text: # Thread: Extremely difficult IMO problem
  
  1. ## Extremely difficult IMO problem
  
  This problem is from the 2003 IMO, the third level of the American Mathematics Competition. I do not even know how to approach this.
  
  "Determine all paris (a,b) such that $\frac{a^2}{2ab^2-b^3+1}$ is a positive integer."
  
  The only assertion I can make is that $2ab^2-b^3+1>0$.
  
  2. Originally Posted by Jameson
  This problem is from the 2003 IMO, the third level of the American Mathematics Competition. I do not even know how to approach this.
  
  "Determine all paris (a,b) such that $\frac{a^2}{2ab^2-b^3+1}$ is a positive integer."
  
  The only assertion I can make is that $2ab^2-b^3+1>0$.
  Are a, b integers as well? If so, then another criterion comes from the rational roots test:
  
  Let n be a positive integer, then let
  $\frac{a^2}{2ab^2-b^3+1} = n$
  
  So $nb^3-2nab^2+(a^2-n)=0$
  
  Thus we know that the only rational b values that satisfy this equation are of the form:
  $b = \frac{factor \, of \, a^2-n}{factor \, of \, n}$
  
  Since b must be an integer, the only possible values for b are those for which factors of n are factors of a^2-n.
  
  Just a thought.
  
  -Dan
  
  3. Hmmm...
  
  I just found a non-integer pair $\left ( \frac{1 \pm \sqrt{15}}{4}, \, \frac{1}{2} \right )$ that seems to work.
  
  I guess they DON'T need to be integers...
  
  -Dan
  
  4. (Shrugs) I guess by letting n be a positive integer and otherwise letting n and b be arbitrary we can always find two "a" values for each (b, n) pair:
  
  $a = nb^2 \pm \sqrt{n^2b^4-nb^3+n}$
  
  as long as the discriminant is real.
  
  -Dan
  
  5. Could you show your work for your last post? I appreciate your insight. The question asks for all pairs, and I'm getting from your work that there could be infinite. How would I go about first finding the number of pairs?
  
  6. Originally Posted by Jameson
  Could you show your work for your last post? I appreciate your insight. The question asks for all pairs, and I'm getting from your work that there could be infinite. How would I go about first finding the number of pairs?
  $\frac{a^2}{2ab^2-b^3+1}=n$
  Thus,
  $a^2=2ab^2n-b^3n+n$
  
  7. Originally Posted by ThePerfectHacker
  topsquark (the physicist)
  Do I detect a note of disdain here? (I think he's just jealous. ) And hey, even the workers at Friendly's have enough respect to call me "The Professor!"
  
  -Dan
  
  8. by the way the actual question is:
  Determine all pairs of positive integers $(a,b)$ such that
  $\frac{a^2}{2ab^2-b^3+1}$
  is a positive integer.
  
  so a and b must be positive integers.
  
  here are a few examples of positive integers which work:
  a = 126, b = 4
  a = 7, b = 2
  a = 116149892142279364320, b = 123456
  
  9. ## Are there any more pairs except these?
  
  Originally Posted by topsquark
  (Shrugs) I guess by letting n be a positive integer and otherwise letting n and b be arbitrary we can always find two "a" values for each (b, n) pair:
  
  $a = nb^2 \pm \sqrt{n^2b^4-nb^3+n}$
  
  as long as the discriminant is real.
  
  -Dan
  I visited this thread yesterday and the question sounded interesting to me.
  Continuing the quoted work, I tried to find the integer values(of b and n) for which discriminant is real and a perfect square(so that a is an integer), and hence the answer to the question.
  I found that a= 8*(k^4) - k
  b= 2k
  
  where k is any positive integer gives the pairs pair satisfying the conditions in the question.
  e.g
  k=1, b=2, a=7
  k=2, b=4, a=126
  k=3, b=6, a=645
  ...................k is any positive real number
  Are there any more pairs?
  If you want to know how I arrived at the generalization, tell me.
  
  10. The more trivial ones.
  (a,b)=(k,2k)
  (a,b)=(2k,1)
  
  11. Originally Posted by ThePerfectHacker
  The more trivial ones.
  (a,b)=(k,2k)
  (a,b)=(2k,1)
  The method I used show that there should be two values of a for each b
  i.e. if b=2k either a = 8*(k^4) - k or a=k(a=k is not trivial)
  The trivial one is (2k,1)
  Can b be a odd no. except 1? I think no. What do you think?
  
  12. We have,
  $\frac{a^2}{2ab^2-b^3+1}=k\geq 1$
  Thus,
  $a^2-2ab^2k+b^3k-k=0$
  Let us find under which conditions this quadradic has a unique solution, that is when,
  $4b^4k^2-4b^3k+4k=0$
  Thus,
  $b^3k-b^2+1=0$
  Thus,
  $k=\frac{b^2-1}{b^3}$
  Is not an integer because the numerator factors into $(b-1)(b+1)$ and $\gcd(b,b+1)=1$ therefore the only case is $b=1$, which leads to our pairs $(2k,1)$.
  Otherwise what I have showed is that if this equation has two solutions and one is a positive integer then the other must be also because it follows by Viete's theorem.
explanation: The document is a discussion thread from a mathematics forum, focusing on a problem from the 2003 International Mathematical Olympiad (IMO). The problem asks to determine all pairs \((a, b)\) such that \(\frac{a^2}{2ab^2 - b^3 + 1}\) is a positive integer. The thread includes multiple posts with different approaches and insights, including discussions on the nature of \(a\) and \(b\) (whether they are integers), potential solutions, and generalizations.
  
