chansey. Nobody likes Chansey. Not even my extremely dead grandmother can endure the pain of perdition. Nor my extremely old and fat tabby cat had endured the pain of frolicking in the realm of pokemon and dragons. [/hide]

Why is the cookie doe not disposed off properly? Because Kamikazee watermelons decided to bomb the cookie violently. Pikachu and Agumon ate cookies like [i]the ones made[/i]

[hide]Once upon a time. The End. Once again an End is near but also far from clucking chickens! There lived a man who could divide by zero! But only on Chuck Norris's calculator! But it's not Norris', it's Jesus'. So to get those infidel Muslims, one must calculate the square root of a monkey, as accurately as Jesus will allow. Ekad! He's almost a constipated penguin. That will slow down the universe to the speed of a bad Bowling Ball, on a muddy field.

Since there was little speed, the bowling ball exploded, and little flowers sprouted and became various Pokemon. Then they exploded too into various Pokemon. A challenger appears! Named Chuck Norris/Jesus. He thinks 1=2. Which is true. The human race wants to lynch the man who proved that 1=2. On the way, Pokemon appeared again! Pikachu attacks! It explodes into little jackdillons that eat dirt. The end.

After the end, God said, "Let epsilon be positive." Gauss remarked that God also said: "People shall be defined as compact disks of love if for all people, there exists a positive delta within every twinkie. Even though twinkies are actually pokemon, who have twinkies which are pokemon and other creatures" The Grammar Police saw that there was no pizza up their butts with more Pokemon food. They started a Pokemon battle.

...he woke up from his long Pokemon battle nap and he saw that his pizza came alive and wanted a cookie. It craved the chocolate chips that killed the dinosaurs. It knew that it could run at the speed of cookies, but alas! It got eaten! It started anew in the bakery, when a new purple attacked him, and he died.

"Respawn point???" he screamed as he lost The Game. He is no good. He should end this blasphemy. He pushed a big slice of roast chicken up the hill, where Nurse Joy awaited with a Chansey. Nobody likes Chansey. Not even my extremely dead grandmother can endure the pain of perdition. Nor my extremely old and fat tabby cat had endured the pain of frolicking in the realm of pokemon and dragons. [/hide]

Why is the cookie doe not disposed off properly? Because Kamikazee watermelons decided to bomb the cookie violently. Pikachu and Agumon ate cookies like the ones made [i]from Banana peels[/i]

[hide]Once upon a time. The End. Once again an End is near but also far from clucking chickens! There lived a man who could divide by zero! But only on Chuck Norris's calculator! But it's not Norris', it's Jesus'. So to get those infidel Muslims, one must calculate the square root of a monkey, as accurately as Jesus will allow. Ekad! He's almost a constipated penguin. That will slow down the universe to the speed of a bad Bowling Ball, on a muddy field.

Since there was little speed, the bowling ball exploded, and little flowers sprouted and became various Pokemon. Then they exploded too into various Pokemon. A challenger appears! Named Chuck Norris/Jesus. He thinks 1=2. Which is true. The human race wants to lynch the man who proved that 1=2. On the way, Pokemon appeared again! Pikachu attacks! It explodes into little jackdillons that eat dirt. The end.

After the end, God said, "Let epsilon be positive." Gauss remarked that God also said: "People shall be defined as compact disks of love if for all people, there exists a positive delta within every twinkie. Even though twinkies are actually pokemon, who have twinkies which are pokemon and other creatures" The Grammar Police saw that there was no pizza up their butts with more Pokemon food. They started a Pokemon battle.

...he woke up from his long Pokemon battle nap and he saw that his pizza came alive and wanted a cookie. It craved the chocolate chips that killed the dinosaurs. It knew that it could run at the speed of cookies, but alas! It got eaten! It started anew in the bakery, when a new purple attacked him, and he died.

"Respawn point???" he screamed as he lost The Game. He is no good. He should end this blasphemy. He pushed a big slice of roast chicken up the hill, where Nurse Joy awaited with a Chansey. Nobody likes Chansey. Not even my extremely dead grandmother can endure the pain of perdition. Nor my extremely old and fat tabby cat had endured the pain of frolicking in the realm of pokemon and dragons.[/hide]

Why is the cookie doe not disposed off properly? Because Kamikazee watermelons decided to bomb the cookie violently. Pikachu and Agumon ate cookies like the ones made from Banana peels [i]and God's kittens[/i]

Once upon a time. The End. Once again an End is near but also far from clucking chickens! There lived a man who could divide by zero! But only on Chuck Norris's calculator! But it's not Norris', it's Jesus'. So to get those infidel Muslims, one must calculate the square root of a monkey, as accurately as Jesus will allow. Ekad! He's almost a constipated penguin. That will slow down the universe to the speed of a bad Bowling Ball, on a muddy field.

Since there was little speed, the bowling ball exploded, and little flowers sprouted and became various Pokemon. Then they exploded too into various Pokemon. A challenger appears! Named Chuck Norris/Jesus. He thinks 1=2. Which is true. The human race wants to lynch the man who proved that 1=2. On the way, Pokemon appeared again! Pikachu attacks! It explodes into little jackdillons that eat dirt. The end.

After the end, God said, "Let epsilon be positive." Gauss remarked that God also said: "People shall be defined as compact disks of love if for all people, there exists a positive delta within every twinkie. Even though twinkies are actually pokemon, who have twinkies which are pokemon and other creatures" The Grammar Police saw that there was no pizza up their butts with more Pokemon food. They started a Pokemon battle.

...he woke up from his long Pokemon battle nap and he saw that his pizza came alive and wanted a cookie. It craved the chocolate chips that killed the dinosaurs. It knew that it could run at the speed of cookies, but alas! It got eaten! It started anew in the bakery, when a new purple attacked him, and he died.

"Respawn point???" he screamed as he lost The Game. He is no good. He should end this blasphemy. He pushed a big slice of roast chicken up the hill, where Nurse Joy awaited with a Chansey. Nobody likes Chansey. Not even my extremely dead grandmother can endure the pain of perdition. Nor my extremely old and fat tabby cat had endured the pain of frolicking in the realm of pokemon and dragons.

Why is the cookie doe not disposed off properly? Because Kamikazee watermelons decided to bomb the cookie violently. Pikachu and Agumon ate cookies like the ones made from Banana peels and God's kittens
[i]and then belched[/i]

[hide]Once upon a time. The End. Once again an End is near but also far from clucking chickens! There lived a man who could divide by zero! But only on Chuck Norris's calculator! But it's not Norris', it's Jesus'. So to get those infidel Muslims, one must calculate the square root of a monkey, as accurately as Jesus will allow. Ekad! He's almost a constipated penguin. That will slow down the universe to the speed of a bad Bowling Ball, on a muddy field.

Since there was little speed, the bowling ball exploded, and little flowers sprouted and became various Pokemon. Then they exploded too into various Pokemon. A challenger appears! Named Chuck Norris/Jesus. He thinks 1=2. Which is true. The human race wants to lynch the man who proved that 1=2. On the way, Pokemon appeared again! Pikachu attacks! It explodes into little jackdillons that eat dirt. The end.

After the end, God said, "Let epsilon be positive." Gauss remarked that God also said: "People shall be defined as compact disks of love if for all people, there exists a positive delta within every twinkie. Even though twinkies are actually pokemon, who have twinkies which are pokemon and other creatures" The Grammar Police saw that there was no pizza up their butts with more Pokemon food. They started a Pokemon battle.

...he woke up from his long Pokemon battle nap and he saw that his pizza came alive and wanted a cookie. It craved the chocolate chips that killed the dinosaurs. It knew that it could run at the speed of cookies, but alas! It got eaten! It started anew in the bakery, when a new purple attacked him, and he died.

"Respawn point???" he screamed as he lost The Game. He is no good. He should end this blasphemy. He pushed a big slice of roast chicken up the hill, where Nurse Joy awaited with a Chansey. Nobody likes Chansey. Not even my extremely dead grandmother can endure the pain of perdition. Nor my extremely old and fat tabby cat had endured the pain of frolicking in the realm of pokemon and dragons. [/hide]

Why is the cookie doe not disposed off properly? Because Kamikazee watermelons decided to bomb the cookie violently. Pikachu and Agumon ate cookies like the ones made from Banana peels and God's kittens
and then belched [i]in fudge and[/i]

[hide="It's dough, and not doe."]

Once upon a time. The End. Once again an End is near but also far from clucking chickens! There lived a man who could divide by zero! But only on Chuck Norris's calculator! But it's not Norris', it's Jesus'. So to get those infidel Muslims, one must calculate the square root of a monkey, as accurately as Jesus will allow. Ekad! He's almost a constipated penguin. That will slow down the universe to the speed of a bad Bowling Ball, on a muddy field.

Since there was little speed, the bowling ball exploded, and little flowers sprouted and became various Pokemon. Then they exploded too into various Pokemon. A challenger appears! Named Chuck Norris/Jesus. He thinks 1=2. Which is true. The human race wants to lynch the man who proved that 1=2. On the way, Pokemon appeared again! Pikachu attacks! It explodes into little jackdillons that eat dirt. The end.

After the end, God said, "Let epsilon be positive." Gauss remarked that God also said: "People shall be defined as compact disks of love if for all people, there exists a positive delta within every twinkie. Even though twinkies are actually pokemon, who have twinkies which are pokemon and other creatures" The Grammar Police saw that there was no pizza up their butts with more Pokemon food. They started a Pokemon battle.

...he woke up from his long Pokemon battle nap and he saw that his pizza came alive and wanted a cookie. It craved the chocolate chips that killed the dinosaurs. It knew that it could run at the speed of cookies, but alas! It got eaten! It started anew in the bakery, when a new purple attacked him, and he died.

"Respawn point???" he screamed as he lost The Game. He is no good. He should end this blasphemy. He pushed a big slice of roast chicken up the hill, where Nurse Joy awaited with a Chansey. Nobody likes Chansey. Not even my extremely dead grandmother can endure the pain of perdition. Nor my extremely old and fat tabby cat had endured the pain of frolicking in the realm of pokemon and dragons.

Why is the cookie dough not disposed off properly? Because Kamikazee watermelons decided to bomb the cookie violently. Pikachu and Agumon ate cookies like the ones made from Banana peels and God's kittens
and then belched in fudge and [i]ended the sentence.[/i][/hide]

[hide]Once upon a time. The End. Once again an End is near but also far from clucking chickens! There lived a man who could divide by zero! But only on Chuck Norris's calculator! But it's not Norris', it's Jesus'. So to get those infidel Muslims, one must calculate the square root of a monkey, as accurately as Jesus will allow. Ekad! He's almost a constipated penguin. That will slow down the universe to the speed of a bad Bowling Ball, on a muddy field.

Since there was little speed, the bowling ball exploded, and little flowers sprouted and became various Pokemon. Then they exploded too into various Pokemon. A challenger appears! Named Chuck Norris/Jesus. He thinks 1=2. Which is true. The human race wants to lynch the man who proved that 1=2. On the way, Pokemon appeared again! Pikachu attacks! It explodes into little jackdillons that eat dirt. The end.

