modular arithmetic
number theory unsolved
number theory

$ F_1 \equal{} F_2 \equal{} 1, F_{n\plus{}2 }\equal{} F_{n\plus{}1} \plus{} F_n; n\geq 1.$
Let $ p \equal{} 10k \pm 3$ prime number
Prove that
 $ p |F_{p\plus{}1}$, but  $ p^2 \nmid F_{p\plus{}1}$

well,that's what i did:
we just need to have $ p \equiv \pm2 \pmod{5}$.
note that in this case we have $ \left(\frac {5}{p}\right) \equal{} \minus{} 1$.
we have
$ F_{p \plus{} 1} \equal{} \frac {1}{\sqrt {5}}.\left((\frac {1 \plus{} \sqrt {5}}{2})^{p \plus{} 1} \minus{} (\frac {1 \minus{} \sqrt {5}}{2})^{p \plus{} 1} \right)$.
so we must only prove that
$ p \mid \frac {(1 \plus{} \sqrt {5})^{p \plus{} 1} \minus{} (1 \minus{} \sqrt {5})^{p \plus{} 1}}{\sqrt {5}} \equal{} \binom{p \plus{} 1}{1}1^p \plus{} \binom{p \plus{} 1}{3}5 \plus{} ... \plus{} \binom{p \plus{} 1}{p \minus{} 2}5^{\frac {p \minus{} 3}{2}} \plus{} \binom{p \plus{} 1}{p}5^{\frac {p \minus{} 1}{2}}$
but it's obvious that the terms between $ \binom{p \plus{} 1}{1}1^p$ and $ \binom{p \plus{} 1}{p}5^{\frac {p \minus{} 1}{2}}$ are all of them divisible by $ p$.
so it remains to prove that $ p \mid \binom{p \plus{} 1}{1}1^p \plus{} \binom{p \plus{} 1}{p}5^{\frac {p \minus{} 1}{2}}$ which is obviously true because of what was mentioned in the third line.

Lemma:$ F_{p\minus{}\left(\frac{p}{5}\right)}\equiv 0\pmod{p}$
we know $ p\equiv \pm 3\pmod{10}$  $ \implies$ $ (\frac {p}{5}) \equal{} \minus{} 1$
$ p|F_{p \plus{} 1}$
But it is not known whether there exists a prime $ p$ such that $ F_{p \minus{} \left(\frac {p}{5}\right)} \equiv 0 \pmod{p^2}$

$ F_n\equal{}\frac{(1\plus{}\sqrt 5)^n\minus{}(1\minus{}\sqrt 5)^n}{2^n\sqrt 5}\equal{}\frac{1}{2^{n\minus{}1}}(C_n^1\plus{}C_n^3*5\plus{}...)$.
Therefore (use $ C_{p\minus{}1}^{2k\plus{}1}\equal{}\minus{}1\mod p$)
$ F_{p\minus{}1}\equal{}\minus{}\frac{(1\plus{}5\plus{}...\plus{}5^{(p\minus{}3)/2})}{2^{p\minus{}2}}\mod p\equal{}\frac{1\minus{}5^{(p\minus{}1)/2}}{2^p}\mod p\equal{}1\mod p.$
$ F_p\equal{}\frac{5^{(p\minus{}1)/2}}{2^{p\minus{}1}}\mod p\equal{}\minus{}1\mod p$. It give $ p|F_{p\plus{}1}$.
Consider $ \frac{F_{mn}}{F_n}\equal{}q_1^{(m\minus{}1)n}\plus{}q_1^{(m\minus{}2)n}q_2^n\plus{}...\plus{}q_2^{(m\minus{}1)n}$, were $ q_{1,2}\equal{}\frac{1\pm \sqrt 5}{2}.$
Let $ n\equal{}p\plus{}1, m\equal{}p\minus{}1$. Then $ q_2^n\equal{}q_1^n\mod p \equal{}\minus{}1\mod p$ (it is easy to calculate).
Therefore $ \frac{F_{p^2\minus{}1}}{F_{p\plus{}1}}\equal{}\minus{}1\minus{}1...\minus{}1\equal{}1\mod p$. Because $ p|F_{p\plus{}1}$ we find $ F_{p\plus{}1}\equal{}F_{p^2\minus{}1}\mod p^2$.
By analogy with calculate $ F_{p\minus{}1}$ we find $ F_{p^2\minus{}1}\equal{}\frac{1\minus{}5^{(p^2\minus{}1)/2}}{2^{p^2\minus{}1}}\mod p^2\equal{}1\minus{}5^{(p^2\minus{}1)/2}\mod p^2.$
Let $ 5^{(p\minus{}1)/2}\equal{}\minus{}1\plus{}ap\mod p^2$, then $ 5^{(p^2\minus{}1)/2}\equal{}(\minus{}1\plus{}ap)(\minus{}1\plus{}ap)^p\equal{}1\minus{}ap\mod p^2$. So $ F_{p\plus{}1}\equal{}1\plus{}5^{(p\minus{}1)/2}\mod p^2$.
And your question equavalent to
$ p^2\not |5^{p\minus{}1}\minus{}1$ if $ p\equal{}\pm 3\mod 10.$