  The thread contains a question and multiple answers or partial solutions to the problem.
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distilabel_metadata.content: # Thread: Extremely difficult IMO problem
  
  1. ## Extremely difficult IMO problem
  
  This problem is from the 2003 IMO, the third level of the American Mathematics Competition. I do not even know how to approach this.
  
  "Determine all paris (a,b) such that $\frac{a^2}{2ab^2-b^3+1}$ is a positive integer."
  
  The only assertion I can make is that $2ab^2-b^3+1>0$.
  
  2. Originally Posted by Jameson
  This problem is from the 2003 IMO, the third level of the American Mathematics Competition. I do not even know how to approach this.
  
  "Determine all paris (a,b) such that $\frac{a^2}{2ab^2-b^3+1}$ is a positive integer."
  
  The only assertion I can make is that $2ab^2-b^3+1>0$.
  Are a, b integers as well? If so, then another criterion comes from the rational roots test:
  
  Let n be a positive integer, then let
  $\frac{a^2}{2ab^2-b^3+1} = n$
  
  So $nb^3-2nab^2+(a^2-n)=0$
  
  Thus we know that the only rational b values that satisfy this equation are of the form:
  $b = \frac{factor \, of \, a^2-n}{factor \, of \, n}$
  
  Since b must be an integer, the only possible values for b are those for which factors of n are factors of a^2-n.
  
  Just a thought.
  
  -Dan
  
  3. Hmmm...
  
  I just found a non-integer pair $\left ( \frac{1 \pm \sqrt{15}}{4}, \, \frac{1}{2} \right )$ that seems to work.
  
  I guess they DON'T need to be integers...
  
  -Dan
  
  4. (Shrugs) I guess by letting n be a positive integer and otherwise letting n and b be arbitrary we can always find two "a" values for each (b, n) pair:
  
  $a = nb^2 \pm \sqrt{n^2b^4-nb^3+n}$
  
  as long as the discriminant is real.
  
  -Dan
  
  5. Could you show your work for your last post? I appreciate your insight. The question asks for all pairs, and I'm getting from your work that there could be infinite. How would I go about first finding the number of pairs?
  
  6. Originally Posted by Jameson
  Could you show your work for your last post? I appreciate your insight. The question asks for all pairs, and I'm getting from your work that there could be infinite. How would I go about first finding the number of pairs?
  $\frac{a^2}{2ab^2-b^3+1}=n$
  Thus,
  $a^2=2ab^2n-b^3n+n$
  
  7. Originally Posted by ThePerfectHacker
  topsquark (the physicist)
  Do I detect a note of disdain here? (I think he's just jealous. ) And hey, even the workers at Friendly's have enough respect to call me "The Professor!"
  
  -Dan
  
  8. by the way the actual question is:
  Determine all pairs of positive integers $(a,b)$ such that
  $\frac{a^2}{2ab^2-b^3+1}$
  is a positive integer.
  
  so a and b must be positive integers.
  
  here are a few examples of positive integers which work:
  a = 126, b = 4
  a = 7, b = 2
  a = 116149892142279364320, b = 123456
  
  9. ## Are there any more pairs except these?
  
  Originally Posted by topsquark
  (Shrugs) I guess by letting n be a positive integer and otherwise letting n and b be arbitrary we can always find two "a" values for each (b, n) pair:
  
  $a = nb^2 \pm \sqrt{n^2b^4-nb^3+n}$
  
  as long as the discriminant is real.
  
  -Dan
  I visited this thread yesterday and the question sounded interesting to me.
  Continuing the quoted work, I tried to find the integer values(of b and n) for which discriminant is real and a perfect square(so that a is an integer), and hence the answer to the question.
  I found that a= 8*(k^4) - k
  b= 2k
  
  where k is any positive integer gives the pairs pair satisfying the conditions in the question.
  e.g
  k=1, b=2, a=7
  k=2, b=4, a=126
  k=3, b=6, a=645
  ...................k is any positive real number
  Are there any more pairs?
  If you want to know how I arrived at the generalization, tell me.
  
  10. The more trivial ones.
  (a,b)=(k,2k)
  (a,b)=(2k,1)
  
  11. Originally Posted by ThePerfectHacker
  The more trivial ones.
  (a,b)=(k,2k)
  (a,b)=(2k,1)
  The method I used show that there should be two values of a for each b
  i.e. if b=2k either a = 8*(k^4) - k or a=k(a=k is not trivial)
  The trivial one is (2k,1)
  Can b be a odd no. except 1? I think no. What do you think?
  
  12. We have,
  $\frac{a^2}{2ab^2-b^3+1}=k\geq 1$
  Thus,
  $a^2-2ab^2k+b^3k-k=0$
  Let us find under which conditions this quadradic has a unique solution, that is when,
  $4b^4k^2-4b^3k+4k=0$
  Thus,
  $b^3k-b^2+1=0$
  Thus,
  $k=\frac{b^2-1}{b^3}$
  Is not an integer because the numerator factors into $(b-1)(b+1)$ and $\gcd(b,b+1)=1$ therefore the only case is $b=1$, which leads to our pairs $(2k,1)$.
  Otherwise what I have showed is that if this equation has two solutions and one is a positive integer then the other must be also because it follows by Viete's theorem.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a discussion thread from a mathematics forum, focusing on a problem from the 2003 International Mathematical Olympiad (IMO). The problem asks to determine all pairs \((a, b)\) such that \(\frac{a^2}{2ab^2 - b^3 + 1}\) is a positive integer. The thread includes multiple posts with different approaches and insights, including discussions on the nature of \(a\) and \(b\) (whether they are integers), potential solutions, and generalizations.
  
  The thread contains a question and multiple answers or partial solutions to the problem.
  
  Label: multiple-question-and-answer