After the end, God said, "Let epsilon be positive." Gauss remarked that God also said: "People shall be defined as compact disks of love if for all people, there exists a positive delta within every twinkie. Even though twinkies are actually pokemon, who have twinkies which are pokemon and other creatures" The Grammar Police saw that there was no pizza up their butts with more Pokemon food. They started a Pokemon battle.

...he woke up from his long Pokemon battle nap and he saw that his pizza came alive and wanted a cookie. It craved the chocolate chips that killed the dinosaurs. It knew that it could run at the speed of cookies, but alas! It got eaten! It started anew in the bakery, when a new purple attacked him, and he died.

"Respawn point???" he screamed as he lost The Game. He is no good. He should end this blasphemy. He pushed a big slice of roast chicken up the hill, where Nurse Joy awaited with a Chansey. Nobody likes Chansey. Not even my extremely dead grandmother can endure the pain of perdition. Nor my extremely old and fat tabby cat had endured the pain of frolicking in the realm of pokemon and dragons.

Why is the cookie dough not disposed off properly? Because Kamikazee watermelons decided to bomb the cookie violently. Pikachu and Agumon ate cookies like the ones made from Banana peels and God's kittens
and then belched in fudge and ended the sentence.[i] But then a[/i][/hide]

Once upon a time. The End. Once again an End is near but also far from clucking chickens! There lived a man who could divide by zero! But only on Chuck Norris's calculator! But it's not Norris', it's Jesus'. So to get those infidel Muslims, one must calculate the square root of a monkey, as accurately as Jesus will allow. Ekad! He's almost a constipated penguin. That will slow down the universe to the speed of a bad Bowling Ball, on a muddy field.

Since there was little speed, the bowling ball exploded, and little flowers sprouted and became various Pokemon. Then they exploded too into various Pokemon. A challenger appears! Named Chuck Norris/Jesus. He thinks 1=2. Which is true. The human race wants to lynch the man who proved that 1=2. On the way, Pokemon appeared again! Pikachu attacks! It explodes into little jackdillons that eat dirt. The end.

After the end, God said, "Let epsilon be positive." Gauss remarked that God also said: "People shall be defined as compact disks of love if for all people, there exists a positive delta within every twinkie. Even though twinkies are actually pokemon, who have twinkies which are pokemon and other creatures" The Grammar Police saw that there was no pizza up their butts with more Pokemon food. They started a Pokemon battle.

...he woke up from his long Pokemon battle nap and he saw that his pizza came alive and wanted a cookie. It craved the chocolate chips that killed the dinosaurs. It knew that it could run at the speed of cookies, but alas! It got eaten! It started anew in the bakery, when a new purple attacked him, and he died.

"Respawn point???" he screamed as he lost The Game. He is no good. He should end this blasphemy. He pushed a big slice of roast chicken up the hill, where Nurse Joy awaited with a Chansey. Nobody likes Chansey. Not even my extremely dead grandmother can endure the pain of perdition. Nor my extremely old and fat tabby cat had endured the pain of frolicking in the realm of pokemon and dragons.

Why is the cookie dough not disposed off properly? Because Kamikazee watermelons decided to bomb the cookie violently. Pikachu and Agumon ate cookies like the ones made from Banana peels and God's kittens and then belched in fudge and ended the sentence. But then a [i]new sentence started[/i]

[hide]Once upon a time. The End. Once again an End is near but also far from clucking chickens! There lived a man who could divide by zero! But only on Chuck Norris's calculator! But it's not Norris', it's Jesus'. So to get those infidel Muslims, one must calculate the square root of a monkey, as accurately as Jesus will allow. Ekad! He's almost a constipated penguin. That will slow down the universe to the speed of a bad Bowling Ball, on a muddy field.

Since there was little speed, the bowling ball exploded, and little flowers sprouted and became various Pokemon. Then they exploded too into various Pokemon. A challenger appears! Named Chuck Norris/Jesus. He thinks 1=2. Which is true. The human race wants to lynch the man who proved that 1=2. On the way, Pokemon appeared again! Pikachu attacks! It explodes into little jackdillons that eat dirt. The end.

After the end, God said, "Let epsilon be positive." Gauss remarked that God also said: "People shall be defined as compact disks of love if for all people, there exists a positive delta within every twinkie. Even though twinkies are actually pokemon, who have twinkies which are pokemon and other creatures" The Grammar Police saw that there was no pizza up their butts with more Pokemon food. They started a Pokemon battle.

...he woke up from his long Pokemon battle nap and he saw that his pizza came alive and wanted a cookie. It craved the chocolate chips that killed the dinosaurs. It knew that it could run at the speed of cookies, but alas! It got eaten! It started anew in the bakery, when a new purple attacked him, and he died.

"Respawn point???" he screamed as he lost The Game. He is no good. He should end this blasphemy. He pushed a big slice of roast chicken up the hill, where Nurse Joy awaited with a Chansey. Nobody likes Chansey. Not even my extremely dead grandmother can endure the pain of perdition. Nor my extremely old and fat tabby cat had endured the pain of frolicking in the realm of pokemon and dragons.[/hide]

Why is the cookie dough not disposed off properly? Because Kamikazee watermelons decided to bomb the cookie violently. Pikachu and Agumon ate cookies like the ones made from Banana peels and God's kittens and then belched in fudge and ended the sentence. But then a new sentence started [i]when a wobble-alien[/i] (see the vending machine topic)

[hide] Once upon a time. The End. Once again an End is near but also far from clucking chickens! There lived a man who could divide by zero! But only on Chuck Norris's calculator! But it's not Norris', it's Jesus'. So to get those infidel Muslims, one must calculate the square root of a monkey, as accurately as Jesus will allow. Ekad! He's almost a constipated penguin. That will slow down the universe to the speed of a bad Bowling Ball, on a muddy field.

Since there was little speed, the bowling ball exploded, and little flowers sprouted and became various Pokemon. Then they exploded too into various Pokemon. A challenger appears! Named Chuck Norris/Jesus. He thinks 1=2. Which is true. The human race wants to lynch the man who proved that 1=2. On the way, Pokemon appeared again! Pikachu attacks! It explodes into little jackdillons that eat dirt. The end.

After the end, God said, "Let epsilon be positive." Gauss remarked that God also said: "People shall be defined as compact disks of love if for all people, there exists a positive delta within every twinkie. Even though twinkies are actually pokemon, who have twinkies which are pokemon and other creatures" The Grammar Police saw that there was no pizza up their butts with more Pokemon food. They started a Pokemon battle.

...he woke up from his long Pokemon battle nap and he saw that his pizza came alive and wanted a cookie. It craved the chocolate chips that killed the dinosaurs. It knew that it could run at the speed of cookies, but alas! It got eaten! It started anew in the bakery, when a new purple attacked him, and he died.

"Respawn point???" he screamed as he lost The Game. He is no good. He should end this blasphemy. He pushed a big slice of roast chicken up the hill, where Nurse Joy awaited with a Chansey. Nobody likes Chansey. Not even my extremely dead grandmother can endure the pain of perdition. Nor my extremely old and fat tabby cat had endured the pain of frolicking in the realm of pokemon and dragons.

Why is the cookie dough not disposed off properly? Because Kamikazee watermelons decided to bomb the cookie violently. Pikachu and Agumon ate cookies like the ones made from Banana peels and God's kittens and then belched in fudge and ended the sentence. But then a new sentence started when a wobble-alien (see the vending machine topic) [i] is found in [/i]
[/hide]


Hello,

we have $ x\plus{}y\equal{}7$ and $ x^2 \minus{}y^2\equal{}21$

Factoring $ x^2 \minus{}y^2\equal{}21$ to get $ (x\minus{}y)(x\plus{}y)\equal{}21$

From there $ x\minus{}y\equal{}3$. Now adding $ x\plus{}y\equal{}7$ to $ x\minus{}y\equal{}3$ yields $ x\equal{}5$, $ y\equal{}2$.

So $ 2x\plus{}3y\equal{}16$

i have used a different method

$ x\plus{}y\equal{}7$
$ x^2\minus{}y^2\equal{}21$
$ y\equal{}7\minus{}x$
$ x^2\minus{}(7\minus{}x)^2$
$ x^2\minus{}x^2\plus{}14x\minus{}49$
$ 14x\equal{}70$
$ x\equal{}5$
$ \implies\ y\equal{}2$
$ 2x\plus{}3y\equal{}2*5\plus{}3*2\equal{}10\plus{}6\equal{}16$

I suppose it is correct :blush:


The hypotenuse of a 45-45-90 triangle is equal to the longer leg of a 30-60-90 triangle. What is the ratio fof the area of the larger triangle to the area of the smaller one?

[hide="Hark, Thucydides!"]Let the side of the 45-45-90 triangle be $ s$; the hypotenuse is then $ s\sqrt {2}$. Since this is equal to the longer leg of the 30-60-90 triangle, the shorter leg is $ \frac {s\sqrt {2}}{\sqrt {3}}$. Finding the area of both, we have that the first triangle has area $ \frac {s^{2}}{2}$, and the second one has area $ \frac {2s^{2}}{2\sqrt {3}} \equal{} \frac {s^{2}}{\sqrt {3}}$. The ratio is just:

$ \frac {s^{2}}{\sqrt {3}}\times \frac {2}{s^{2}} \equal{} \frac {2}{\sqrt {3}} \equal{} \boxed{\frac {2\sqrt {3}}{3}}$.

(Note I determined which triangle was "larger" by comparing area.)[/hide]


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let G be a finitely generated abelian group and let T_p be the subgroup of all elements having order some power of a prime p. Suppose

      $ T_p \simeq \mathbb{Z}_{p^{r_1}} \times \mathbb{Z}_{p^{r_2}} \times \cdots \times \mathbb{Z}_{p^{r_m }} \simeq \mathbb{Z}_{p^{s_1}}  \times \mathbb{Z}_{p^{s_2}} \times \cdots \times \mathbb{Z}_{p^{s_m }}$

where       $ 1 \leq r_1 \leq \cdots \leq r_m$ and similarly for the s_i.

Prove that n = m and s_i = r_i for all i. Hint: first prove r_1 = s_1 and then use induction.

I can prove n = m but I am have tons of difficulty proving r_1 = s_1. How do you use the minimality of r_1 and s_1? We know the right-hand group contains a copy of the $ \mathbb{Z}_{p^{r_1}}$, but how do we know that must be its first factor?

Look at the set of cyclic subgroups of $ T_p$ that are maximal, that is, that are not contained in any other cyclic subgroups.  Show that the the minimum number of elements in such a group has order exactly $ p^{r_1}$.


Hi, a little question.

What is the largest 2 digit prime factor that divides 200C100?

Why must the two digit prime  must satisfy 3p<200.

[quote="mdk"]What is the largest 2 digit prime factor that divides 200C100?

Why must the two digit prime  must satisfy 3p<200.[/quote]

${200 \choose 100}= \frac{200(199)(198)(197)...(101)}{100(99)(98)(97)...(1)}$

If $3p > 200$ then $p$ appears at most once in the numerator but must also appear once in the denominator, which means that the two factors cancel.


calculus
integration
analytic geometry
combinatorics unsolved
combinatorics

The set $S = \{ (a,b) \mid 1 \leq a, b \leq 5, a,b \in \mathbb{Z}\}$ be a set of points in the plane with integeral coordinates. $T$ is another set of points with integeral coordinates in the plane. If for any point $P \in S$, there is always another point $Q \in T$, $P \neq Q$, such that there is no other integeral points on segment $PQ$.  Find the least value of the number of elements of $T$.

Please check for my idea:
First,$|T|>1$, if $|T|=1$ and let the point be $U=(x,y)$; there be at least one point $V=(x',y')$ in $S$ such that $x$ and $x'$ ,$y$ and $y'$ have same parity so the mid-point of $UV$ is an integeral point.
For $|T|=2$, two points be $(4,3)$ and $(3,4)$ work.

(I assume that extra condition $P \neq Q$ in the question mean $S$,$T$ can have overlap element(s))


let $ n \in N$ and $ n\geq 1$.Prove that:
If $ a \equal{}2 \plus{} 2 .\sqrt []{28n^2 \plus{} 1}\in Z$ so $ a$ is perfect square number


Solve the third equation for $y$ and substitute into the second one. When you expand everything out, you get the following cubic:

$x^3-512x-1536$

That has three real roots, which give you the three $x$ values I posted above. Then you have your choice of how to solve for $y$. I checked the values and they worked (I sure hope I did it correctly though).

I think you did since the answers were supposed to come out nicely :)

[quote="eminem"]how did you find the roots of  $x^3-512x-1536$ ???[/quote]

I haven't tried this question yet but i think you can do it by substituting y in the first equation as well then u get two equations in terms of x.Not sure.

[quote="eminem"]how did you find the roots of  $x^3-512x-1536$ ???[/quote]

Sorta just trial and error...

What I just did was $x^3-512x-3\cdot512.$  If we have a rational root it must be $\pm$ a divisor of $3\cdot512.$  Also note that $512=8^3.$  After trying $8$ and $2\cdot 8$ I found that $3\cdot 8$ worked.  Then the other roots aren't too hard to find.

Of course...the only reason that worked is because the problem came out nicely...this is not at all a good assumption except on competitions without calculator ;)


Solution:
Let a = 2x/(x^2 + 1), b = 2y/(y^2 + 1) and c = 2z/(x^2 + 1) then
0 <= a , b , c <= 1 ( by am-gm ineq ). The problem reduces to proving:

a/(b + c + 1) + b/(c + a + 1) + c/(a + b + 1) + (1 - a)(1 - b)(1 - c) <= 1 for 0<= a , b , c <= 1. (*)

This is a classic problem that can be solved by using convexity of the lhs of (*). The maximum of lhs is 1 and conclustion follows.
____________________________________________________________


Find general term $ u_n$ of 2 sequence below:
    1) let $ u_1$=2;$ u_{n\plus{}1}\equal{}u^3_n\plus{}3u_n\minus{}3$
    2)let $ u_1\equal{}\frac{\sqrt6}{2}$; $ u_n\plus{}1\equal{}24u^3_n\minus{}12\sqrt6u^2_n\plus{}15u_n\minus{}\sqrt6$

hello, for 2) i have found
$ u_n\equal{}\frac{1\plus{}\sinh\left(3^{n\minus{}1}arsinh(2)\right)}{\sqrt{6}}$.
Sonnhard.

[quote="king.of.dragon"]can u tell more details?result isn't important[/quote]

Work out the formula for $ \sinh(3t)$ in terms of $ \sinh(t)$, then compare to the original recurrence.

For the first one... examine the cubics you get for $ \cos(3t)$ in terms of $ \cos(t)$, for $ \sin(3t)$, for $ \cosh(3t)$, for $ \sinh(3t)$.  If one of them fits, you are in!  Most cubics won't give you a nice formula for this recursion, but since somebody posed this, maybe this one will.


Hello, can anyone help me with this problem?

4^x-4^(x-1)=12

 It seems simple enough, but I can't seem to get a handle on starting it. When I try to use the rules of logarithms, I get a bizarre answer, or else my x variable gets subtracted altogether. I know the answer is two by guess and check, but I want to know how this is attained. I'd greatly appreciate any help.

you don't even need laws of logarithms  ;)

You need to know $4^{x-1} \cdot 4^1 = 4^{x-1+1} = 4^x$

so knowing the laws of logs (or exponents) would definitely help

Right, I realize  4^(x-1) * (4-1) = 12, and 4^(x-1) * 4^1 = 4^x, but how do these facts help me solve the problem? Multiplying the equation by 4^1 only gives me 4^(x+1)-4^x=48, which isn't helpful.

[quote="Ijo07"]Right, I realize  4^(x-1) * (4-1) = 12[/quote]

OK to solve an equation we need to isolate the variable so we have $4^{x-1}\cdot (4-1)=12\iff 4^{x-1}=4$. So $x-1=1\implies x=2$.

Alright, I guess what I'm asking is how do you get from my original equation to 4^(x-1) *(4-1)=12?

Thanks much, now I understand. I guess the part I was missing was factoring out the 4^(x-1).


Prove that every single positive integer has a unique representation in any arbitrary base.

Suppose some number has 2 different representations; wolog they differ in the first digit.  That difference is atleast $b^n$; the other digits difference is at most $(b-1)\sum_{k=0}^{n-1} b^k = b^n - 1$.  Thus these differences can never be equal.


Two positive numbers are such that their difference is 6 and the difference of their squares is 48. What is their sum?

$ \frac {a^2 \minus{} b^2}{a \minus{} b} \equal{} \frac {48}{6} \equal{} 8$.

$ \frac {a^2 \minus{} b^2}{a \minus{} b} \equal{} \frac {(a \plus{} b)\cancel{(a \minus{} b)}}{\cancel{a \minus{} b}} \equal{} a \plus{} b$.

Thus, $ \boxed{8}$.


LaTeX
function
calculus
derivative
logarithms
real analysis
real analysis unsolved

I have a question on a differential equation. I don't know how to use LaTeX (I am new here, sort of) so I just scanned the question.

[img]http://i6.photobucket.com/albums/y208/DemonicImpact/Question.jpg[/img]

Basically, I don't know how to solve this differential equation and put it in to "y equals" form... When I solve it I get

ln( (y^2-1)^(1/2) + y ) = x

but is there a way to get it into y equals form?

Thanks.

You don't have to get it into $ y\equal{}y(x)$ form.  As long as that function gives the right derivatives, you're fine (I didn't check it so I don't know if it's right).

Well, that's a known form, so one solution is to just recognize it from things you've done ;).  Or, we can try to figure it out: both the logarithm and the $ \sqrt {y^2 \minus{} 1}$ suggest a hyperbolic substitution might be in order.

Thanks. It became really simple after I did a hyperbolic substitution.


can anyone help me with this problem( please show work!)

find the mystery number _ _, _ _ _ ._ _ _ _ _

clues:
1 the digits in the ten thousands, thousands ,and tenths places are the same
2 the hundreds digit is 1 more than the thousands digit
3 the digit in the ten thousandths place is the sum of the digits in the thousands and ten thousands places
4 no odd digit occurs more than once
5 the digit in the hundred thousandths place is the sum of the digits in the tenths and hundredths laces
6 the part of the number to the right of the decimal has no repeated digits.
7 the only odd digits appear in the thousandths, hundreds, and ones places
8 the digit in the ones place is the sum of the digits in the tens and hundreds places
9 the digits in the tens and hundredths places are the same and are half of the digit in the tenths place
10 the digit in the tenths and thousandths places have exactly 3 factors

[hide="Solution"]
Go through all the clues writing down facts you learn.  At first, 1 & 2 don't help much.  From #3 we know that the ten thousands and and thousands (as well as tenths from #1) have to be 4 or less since they sum to a digit (0-9).  #4 tells us that these 3 digits are also not odd, so they must be 4,2, or 0.

From #9 we know that the ten thousands, thousands and tenths must be [b]4[/b].  (0 or 2 divided by 2 will not yield an even number.)

Knowing these, from #2, then the hundreds digit is [b]5[/b].

From #3 the ten thousandths place is [b]8 [/b](sum of the thousands and ten thousands).

#9 shows then that the tens and the hundredths are [b]2[/b] (half of the tenths place)

The thousands place is odd (#7) and because of #10 we know that the thousandths must be odd and have only 3 factors.  The factors would be one and itself plus another number.  Therefore the thousandths place must be a perfect square, or [b]9[/b].

Use #8 to find the ones.  Sum of tens and hundreds is [b]7[/b].

The completed number would be as follows: $ \boxed{44,527.4298}$

[/hide]


If $ X$ is any space, $ A$ is closed subset of $ X$, and $ p \notin X$, the space $ X \plus{}f\{p\}$ resulting from the function $ f: A \to \{p\}$ is homeomorphic to the quotient space of $ X$ obtained by identifying $ A$ to a single point.

Well... let $ Y \equal{} X \plus{}_f \{p\}$. Then $ Y$ is the pushout of the maps $ A \to X$ and $ A \to \{p\}$; so a map $ Y \to Z$ is the same as a pair
of maps $ X \to Z$ and $ \{p\} \to Z$ such that $ \{p\}$ and $ A$ go to the same point in $ Z$.

This is the same universal mapping property that $ X/A$ enjoys, so $ X/A$ and $ Y$ are homeomorphic.

Thank you very much.

What if we let $ X \equal{} I$, $ Y \equal{} [2,3]$, $ A \equal{} \{0,1\}$, and let $ f: A \to Y$ be defined by $ f(0) \equal{} 2$ and $ f(1) \equal{} 3$. Then show that $ X \plus{}fY$ is homeomorphic to $ S^{1}$.

You're basically attaching $ X$ to $ Y$ along the map $ f$, so you're gluing $ 0$ with $ 2$ and $ 3$ with $ 1$. Certainly if you do that you'll get something homeo. to $ S^1$ :).

It's pretty standard to look at $ S^1$ as $ [0,1]$ with $ 0$ and $ 1$ identified, which makes this pretty easy.


If there was a neurotransmitter (fast working) in our system - idwcwdmm for short, that, when reached it's receptor caused this breakdown in motivation, what would the results be?
How would the tasks be completed
what role would either the agonists or antagonists play and why?

what if I had to:

show the mode for both x and y

the median for both x and y

the standard deviation for both x and y

and I had to graph the relationships between x and y

X is independent variable on a scale of 0-100 and the number is meant to signify the level of i dontwannacantwontdontmakeme (100 is a made up number meant to equal the highest percentage levels ever found

Y is dependent variable, percentage showing failure to complete, random and wide range of tasks.

the following 10 subjects (randomly selected and meant to represent a much larger segment of the population

     x         y
1  45         50
2  65         65
3  70         70
4  75         75
5  80         80
6  85         80
7  85         85
8  85         85
9  90         90
10 100     100

does anyone get this or is anyone able to solve this? Its a challenge and I'm in over my head for sure. And good luck to you all that understand and take more complex math classes. I could never do it.  Rock on!

use google to search what those terms mean...if you try this, and still cannot understand, ask again...and actually do this...we are not here to do your homework...we are just people who like doing nontrivial math...sadly this is trivial math...i don't mind helping people once in a while, but these types of questions are just definitions

I know that the neurotransmitter question basically means that you have something that works faster than others, and because of this, sends certain signals faster than others. And an agonist will facilitate that action, and the antagonist inhibits it. So, if you want someone to do something, you'd hope the antagonist is working first, and then the agonist kicks in when you want them to stop. But even with knowing that, I can;t get past that part. I wasn't asking for anyone to do my homework, I am seriously stumped and was looking for direction and explaination. Thank you for trying to help me though. Have a good day.


Let be given a function f(x) satisfying f"(x) > 0 for all x > 0 and the graphics of f(x) has skew asymptote y = ax + b when x -> oo.
1) Prove that function g(x) = f(x) - ax - b has g'(x) <=0 for all x>0.
2) Prove that graphics of f(x) (for x>0) lies over skew asymptote.

<i>Source: VUNC 2001</i>

Namdung

1)lim(x--> \infty){f'(x)} = a .... f''(x) > 0 ===> f'(x) is increasing ===> f'(x) < a ===> g'(x)=f'(x)-a < 0
2)If y = ax + b  is asymptote of f when x-->\infty , then lim(x--> \infty){g(x)} = 0 ... g'(x) < 0 ===> g(x) is decreasing ===> g(x) > g(\infty)=0 ===> f(x) > a*x+b


What are you dividing by two?  Your explanation isn't very clear.

You could just use simple logic, the square roots of a+b and a-b have the same parity, we let a+b=9 and a-b=1, so a=5.

[quote="Olga1146B"]i would like to challenge you all to find the answer for this problem.

Define a Benson Rectangular number as a positive integer "a" that if increased and decreased by the same positive integer "b", yields two positive integers "a+b" and "a-b", that are both perfect squares. what is the smallest Benson Rectangular number?????????
 :cool:

 [hide] the answer is five, b/c we give the problems natural numbers "m" and "n". divided by two we get ten over two there fore becoming the answer five [/hide][/quote]
You just find two square numbers that have a even difference.
Smallest are 1 and 9.
So since their difference is 8, the number would be 4 away from 1 and 9.
So 5.

[quote="sanghnlee"]wait what is a benson rectangle???
 :lol:[/quote]
You don't need to know.
"They just said that a Benson [b]Rectangular Number[/b] is..."

i am pretty sure that i know the answer considering that i know the person who wrote this and she is my best friend and i just might of looked at her paper when she was doing the problem so i am pretty sure that the answer is...

[hide]5. but that is just a guess. jk. i know that is the answer cuz i cheated. oops. i am a cheater. a cheater. A CHEATER!!! hee hee. i love ya ashley.[/hide]

Sorry, but I'm starting to believe you are just copying answers. Welcome to AoPS, but dont post just to raise your post count. If you dont have anything worthy of posting, it's usually best you dont post it.

[quote="helga7032Z"]what i just wrote is worthy of posting so i don't really see anything wrong with that. do you? i am sorry if you were somehow offended by my talking about my best friend on the internet.  :) thanks for welcoming me though. i really appreciate that. heehee. heehee.[/quote]

This is a math fourm.
Used by people to help them on math.
People do not want to go through five pages of spam.

And math is not about getting the right answer, it's having a good method to approach the problem.  Otherwise this site probably wouldn't exist.

actually my friend helga is correct thx anyway
ur right the answer is 5.

Lol I'm this is old as f but yes if you actually asking a question and answering that's fine but helga if you wanna chat with your friend PM or message her I don't wanna scroll through 50 pages of " Hi omfg so long no see" and stuff like that just to find out if the answer is correct thanks alot


number theory
greatest common divisor
number theory unsolved

Determine $ a,m,n \in N$, $ a\ge 2,m\ge2$ such that $ a^m\plus{}1 | a^n\plus{}203$

here is my solution i dont think it is right.please tell me why;[hide]it is not hard to prove that $ a^m\plus{}1$ divides $ a^n\minus{}203a^m$.then $ a^m\plus{}1$ divides both $ a^n\plus{}203$ and $ a^n\minus{}203$.but gcd of $ a^n\plus{}203$ and $ a^n\minus{}203$ is 1 when 203 does not divide a. so there are no solutions in natural numbers[/hide]

But there are solutions e.g. $ (a,m,n) \equal{} (2,2,1)$

I did it with much casework^^ Has anybody got another proof?

but surely apart from $ 2,2,1$ there are no more solutions

Let $ k\equal{}[n/m],l\equal{}n\minus{}km$, then $ 0\le l<m$ and $ a^m\plus{}1|a^n\plus{}203\to a^m\plus{}1|(\minus{}1)^ka^l\plus{}203$.
Obviosly, if $ a>203$, then  $ a^m\plus{}1\not |(\minus{}1)^ka^l\plus{}203, l<m$, because $ (\minus{}1)^ka^l\plus{}203\not \equal{}0$ and $ |(\minus{}1)^k\plus{}203|<a^m\plus{}1$.
If $ a\equal{}203$, then $ l\equal{}1,k\minus{}odd$ and solution $ a\equal{}203,n\equal{}km\plus{}1$, were $ m\ge 2,k\minus{}odd$.
If $ a<203$, then $ a^l\not \equal{}(\minus{}1)^{k\plus{}1}203$, therefore must be $ |a^l\plus{}(\minus{}1)^k203|\ge a^m\plus{}1$
Consider case, when $ l\equal{}0$. Then $ a^m\plus{}1|202$ (k-odd) or $ a^m\plus{}1|204$ (k-even). First case give $ a\equal{}10,m\equal{}2, n\equal{}2k,k\minus{}odd$, second case give $ a\equal{}2,m\equal{}4, n\equal{}8k$ or $ a\equal{}4,m\equal{}2,n\equal{}4k$. Let $ l\ge 1$
If $ a\ge 15$, then $ a^m\plus{}1\ge a^{l\plus{}1}\plus{}1>a^l\plus{}203$. Therefore we can chek only $ a<15$
If $ a>7$, then $ m\equal{}2,l\equal{}1$, but $ a^2\plus{}1\not|203\pm a$ for $ a>7$,
Obviosly $ a\equal{}7$ does not work (chek, by mod 7). If $ m\equal{}3$, then $ a\not \equal{}5,6$, because $ 7|a^3\plus{}1, 7\not|203\pm a^l$.
$ m\equal{}2,l\equal{}1$ give solution $ a\equal{}5,m\equal{}2,n\equal{}4k\plus{}1$ ($ l\equal{}1,k\minus{}even$ )


Find $ a,b,c \in \mathbb{Z}$ such that:
$ a^{2}\plus{}1 \in \mathbb{P}$
$ b^{2}\plus{}1 \in \mathbb{P}$ and
$ c^{2}\plus{}1\equal{}(a^{2}\plus{}1)(b^{2}\plus{}1)$
 :wink:

This nice problem comes from Poland M.C. 2002 and it was already discussed.Please visit: [url]http://www.mathlinks.ro/viewtopic.php?p=386500#p386500[/url]


How many bit string of length 10 contain at least three 1s and at least three 0s?

I am assuming bit strings are binary sequences.

What we don't want is

10 ones, 0 zeros --> 1 way
9 ones, 1 zero --> 10 ways
8 ones, 2 zeros --> 45 ways
2 ones, 8 zeros --> 45 ways
1 one, 9 zeros --> 10 ways
0 ones, 10 zeros. --> 1 way.

So since there are $ 2^10\equal{}1024$ total sequences, there are $ 1024\minus{}112\equal{}912$ total sequences with at least 3 ones and at least 3 zeros.

[hide="Without casework"]First let's see how we'd count all the cases where there are three zeroes. There would be $ \binom{10}{3}$ ways to choose the terms that would be $ 0$ and the rest would be $ 1$. This would likewise continue for the other possible numbers of zeroes (up to $ 7$ zeroes, after which we don't have enough ones).

We essentially have $ \sum_{k\equal{}3}^{7}\binom{10}{k}$, which is equal to $ 2(120\plus{}210)\plus{}252\equal{}\boxed{912}$.[/hide]

Maybe it's not true to say "without casework." The casework's sort of still there...


function
trigonometry
logarithms
integration
real analysis
real analysis unsolved

Can we prove that the following function
\[ f(x)\equal{}\begin{cases}
\dfrac{\sin x}{x}, & x \ne 0,\\
1,& x\equal{}0,
\end{cases}
\]
is a fundamental function?  :idea:  :roll:

[quote="JoeBlow"]What is a fundamental function?[/quote]

Like a finite +, -, *, / of the following functions $ \sin, \cos, \tan, \log, exp, ax\plus{}b$ and their inverses.

[quote]$ f(x) \equal{} \begin{cases} \dfrac{\sin x}{x}, & x \ne 0, \\
1, & x \equal{} 0, \end{cases}$[/quote]

What is your motivation for this question? to evaluate $ \int_{x_{0}}^{x}f(t)dt$? :D


prove that for every positive integer $ a>1$ there exist infinitely many positive integers n such that $ n \mid a^n\plus{}1$

There is a good proof in the Sierpinski's book...
Problem 1\22 :wink:

Fermat's theorem=>
$ a^n\equiv a(\mod n)$

See that there exist infinitely many integers such that $ a\equiv \minus{}1(\mod n)$
QED

I feel queaky about this sol myself so I apologise if it is completely incorrect or absurd.

[quote="Potla"]Fermat's theorem=>
$ a^n\equiv a(\mod n)$

See that there exist infinitely many integers such that $ a\equiv \minus{} 1(\mod n)$
QED

I feel queaky about this sol myself so I apologise if it is completely incorrect or absurd.[/quote]
There doesn't exists infinitly many primes such that $ p \mid a \plus{} 1$.

I do not have Sierpinski's book and it will be great if someone can post his proof. Here is mine.

We will define series $ \{n_i\}$ such that $ n_i|a^{n_i}\plus{}1$.

1) $ 2|a$.
$ n_1\equal{}1$, $ n_{i\plus{}1}\equal{}a^{n_i}\plus{}1>n_i$.
Obviously $ n_{i\plus{}1}$ is odd and we have
$ n_i|a^{n_i}\plus{}1\equal{}n_{i\plus{}1}\Rightarrow a^{n_i}\plus{}1|a^{n_{i\plus{}1}}\plus{}1$, or

$ n_{i\plus{}1}|a^{n_{i\plus{}1}}\plus{}1$

2) $ 2\not|a$, $ a\plus{}1\equal{}2^t*b$, where $ b>1$ is odd.
$ n_1\equal{}1$. $ n_{i\plus{}1}\equal{}\frac{a^{n_i}\plus{}1}{2^t}$.
Since $ \frac{a^{n_i}\plus{}1}{a\plus{}1}$ is odd, $ n_{i\plus{}1}$ is odd and also
$ n_{i\plus{}1}>\frac{a^{n_i}\plus{}1}{a\plus{}1}>n_i$.
Since both $ n_i$ and $ n_{i\plus{}1}$ are odd numbers,
$ n_i|a^{n_i}\plus{}1\Rightarrow n_i|n_{i\plus{}1}$.
Then $ n_{i\plus{}1}|a^{n_i}\plus{}1|a^{n_{i\plus{}1}}\plus{}1$.

3) $ 2\not|a$, $ a\equal{}2^t\minus{}1$.
Since we can write $ a^2\plus{}1\equal{}2^m*b$, for odd $ b>1$, we can find infinitely many odd $ n$ such that $ n|a^{2n}\plus{}1$, by 2).
Since $ a$ is odd, we have $ 2n|a^{2n}\plus{}1$ for infinitely many $ n$.


How many of you were at the northern California Mathcounts state competition. I was there. I love those rubber brains

wooo

I was the guy who failed the most.

To people who are about to complain, I actually did, probably.

bleh, after the test i was expecting a 34, got a 30, didnt even make the list of top 45 in northern california
but at least i didnt fail like i did at chapters (30 :ninja: )

4th in chapters(my chapter). YAY. I'm still confused about my score. It was the same as the person in second place. How does the tiebreaker work?

[quote="ashpotter"]Who was that guy in the fanny pack who got really far in the countdown round? I failed more than any of you!!! (we failed more than anyone else, I got a 19. The highest score in our team was 23. We fail)[/quote]

That would be Evan O'Dorney.

I'm surprised he's good at MathCounts given how crazily beastly he is at other stuff (including Olympiads).

[quote="Ubemaya"]
That would be Evan O'Dorney.

I'm surprised he's good at MathCounts given how crazily beastly he is at other stuff (including Olympiads).[/quote]
Don't you mean you're not suprised?

No, I mean I am surprised.

MathCounts is on a completely different level from Olympiads.

yeah, but that doesn't mean there's supposed to be a negative correlation....
quite a lot of people who were really good at mathcounts happen to go on to excel at olympiads

Yeah, but MATHCOUNTS = speed while Olympiads = supersmartness and creativity

Someone ask about the tiebreaker?

Basically if there's a tie, they look at your questions in backwards order or something.  The person who gets the most right in the back (the hardest questions) receives a higher spot.

Here's the tiebreaker for individuals:

1. Sprint score
2?. Score from the last 10 problems on the sprint round (not sure about this one)
3. Last 10 problems on the sprint round, compared in reverse order one at a time
4. Target round problems compared in reverse order one at a time
5. First 20 problems on the sprint round compared one at a time
6. Tiebreaker round

And for teams:

1. Team round score
2. Sum of sprint scores
3. Team round problems compared in reverse order
4. Tiebreaker round

The tiebreaker round is a set of two problems.  You're given both at the same time and allowed 6 minutes to turn in answers.  If it's a team tiebreaker, the team works together.  You have only one chance to turn in your answer, but you can do so at any time in the 6 minutes.  If both people/teams get the same score, the first to turn in answers wins.  If you've been to ARML, it's nearly identical to the ARML tiebreaker except that there are two problems.

I think they write four tiebreaker questions, but I've never seen any used.  It would probably only come up in the event that there were multiple perfect scores.

I believe the tiebreaker for individuals is:
1. Sprint score.
2. Total score of last 10 Sprint problems.
3. Last Sprint problems one by one, starting from #30.
4. Total Target.
5. Middle 10 Sprint problems, total.
6. Middle 10 Sprint problems, one by one.
7. First 10 Sprint problems, one by one.

I'm not quite sure though... I read the rules a looonnnnggggg time ago.

I was in a three-way tiebreaker last year for Chapters. It was basically like a Target Round, except it wasn't timed (calculators were allowed, although they're slower than computing mentally). When you finished, you put down your paper and raised your hand. The person with the highest score wins, and they break ties between same scores with whoever finished faster. Although of course, my chapter does it differently than it's supposed to be done, I think.

Hey, Ubemaya, that must have been fun.  What was it like?

[quote="ashpotter"]i never figured out what zomg stands for.

How did you like those brains?[/quote]

You guys got those sticky brains too? I got one in '05. They were pretty awesome.

[quote="ashpotter"]mine lost it's stickyness *sniff* *sniff*

What was really cool about it was that WHATEVER shape you bent it in, it will ALWAYS go back to it's original position. Eventually.[/quote]

Unless you rip it in half.  :rotfl:
I personally liked the AoPS book (or excerpts).

So, how do you know who advanced to nationals?

For those of us who weren't there, there's a pdf linked from [url=http://www.mathcounts.org/webarticles/anmviewer.asp?a=536&z=5]here[/url] with the teams from every state.

The CA team this year was the youngest team there, with two 7th graders and a 6th grader.

Yes, indeed I failed chapter, only doing ever so slightly better than i_like_pie. In other words, I got nowhere near state, and I was nearly like 20 places behind vishalarul. And like 25 places behind Ubemaya, and etc.

I failed Chapter too. My Countdown was disastrous. I left such a bad impression that at States everyone was betting on Nathan to win.

[quote="xscapezaer"]Omfg, Ubemaya, you got second in our chapter (as you display in your signature) and you got somewhere very near top in countdown too...

Btw, nathan who?[/quote]
wow
just wow

[quote="xscapezaer"]Sorry, was that like a stupid thing to ask or something?[/quote]
YES

[quote="archimedes1"]Ubemaya.
Get used to it.[/quote]
dude wtf
how about we just STOP TAKING HIM SERIOUSLY
OK?

I think you're taking him too seriously.


parameterization
graph theory
combinatorics proposed
combinatorics

Some teams participated in a tennis tournament in which there was at most one match between any two teams and each match had a winner (no draw). The referees decided to give each team, a score which is an integer number. Of course they want to make sure that for any two teams which played with each other during the tournament, the score of the winner is larger than the score of the loser. What is the minimum number of different score values that they need?

What if the tournament contains a Hamiltonian cycle? Then one cannot assign to winners scores larger than to losers, since $ A$ beat $ B$ who beat $ C$ ... who beat $ Z$ who beat $ A$.

In this case (i.e. when the tournament contains a directed cycle) the answer is infinite. In fact the answer is in terms of some parameter of the tournament directed graph.


For what primes $p,q>2$ the numbers $pq\cdot 2^{n}+1$ are composite(i.e. not prime) for all $n\ge 0$ ?


I've made this:

[img]http://img179.imageshack.us/img179/3773/sdkfjsdz8.png[/img]

Is there a way to remove "1 Unique Chapter" and set it as "Unique Chapter"?

Also the first "1 Unique Chapter" has a hyperlink, how could I set it like text anymore? (It's fine having hyperlinks for subsections.)

Not enough info as you don't say what class you are using, whether you are using \chapter or \section and how you added the hyperlinks, which are not there by default. Need a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]minimal example[/url].

Ohhh, sorry, I'm using the following preamble:

\documentclass{article}
\usepackage{fix-cm}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage[bookmarks=false,colorlinks=true,linkcolor={blue},pdfstartview={XYZ null null 1.22}]{hyperref}
\usepackage[dvips]{graphicx}
\usepackage{color}
\usepackage[latin1]{inputenc}
\pdfpagewidth 8.5in
\pdfpageheight 12in
\setlength\topmargin{0in}
\setlength\headheight{0in}
\setlength\headsep{-1in}
\setlength\textheight{10.2in}
\setlength\textwidth{6.7in}
\setlength\oddsidemargin{-0.2in}
\setlength\evensidemargin{0in}
\setlength\parindent{0.25in}
\setlength\parskip{0.25in}

Oh dear. That's not a minimal example since it isn't a complete document and is certainl not minimal. It doesn't show how you got the hyperlinks or what the table of contents is showing. So please re-read the instructions at [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]Minimal example please![/url].

Also using all those settings for page layout are not necessary and can cause problems. Use the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=geometry]geometry package[/url] instead which is very simple to use and will avoid any possible inconsistencies.

The tex file you sent me was not a [b]minimal[/b] example as it had a lot of irrelevant stuff in it. As explained in the article it is a good learning exercise to construct a minimal example.
1. hyperref should be the last package loaded as it overwrites some commands;
2. Use the starred command \section* to not be numbered, but this removes it from the table of contents. To put it back you could use \addcontentsline, but this is then linked, so use the alternative \addtocontents{toc}{Unique Chapter} which isn't linked and the Unique Chapter can be made bold, larger or whatever you fancy.
The result is
[code]\documentclass{article}
\usepackage[bookmarks=false,colorlinks=true,linkcolor={blue},pdfstartview={XYZ null null 1.22}]{hyperref}
\begin{document}
\tableofcontents
\section*{Unique Chapter}
\addtocontents{toc}{Unique Chapter}
\subsection{workin' on it}
\end{document}[/code]Note that this is a minimal example - it is a complete working document which shows the features required but does not include any extraneous material.

Yes, exactly, that's what I wanted.

Sorry for not giving the minimal example, but I couldn't express very well my ideas.

I've got one more question: the numeration of subsections starts with 0.1, how can I make it to start with 1?


The equation $\cos^2{x}-\sin{x}+a=0$ has root(s) when $x\in(0,\frac{\pi}{2})$, find $a$.

This is impossible.
$x=0$ results in $1-0+a=0$, implying $a=-1$
$x=\frac{\pi}{2}$ results in $0-1+a=0$ implying $a=1$
How can $a=1=-1$ be true.   :blush:

[hide]$\cos^2x-\sin x+a=0$
$1-\sin^2x-\sin x+a=0$
$0=\sin^2x+\sin x-\left(a+1\right)$

$0< x< \frac{\pi}{2}$
$0<\sin x< 1 $
$\sin x=\frac{-1\pm\sqrt{1^2-4\cdot 1\cdot\left(-a-1\right)}}{2}=\frac{-1\pm\sqrt{4a+5}}{2}$
Here only the plus-signed root can be within the interval.
$1<4a+5<9$
$-1<a<1$
Since the values of the roots depend on the value of $a$, I don't think there is a specific number to find.[/hide]

Lemme try ...

[hide]
cos^2x+sin^2x=1, from this, cos^2x = 1-sin^2x

1-sin^2x-sinx+a=0

sin^2x+sinx-1=a

Let y=sin x

y^2+y-1 = a

Interesting enough, one of hte solutions to this equation when a = 0 is the reciprocal of the golden ratio... but that's irrelevant.

We can factor y^2+y-1 into some fractions but that's not going to help us get the value of a.

Yeah, I agree with everyone else, the solutions depend on the value of a because that will affect the discriminant when solving using the quadratic formula.

[/hide]

Set $\sin x=t$, we can rewrite our equation to $t^2+t-1=a$. Since $0<x<\frac{\pi}{2}$, We should determin the condition such that $y=t^2+t-1\ (0<t<1)$ and $y=a$ have a least of one intersection point.

[quote="seamusoboyle"]I think its that the roots in are in between those values, you just proved the roots cant be both 1 & 0....[/quote]
No, I proved your function cannot exist on that interval.  If you were to reword your question as find the roots of $f(x)=cos^2x-sinx+a$ on $x \in (0,\frac{\pi}{2})$ then you would have roots but you specified f(x)=0.  If this is what you meant however then the equation may be solved like
[hide="This"]
$-sin^2(x)-sin(x)+(1+a)=0$ Let u=sin(x) Then
$-u^2-u+(1+a)=0$
$u=\frac{1\pm \sqrt{5+4a}}{-2}$
$x=sin^{-1}(\frac{1\pm \sqrt{5+4a}}{-2})$
The solutions of this on $x\in(0,\frac{\pi}{2})$ are those that you are looking for.
[/hide]


Let be given a right triangle $ABC$ at $C$ have: $b+c\neq 1,\,c-b\neq 1$
Prove that: $\log_{b+c}a+\log_{c-b}a=2\log_{b+c}a.\log_{c-b}a$

[hide]Divide $2\log_{b+c}a$.  Then subtract so the left hand side is left with 1/2.
$1/2=\log_{c-b}a-\frac{\log_{c-b}a}{2\log_{b+c}a}$.
Factor out $\log_{c-b}a$ and then rewrite under one fraction.
$1/2=\log_{c-b}a(\frac{\log_{b+c}(\frac{a^{2}}{b+c})}{\log_{b+c}(a^{2})})$
By the change of base formula, we have $1/2=\log_{c-b}a(1-\log_{a^{2}}(b+c))$
Since $1/\log_{a}b=\log_{b}a$ by the change of base formula, we have $1/2*\log_{a}(c-b)=1-\log_{a^{2}}(b+c)$.
If we changed the log base on the left hand side to $a^{2}$, we would need to multiply the equation on that side by 2 to maintain equality, which removes the one half.  Then add the log from the right hand side.
$\log_{a^{2}}(c-b)+\log_{a^{2}}(c+b)=\log_{a^{2}}(c^{2}-b^{2})=\log_{a^{2}}(a^{2})=1$ by the Pythagorian theorem.[/hide]

Thanks for your solution. And here is my proof  :D
We have:
\begin{eqnarray*}\log_{b+c}a+\log_{c-b}a&=&\frac{1}{\log_{a}(b+c)}+\frac{1}{\log_{a}(c-b)}\\ &=&\frac{\log_{a}(b+c)+\log_{a}(c-b)}{\log_{a}(b+c)\cdot\log_{a}(c-b)}\\ &=&\frac{\log_{a}a^{2}}{\frac{1}{\log_{b+c}a}\cdot\frac{1}{\log_{c-b}a}}\\ &=&2\log_{b+c}a\cdot\log_{c-b}a \end{eqnarray*}

[hide]$\implies \frac{1}{\log_{a}(b+c)}+\frac{1}{\log_{a}(c-b)}=\frac{\log_{a}(b+c)+\log_{a}(c-b)}{\log_{a}(b+c)\log_{a}(c-b)}=\frac{2}{\log_{a}(b+c)\log_{a}(c-b)}$.
$\implies \log_{a}(b+c)+\log_{a}(c-b)=\log_{a}(c^{2}-b^{2})=2$
$\implies a^{2}=c^{2}-b^{2}$, which is true from Pythagorean Theorem.[/hide]

$\left\{\begin{array}{c}\triangle ABC\ ,\ A=90^{\circ}\\\ a\not\in\{ 1\pm\ b\}\end{array}\right\|\Longrightarrow \log_{a+b}c+\log_{a-b}c=2\log_{a+b}c\cdot\log_{a-b}c\ \ (*)\ .$

[b]Proof.[/b]

$(*)\Longleftrightarrow$ $\left\{\begin{array}{c}\frac{1}{\log_{a-b}c}+\frac{1}{\log_{a+b}c}=2\\\\ \log_{x}y\cdot\log_{y}x=1\end{array}\right\|$ $\Longleftrightarrow$ $\log_{c}(a-b)+\log_{c}(a+b)=2$

$(*)$ $\Longleftrightarrow$ $\log_{c}\left(a^{2}-b^{2}\right)=2$ $\Longleftrightarrow$ $a^{2}-b^{2}=c^{2}$ $\Longleftrightarrow$ $a^{2}=b^{2}+c^{2}\ .$


function
calculus
derivative
quadratics
linear algebra
matrix
real analysis

How to determine whether a point is a maximum or a minimum at point $(a,b)$, if $D=\frac{\partial f}{\partial x \partial x} \frac{\partial f}{\partial y \partial y} - (\frac{\partial f}{\partial x \partial y})^2$ at $(a,b)$ is $0$? thanks in advance

That should be $D=\frac{\partial^2f}{\partial x^2} \cdot \frac{\partial^2f}{\partial y^2} - \left(\frac{\partial^2f}{\partial x\partial y}\right)^2.$

This is the inconclusive case for the second derivative test. As in one variable, you can't expect the second derivative to settle everything. (The quadratic form determined by the matrix of second derivatives is not positive definite, negative definite, nor conclusively indefinite. It may be positive semidefinite, negative semidefinite, or zero.) You will have to settle the case using criteria other than second derivatives.

There's still a little bit that the second derivatives can tell you. (This is the "semidefinite" case referred to above.) If either $\frac{\partial^2f}{\partial x^2}>0$ or $\frac{\partial^2f}{\partial y^2}>0$ at the critical point, then we cannot have a local maximum. Similarly, if either $\frac{\partial^2f}{\partial x^2}<0$ or $\frac{\partial^2f}{\partial y^2}<0$ at the critical point, then we cannot have a local minimum.

Here's a family of examples to illustrate what might happen. In all cases, $(0,0)$ is a critical point. $\frac{\partial^2f}{\partial x^2}>0$ at that critical point, but all other second derivatives are zero, hence $D=0.$ As I said, this cannot be a local maximum. But let's look at what it could be:

Case 1: $f(x,y)=x^2.$

Non-isolated local minimum. (Think of it as a flat valley floor along the $y$-axis.)

Case 2: $f(x,y)=x^2+y^4.$

Simple local minimum.

Case 3: $f(x,y)=x^2-y^4.$

Saddle point.

Case 4: $f(x,y)=x^2+y^3.$

Not an extremum. Take the valley floor from Case 1 and tilt it, but leave a flat inflection point.

But what happens when $D<0$? Is this situation possible?

yes $f=x^2-y^2$

does the trick

if that happens, then you can be SURE that is not an extremum


[quote="arqady"][quote="M4RI0"]Let $x, y, z$ be positive numbers such
that $xyz = 1$, prove the inequality

$\frac{x^2}{x + y + y^3z}+ \frac{y^2}{y + z + z^3x} + \frac{z^2}{z + x + x^3z} \leq 1.$[/quote]
$\sum_{cyc}\frac{x^2}{x + y + y^3z}=\sum_{cyc}\frac{x^2}{x + y + \frac{y^2}{x}}=\sum_{cyc}\frac{x^3}{x^2 + xy + y^2}=\sum_{cyc}\frac{x^3-y^3+y^3}{x^2 + xy + y^2}=$
$=\sum_{cyc}(x-y)+\sum_{cyc}\frac{y^3}{x^2 + xy + y^2}=\sum_{cyc}\frac{y^3}{x^2 + xy + y^2}.$
Hence, $\sum_{cyc}\frac{x^2}{x + y + y^3z}=\frac{1}{2}\cdot\sum_{cyc}\frac{x^3+y^3}{x^2 + xy + y^2}=\frac{1}{2}\cdot\sum_{cyc}(x+y)\cdot\frac{x^2-xy+y^2}{x^2+xy+y^2}\geq$
$\geq\frac{1}{2}\cdot\sum_{cyc}\frac{1}{3}\cdot(x+y)=\frac{1}{6}\sum_{cyc}(x+y)=\frac{x+y+z}{3}\geq1.$ :? :D  :)[/quote]

Just want to say that:
$\sum \frac{x^3}{x^2+xy+y^2} \geq \sum \frac{2x-y}{3}$ and the rest is a piece of cake.

[quote="M4RI0"]Let $x, y, z$ be positive numbers such
that $xyz = 1$, prove the inequality

$\frac{x^2}{x + y + y^3z}+ \frac{y^2}{y + z + z^3x} + \frac{z^2}{z + x + x^3z} \leq 1.$[/quote]

[color=darkred]I think you are wrong with the sign.[/color]

Davron Latipov

[quote="arqady"][quote="Gibbenergy"]

Just want to say that:
$\sum \frac{x^3}{x^2+xy+y^2} \geq \sum \frac{2x-y}{3}$ and the rest is a piece of cake.[/quote]
One more opportunity:
$\sum_{cyc} \frac{x^3}{x^2+xy+y^2}=\sum_{cyc}\frac{x^4}{x^3+x^2y+xy^2}\geq\frac{(x^2+y^2+z^2)^2}{\sum(x^3+x^2y+x^2z)}=\frac{x^2+y^2+z^2}{x+y+z}.$
But $\frac{x^2+y^2+z^2}{x+y+z}\geq1\Leftrightarrow x^2+y^2+z^2\geq x+y+z.$ This is obviously true.[/quote]are you sure that is true with t<1

[quote="eagerkill"][quote="arqady"]$x^2+y^2+z^2\geq x+y+z.$ This is obviously true.[/quote]are you sure that is true with t<1[/quote]
$x^2+y^2+z^2\geq x+y+z\Leftrightarrow x^2+y^2+z^2\geq\sqrt[3]{xyz}(x+y+z).$
If $x=a^3,$ $y=b^3,$ $z=c^3$ then $x^2+y^2+z^2\geq\sqrt[3]{xyz}(x+y+z)\Leftrightarrow$
$\Leftrightarrow a^6+b^6+c^6\geq a^4bc+b^4ac+c^4ab.$ This is Muirhead.


Max picks two different cards from a standard 52-card deck. What is the probability that the cards are of different suits?

This is simply $ 1 \times \frac{51 \minus{} 12}{51}$, or $ \boxed{\frac{39}{51}}$.

[quote="ernie"]This is simply $ 1 \times \frac {51 \minus{} 12}{51}$, or $ \boxed{\frac {39}{51}}$.[/quote]

Which simplifies down to $ \frac{13}{17}$.  Thanks!

[hide="Alternative Solution"]Although not the best way, we can solve this problem using complementary counting.  There are $\dbinom{13}{2}$ ways to select two distinct cards within a suit, so there are $4\times\dbinom{13}{2}$ ways to choose two distinct cards that are within a suit.  There are $\dbinom{52}{2}$ ways to select any two distinct cards.  Thus, the desired probability is $1-\dfrac{4\times\dbinom{13}{2}}{\dbinom{52}{2}}=\boxed{\frac{13}{17}}$.[/hide]

[quote name="hoyam" url="/community/p17970782"]
[quote=bestzack66]I put in $\frac{3}{4}$ at first :wallbash_red:

I realized my mistake that I didn't subtract 1 in the denom[/quote]

$5$ year revive?
[/quote]

Uhhhh in gamebot threads im allowed to

[quote=RTB][quote=bestzack66]I put in $\frac{3}{4}$ at first :wallbash_red:

I realized my mistake that I didn't subtract 1 in the denom[/quote]
Thank goodness I'm not the only one who did that.[/quote]

bruh same smh


as of now my profile says i have 199 posts, but i've done more! wats wrong?!?!

posts in Games& Fun Factory Don't count.

No, they don't. Haven't you ever noticed people with 0 posts posting all the time in G&FF?

[url]http://www.mathlinks.ro/viewtopic.php?t=185890[/url]

GFF is now a raging wallow of time wasting and now comments there have the distinguishment of being slightly better than YouTube's, which is herroundous for a website whose remaining portion has an average IQ of 135. It would be okay if so many of the users just went to a different site for spamming it. I see no point in coming to AoPS for the GFF section, as many AoPSers seem to do (see people with 0 posts).

Seriously, the only good place for nonmath discussion is the Round Table, and if it weren't for ZetaX's moderation, it's starting to go down hill as well. The posts generally are far more intelligent than those of Polls/GFF , but the threads still have spammers.

It's sad that most community forums are spammed also (see the NC forum)...

so if u post in say the arizona forum your posts dont count?

I shall only repeat what has been posted.

[quote]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=185890[/quote]


im doin some hw and its a large review and one of the problems asks...

Explain from an atomic viewpoint why charge is usually transferred by electrons.

Electons are on the outside of the atom and are always moving by the quantum theory.  The protons are at the center of the atom and don't really go anywhere else, generally .  So, electrons are just closer to "the outside world" and apparently interact with it more than the protons do.


if it's for a course, i read the text.  doing the questions will usually help as well.
reading the text is very helpful though.

i just sit down and read. and then do problems related to waht i just read. practice is always good when you're self teaching.

[quote="RiemannZetaFunction"]I was just wondring how everybody here self teaches themself? List any tips or secrets or just genreal methods to help benafit everybody here.[/quote]

You need to want to improve your knowledge and another important thing is to practise a lot....

Yes, i think one of the most important things in self teaching is determination and your ability to sit down and figure this stuff out without help from a trained professional (im making teachers look good here) :)

Don't be afraid to re-read stuff over and over and over again... sit down and understand every theorem you are going to use and every concept you are trying to learn (assuming this is math and/or physics.). Self teaching is very satisfying  :)

I prefer to be self-taught.  I'm pretty bad at listening to others explain stuff, I just zone out and don't pay attention.  I absorb more by reading than listening.

Take advantage of flexibility.

If you are self-teaching yourself something you are really interested in, take the time to really explore and ponder the subject and you'll find yourself with a very high level of understanding. You'll get to the point where you are always a step or two ahead of the book, and although you'll start off slow, you'll finish very quickly.

If you are self-teaching yourself something you aren't quite that interested in, or if you need to adhere strictly to a specific syllabus and timeframe, then all you can really do is plow through it with determination, and make sure to keep a clear head. Do whatever it takes to keep a clear head.


limit
real analysis
real analysis unsolved

Let $x>1.$ Find the sum of the series \[S = \sum_{n=0}^{\infty}\frac{2^{n}}{x^{2^{n}}+1}\]


If $x^{2}+2px+15=0$ has real roots for $x$, then the absolute value of the sum of these two roots for x must be greater than $k\sqrt{w}$,
where $k\sqrt{w}$ is in simplest radical form and where $k\sqrt{w}$ is as small as possible to meet the given conditions.
Find the value of $k+w$.

[hide]answer: 17[/hide]

Please help..

[quote="rem"]What is $p$?[/quote]
$p$ is not given.

[quote="Power"]If $x^{2}+2px+15=0$ has real roots for $x$, then the absolute value of the sum of these two roots for x must be greater than $k\sqrt{w}$,
where $k\sqrt{w}$ is in simplest radical form and where $k\sqrt{w}$ is as small as possible to meet the given conditions.
Find the value of $k+w$.
[/quote]

[hide="Solution"]
Vieta's Formula $\implies \left| \sum x_{i}\right| = |2p|$

So we need to find all $p$ that are possible. Using the discriminant:
\[(2p)^{2}-4 (1)(15) \geq 0 \implies |p| \geq \sqrt{15}\implies \boxed{ |2p| \geq 2 \sqrt{15}\implies k+w \geq 17 }\]
[/hide]


Can someone please help me out on this.

What is the median of an equilateral triangle mesuring 10 cm each side...

Please tell me how you found it. Thanks!

[quote="i_like_pie"]A median of a triangle is the line segment going from a vertex to the midpoint of the opposite side. I think you're asking for the centroid, which is the point at which the three medians meet. In an equilateral triangle, the centroid, circumcenter (perpendicular bisectors), incenter (angle bisectors), and orthocenter (altitudes) are all the same point.[/quote]

My friend just told me the answer is 8.6. How did he find that? That's the answer of the problem I'm looking to find.

Oh! You want the [i]length[/i] of the median.

In an equilateral triangle, the length of the median is $\frac{s\sqrt{3}}{2}.$ In this case, it is $5\sqrt{3},$ or approximately $8.66.$


function
calculus
derivative
induction
IMO Shortlist
algebra solved
algebra

A sequence {u_n} of real numbers is given by u_1 and, for n \geq 1, by the recurrence relation 4*u_(n+1) = (64*u_n+15)^(1/3). Describe the behaviour of u_n for n->00.

[i]IMO shortlist 1981[/i]

easy exercise.
consider the function: f(x)=64x3-64x-15. We can see that
f(x)=64x3-4x-60x-15=4x(4x+1)(4x-1)-15(4x+1)=(4x+1)(16x2-4x+15)=64(x+1/4)(x- (1+\sqrt 61)/8)(x-(1-\sqrt 61)/8) and using derivatives we have f'(x)=192x2-64 we see that the function is increasing on (-\infty , -1/ \sqrt 3) and (1/ \sqrt 3 , +\infty ) and decreasing on (-1/ \sqrt 3, 1/ \sqrt 3) and positive and negative on invervals given by the roots of f(x).

and now depending on the choice of u1 we either have
case 1:  u1 > (1+\sqrt 61)/8 > 1> 1/ \sqrt 3 => by induction the sequence un is increasing, and it is not bounded, otherwise it would have to be convergent, but it is already larger than any of the possible three limits.

the other cases can be easily treated simillarly, and all of them will result in limits one of the three roots.


I tried posting this in Pre-Olympiad, but I couldn't find any solutions.

Find all 9-digit numbers such that the numbers formed by looking at the first, middle, and last three digits are all squares and the number itself is square.

I have found 6 solutions... There may be more...
[hide="6"]100400400, 121484484, 144576576, 169676676, 196784784, 225900900 - squares of 10020, 11022, 12024, 13026, 14028, 15030
These we made by my very weird "formula":
x is a digit: (x)0(2x) So if x=10, then it is 10020 and the square is then 100400400
This only works with x=10-15, inclusive because $(2x)^{2}$ = 900 and there is no other even square that is three digits and greater than that. Also $10^{2}$ is the smallest three digit square.
If that makes any sense at all....... :P [/hide]
[hide="EDIT:"]You can reverse x and 2x to get 6 more possibilities.[/hide]


I think we wants us to find the minimum value of $ a \plus{} b \plus{} c \minus{} abc$ given that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$. I thought optimization problems did not belong in high school basics because they require nontrivial proof?

This topic is definitely not Intermediate level.

Can anyone give me some hints for this problem. I tried but I still don't know how to do this.

[quote="devil_maths"]let $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$
find min A=$ a \plus{} b \plus{} c \minus{} abc$[/quote]

[hide="Partial Solution"]
By QM-AM-GM inequality, we have $ \sqrt {\frac {a^2 \plus{} b^2 \plus{} c^2}{3}}\ge \frac {a \plus{} b \plus{} c}{3}\ge \sqrt [3]{abc}$

Doabli: Thanks for pointing out.  :)

Or $ 27\ge (a \plus{} b \plus{} c)^3\ge 27abc$

We will have the smallest value of $ a \plus{} b \plus{} c \minus{} abc$ when $ a \plus{} b \plus{} c$ becomes smallest and $ abc$ becomes largest.

The smallest value of $ a \plus{} b \plus{} c$ is $ 3\sqrt [3]{abc}$, and largest value of $ abc$ is $ (a \plus{} b \plus{} c)^3$[/hide]

[quote="FantasyLover"]
Or $ 27\ge (a \plus{} b \plus{} c)^3\ge abc$
[/quote]
I think this is $ 27\ge (a \plus{} b \plus{} c)^3\ge 27abc$. And we don't have the condition that $ a,b,c \ge 0$ so I don't think we can apply RMS-AM-GM. And I also have the problem about this, $ a\plus{}b\plus{}c$ max when $ a\equal{}b\equal{}c$, $ abc$ min when $ a\equal{}b\equal{}\minus{}c$.


calculus
integration
geometry
inradius
area of a triangle
Heron\u0027s formula

find all integer a,b,c such that:
$ a\plus{}b\plus{}c\equal{}abc$

$ (a,b,c)\equal{}(0,0,0)$, $ \pm(1,2,3)$ come to mind but I'm not sure if they're the only ones.

[hide="Solution."]Since $ 2 \plus{} 2 \plus{} 2 < 2 \cdot 2 \cdot 2$, that means that one of the three variables must be less than $ 2$; similarly, since $ \minus{}2 \minus{} 2 \minus{} 2 > \minus{}2 \cdot \minus{}2 \cdot \minus{}2$, that means that one of the three variables must be greater than $ \minus{}2$.  Say that $ x$ is that variable;  it can only take on three values: $ 0$, $ 1$, and $ \minus{}1$.

If it's $ 0$, that yields $ y \plus{} z \equal{} 0$.  That gives any three integers of the form $ (0, x, \minus{}x)$ as a valid solution, in any permutation.

If it is equal to $ 1$, that yields $ y \plus{} z \plus{} 1 \equal{} yz$.  Solving for $ y$ yields $ y \equal{} \frac{z\plus{}1}{z\minus{}1}$.  There are only four possible values for $ z$ that result in an integral value for $ y$: $ \minus{}1$, $ 0$, $ 2$, and $ 3$.  This results in two possible solutions: $ (1, 0, \minus{}1)$, $ (1, \minus{}1, 0)$, $ (1, 3, 2)$, and $ (1, 2, 3)$.  Of these, the first two have already been considered previously; the last two are identical, but new.

If it is equal to $ \minus{}1$, then $ y \plus{} z \minus{} 1 \equal{} \minus{}yz$.  Rewriting that in terms of $ y$ yields $ y \equal{} \minus{}\frac{z\minus{}1}{z\plus{}1}$.  This also gives four possible values for $ z$, which are exactly the opposite of the ones above: $ \minus{}3$, $ \minus{}2$, $ 0$, $ 1$.  The resultant solutions are $ (\minus{}1, 0, 1)$, $ (\minus{}1, 1, 0)$, $ (\minus{}1, \minus{}3, \minus{}2)$, and $ (\minus{}1, \minus{}2, \minus{}3)$.

So there are two general solutions: permutations of $ (0, x, \minus{}x)$ and permutations of $ \pm(1, 2, 3)$.[/hide]

I guess only $ a \equal{} b \equal{} c \equal{} 0$ is the solution. I got no solid solution, but maybe AM - GM can be used for this?

Did you pay any attention to the previous posts?  The solutions are permutations of $ (0,x,\minus{}x)$, and $ \pm (1,2,3)$.

First assume that $ a,b,c$ are all positive integers. $ b\plus{}c \equal{} a(bc\minus{}1)$ so $ bc\minus{}1|b\plus{}c$.
$ bc\minus{}1\leq b\plus{}c$ $ \equal{}>$ $ bc\leq b\plus{}c\plus{}1$ $ \equal{}>$ $ 2bc\leq (b\plus{}1)(c\plus{}1)$ $ \equal{}>$ $ 2\leq (\dfrac{b\plus{}1}{b})(\dfrac{c\plus{}1}{c})$.
Now it is easy to generate the possible values of $ (b,c)$ (wlog letting $ b\geq c$): $ (1,1), (2,1), (2,2), (3,1), (3,2)$ and we can exclude $ (1,1), (2,2)$ as they don't work.

Noting that this case is analogous to when exactly two of the numbers are positive and when all three are negative, we can generate the triples: $ (1,2,3), (\minus{}1,\minus{}2,\minus{}3)$ and permutations of these.

Now suppose exactly $ b$ and $ c$ are negative. Then $ \frac{b\plus{}c}{bc\minus{}1}$ is a positive number. However, $ bc\minus{}1$ will always be nonnegative yet $ b\plus{}c$ is always negative, leading to a contradiction.

Finally, letting one of the variables be $ 0$ we easily get permutations of the triples $ (0, b, \minus{}b)$.

Here's an interesting interpretation:  given a triangle with side lengths $ x, y, z$, let $ a \equal{} x \plus{} y \minus{} z, b \equal{} x \minus{} y \plus{} z, c \equal{} \minus{}x \plus{} y \plus{} z$.  Then Heron's formula gives

$ abc(a \plus{} b \plus{} c) \equal{} 16K^2$

where $ K$ is the area of the triangle.  If the triangle has inradius $ r$, then $ K \equal{} rs \equal{} \frac{r}{2} \cdot (a \plus{} b \plus{} c)$, which gives

$ abc \equal{} 4r^2 (a \plus{} b \plus{} c)$.

Hence the integer solutions here determine the integer triangles with inradius $ r \equal{} \frac{1}{2}$.  Two follow-ups:

- What are the integer triangles with inradius $ 1$?  $ 2$?
- What are the [i]rational[/i] solutions?

[hide="Remark"] I find the rewrite $ 1 \equal{} 4r^2 \left( \frac{1}{ab} \plus{} \frac{1}{bc} \plus{} \frac{1}{ca} \right)$ illuminating.  What can you say about the possible values of the biggest of the three terms?  The smallest? [/hide]


calculus
integration
function
inequalities
calculus computations

im working with an integral and im hoping to find a nice function f such that for big x,

$ f(x)\le \int_{xT}^\infty e^{ \minus{} t}t^{ \minus{} a} dt$ where $ a,T > 0$ are fixed

of course, exact values for the integral would be cool too i guess.

Integrate by parts.

$ g(x)\equal{}\int_{xT}^{\infty}e^{\minus{}t}t^{\minus{}a}\,dt\equal{}\minus{}e^{\minus{}t}t^{\minus{}a}\bigg{|}_{xT}^{\infty}\minus{}a\int_{xT}^{\infty}e^{\minus{}t}t^{\minus{}a\minus{}1}\,dt$

So that gives us the inequality
\[ g(x)<e^{\minus{}xT}(xT)^{\minus{}a}\]
That already almost answers your first question, but with the inequality sign opposite from what you wanted. But if we could integrate by parts once, why not do it again?
\[ e^{\minus{}xT}\left[(xT)^{\minus{}a}\minus{}a(xT)^{\minus{}a\minus{}1}\right]<g(x)<e^{\minus{}xT}(xT)^{\minus{}a}\]
But why not keep right on integrating by parts?
\[ e^{\minus{}xT}\left[(xT)^{\minus{}a}\minus{}a(xT)^{\minus{}a\minus{}1}\right]<g(x)<e^{\minus{}xT}\left[(xT)^{\minus{}a}\minus{}a(xT)^{\minus{}a\minus{}1}\plus{}a(a\plus{}1)(xT)^{\minus{}a\minus{}2}\right]\]
\[ e^{\minus{}xT}\left[(xT)^{\minus{}a}\minus{}a(xT)^{\minus{}a\minus{}1}\plus{}a(a\plus{}1)(xT)^{\minus{}a\minus{}2}\minus{}a(a\plus{}1)(a\plus{}2)(xT)^{\minus{}a\minus{}3}\right]<g(x)<e^{\minus{}xT}\left[(xT)^{\minus{}a}\minus{}a(xT)^{\minus{}a\minus{}1}\plus{}a(a\plus{}1)(xT)^{\minus{}a\minus{}2}\right]\]
We're building an [i]asymptotic series[/i] here. It shows behavior that is typical of such an asymptotic series, including that it diverges as a series for all $ x,$ but is nonetheless a highly valuable approximation.


Exclude either $x$ or $y$ from the following system
\[
\left\{ \begin{array}{l}
 2x^3  + 3xy^2  - 4x^2 y^2  + y + 1 = 0 \\
 3x + 5x^3 y^2  - 2xy^5  - xy + 2 = 0 \\
 \end{array} \right.
\]

[i](10 points)[/i]

Is there a specific method to tackle such problems...???
 :)

The first equation of the system is already a quadratic equation with respect to $y$.


The Fibbonacci Sequence 1,1,2,3,5,8,13,21... starts with two 1s, and eachterm afterwards is the sum of two predeccessors. Which one of the ten digits is the last to appear in the units of a number in the Fibbonacci Sequence?
A, 0
B, 4
C, 6
D, 7
E, 9

So which is the correct answer and why?Could anyone help me?

[hide]The answer is [b]C  6[/b] because it means which one of them is the last one to appear in the units digit.
1  1  2  3  5  8  13  21  34  55  89  144  233  377  610  987  1597  2584  4181  6765  1094[b]6[/b]
6 is the last one of the 10 digits to appear in the units position of a number in the Fibbonacci Sequence
In a contest you would use just calculate the units digit and use proses of elimination after 0[/hide]

No problem.  These help me get faster on math contests.


function inequalities quadratics trigonometry conics parabola calculus

Find the maximum and minimum values of the following function:
$f(x) = \sqrt{x^4-3x^2-6x+13}-\sqrt{x^4-x^2+1}$

An attempt to solve the problem analytically results in the $7$-th degree equation $12x^7-40x^6+6x^5+43x^4-24x^3-13x^2+18x+9=0$. I doubt it is solvable in radicals, though everybody is welcome to try. Using the computer, it is easy to find that the maximum $3.162278\dots$ is attained at $-0.847127\dots$ and the minimum $-1.547493\dots$ is attained at $2.794621\dots$.

3.162278 looks familiar- that's $\sqrt{10}$ to that many digits. Hmm...

And $\sqrt{10}$ it is! If you try to prove the corresponding inequality, you can see that it reduces to a quadratic one and the point of maximum is the negative root of the equation $3x^2-x-3=0$. On the other hand, the point of minimum is a root of the equation $4x^5-12x^4+2x^3+3x^2-5x-3=0$, which I cannot solve or factor...

Just idea!  $2\sqrt{2}f(\cos \theta)=2\sqrt{2\cos ^ 4 \theta-6\cos ^ 2 \theta-12\cos \theta +26}-\sqrt{\cos 4\theta}.$

Rewrite $f(x)=\sqrt{(x^2-3)^2+(x^2-2)^2}-\sqrt{x^2+(x^2-1)^2}$.
Denote $A =(3,2) , B = (0,1)$ and $P = (x,x^2)$.
Then $f(x) = PA -PB$ where P is a variable point on the parabola $y=x^2$.
But $PA-PB \leq  AB=\sqrt{10}$ for all possible positions of P.

By differentiation and set $f'(x)=0$, we can also arrive at the same result for the maximum.
However, it seems that the minimum is attained at some "ugly" value of x.  :P


If $n>1$, find the two smallest integral values of $n$ for which $x^{2}+x+1$ is a factor of $(x+1)^{n}-x^{n}-1$, over the set of polynomials with integer coefficients.

solve $x^{2}+x+1=0$
$x=cis(120)$ or $x=cis(240)$

To ensure $x^{2}+x+1=0$ is a factor of $(x+1)^{n}-x^{n}-1$

$(x+1)^{n}-x^{n}-1=0$ should have two roots $cis(120)$ and  $cis(240)$, but since they are conjugates, having one of the root means the other one is also the root.

we know that $cis(120)+1=cis(60)$ and $cis(240)+1=cis(300)$

Therefore..

$(cis(60))^{n}-(cis(120))^{n}-1=0$
$cis(60n)-cis(120n)-1=0$
the only way this satisfies is when:
$60n=360p+60$ and $120n=360q+120$ and p and q are integers
$n=6p+1$ and $n=3q+1$
which implies $n$ divide by 6 has remainder of 1...

or..

$60n=360p+300$ and $120n=360q+240$
$n=6p+5$ and $n=3q+2$
which implies $n$ divide by 6 has remainder of 5..

Therefore, the two smallest $n$ are $\boxed{5,7}$

[quote="pkothari13"][quote="Allan Z"]the only way this satisfies is when:
$60n=360p+60$ and $120n=360q+120$ and p and q are integers

or..

$60n=360p+300$ and $120n=360q+240$
[/quote]

Could you explain how that works? I don't seem to understand.[/quote]

um.. because you wanna make the difference of two imaginary numbers to be $1$..
the only way it can happen is
$cis(60)-cis(120)=\frac{1+i\sqrt{3}}{2}-\frac{-1+i\sqrt{3}}{2}=1$

or
$cis(300)-cis(240)=1$

I don't think my solution is the best though, but I can't think of anything else :mad:

Here's my solution (without looking at the other posts).

[b]Solution.[/b] If $x^{2}+x+1|(x+1)^{n}-x^{n}-1$, then $(\omega+1)^{n}-\omega^{n}-1=0$, where $\omega=e^{i\frac{2\pi}{3}}$ is the third root of unity.  Since $\omega^{2}+\omega+1=0$, we have $(-1)^{n}\omega^{2n}-\omega^{n}-1=0\implies (-1)^{n+1}\omega^{2n}+\omega^{n}+1=0$.

[b]Lemma.[/b] Let $\zeta$ be the $n$-th root of unity.  Then $\sum_{j=0}^{n-1}\zeta^{jk}=n$ if $n|k$, and equal to $0$ otherwise.

[b]Proof.[/b]  A direct application of geometric summation. The proof is left as an exercise.

Ipso facto of the above [i]lemma[/i], $n$ must be odd and $3\not| n$.  The two smallest such $n$ are $5$ and $7$. $\blacksquare$

[b]Remark.[/b] The method above can easily be generalized.


The area of a convex quadrilateral $ABCD$ is 1 square unit.
Show that $AB+BC+CD+DA+AC+BD \geq 4+\sqrt 8$.
When does the equality sign hold?

Lets show the perimeter is $\ge 4$ and the sum of the diagonals is $\ge \sqrt{8}$.

In the first part, fix AC, area ABC, and area ADC.  It's easy to show we minimize the perimeter when AB = AC, AD = AC, so that B, D lie on the perp bisector of AC.  Letting M be the midpoint of AC, and now varying areas ABC, ADC (read: BM, DM),  from Cauchy, the perimeter is twice $\sqrt{MA^2 + BM^2} + \sqrt{MA^2 + DM^2} \ge \frac{1}{\sqrt{2}} (BM + MA + MD + MA) = \frac{1}{\sqrt{2}}$ times the sum of the diagonals.   So it suffices to show only the second part now.

In the second part, drop perpendiculars from $A, C$ to $BD$ to meet $BD$ at $X, Y$.  Then $AC + BD \ge 2\sqrt{AC*BD} \ge 2\sqrt{(AX+CY)*BD} = 2\sqrt{2}$, the first ineq being AMGM, the second ineq being triangle ineq, and the 3rd eq appealing to the area.

Checking when equality holds during our work above, its clear it holds when $ABCD$ is a square.