[quote="Rust"]And your question equavalent to
$ p^2\not |5^{p \minus{} 1} \minus{} 1$ if $ p \equal{} \pm 3\mod 10.$[/quote]
That's not quite true. See sequence http://oeis.org/A123692
In particular, for $ p \equal{} 40487,\;1645333507,$ and $ 6692367337$, we have $ p^2\mid (5^{p \minus{} 1} \minus{} 1)$ but $ p^2\not| F_{p \plus{} 1}.$

btw, a prime $ p$ such that $ p^2\mid F_{p \plus{} 1}$ would be called [url=http://en.wikipedia.org/wiki/Wall-Sun-Sun_prime]Wall-Sun-Sun prime[/url]. But no such prime is yet known and the smallest one (if it exists) must be greater than $ 10^{14}$.


real analysis
Support
algebra
polynomial
real analysis unsolved

Let $f$ be a distribution with compact support and let $P$ be a polynomial.

Show that $(P*f)$ is a polynomial where:

$(P*f,\phi) = (P(x)f(y), \phi(x+y))$

and $\phi$ is any test function.

[quote="Kent Merryfield"]A first step: $(u*v)'=(u')*v=u*(v').$

If $P$ is a polynomial, then $(P*f)^{[n+1]}=(P^{[n+1]})*f=0*f=0.$[/quote]

Makes sense, but where does the fact that f has compact support come into play? Is it simply needed to show that the convolution (as I defined it above) makes sense?


[quote="cuopbientcd"]I want to introduce with you about Vietnam my country
but because my English is bad so i will send to you few links
http://anhso.net/lietke.asp?id=89@5
http://anhso.net/lietke.asp?id=90@5
http://anhso.net/lietke.asp?id=93@5 :lol:  :lol:[/quote]

uhmmm,but posting in this forum is unsuitable.

Given that next summer's IMO is in Vietnam, it's not entirely inappropriate. Some of the pictures were very interesting, so thanks for the links, cuopbientcd. I hope I am able to go to Vietnam next July myself- though that doesn't look entirely likely at the moment.  :(

ALL go to VietNam  :D
Sorry for my English , but i think our students have good English   :)

You may find Vietnam news here, it's necessary for u to know a little about where u r going to:

http://english.vietnamnet.vn/news/

:(  I/m a new member so I can't find the Vietnam forums where are they . (If you think I spam so you can del this reply)
Em là thành viên mới ko biết phần dành cho việt nam ở đâu, mong các anh gchỉ giúp giùm

see here http://www.mathlinks.ro/Forum/index.php?f=256  :wink:


Sean $ n$ y $ k$ enteros positivos. Demuestre que:

$ \displaystyle n| \sum_{d|n} \varphi (\frac nd)k^d$.

(Nota: $ \varphi (n)$ es el numero naturales menores que $ n$ que son primos relativos con $ n$)

El problema está resuelto para n=1, porque 1 divide a $ 1\cdot k^1 \equal{} k$. Desde ahora, suponga que n>1. Es conveniente recordar algunas cosas:[hide][list][*]Si $ f,g: \mathbb Z^ \plus{} \to\mathbb Z$ son funciones, entonces la

No quedo completo, podria completarlo? Yo tengo una solución bastante sorprendente, peropues quiero ver si sale una con teoria de numeros standard.

Para enteros positivos $ n, k$ sea
\[ F_n(k) = \sum_{d|n} \varphi \Big(\frac {n}{d}\Big)k^d\]
Supongamos primero que $ n = p^{\alpha}$, con $ p$ primo y $ \alpha\geq 1$.
\begin{eqnarray*}
F_n(k)  & =  & \sum_{i=0}^{\alpha}\varphi(p^i)k^{p^{\alpha-i}}\\
& = & k^{p^{\alpha}} + \sum_{i=1}^{\alpha}(p^i -  p^{i-1})k^{p^{\alpha-i}}\\
& = & p^{\alpha}k + \sum_{i=0}^{\alpha - 1}p^i(k^{p^{\alpha-i}} - k^{p^{\alpha-i-1}})\\
 & = & p^{\alpha}k + \sum_{i=0}^{\alpha - 1}p^i k^{p^{\alpha-i-1}}(k^{\varphi (p^{\alpha-i})}-1)
\end{eqnarray*}
Por Euler-Fermat y ya que $ p^{\alpha-i-1}\geq \alpha - i$, $ p^{\alpha}$ divide a cada uno de los sumandos de la última sumatoria. En particular, $ n=p^{\alpha}|F_n(k)$.
Ahora veamos el caso general. Si $ n = p^{\alpha}m$ con $ (p,m)=1$,
\begin{eqnarray*}
F_n(k)  & =  & \sum_{e|m}\sum_{i=0}^{\alpha}\varphi\Big(p^{i}\frac{m}{e}\Big)k^{p^{\alpha-i}e}\\
 & = & \sum_{e|m}\varphi\Big(\frac{m}{e}\Big)\sum_{i=0}^{\alpha}\varphi(p^{i})(k^e)^{p^{\alpha-i}}\\
 & = & \sum_{e|m}\varphi\Big(\frac{m}{e}\Big)F_{p^{\alpha}}(k^e)
\end{eqnarray*}
Así, por el caso anterior, $ p^{\alpha}| F_n(k)$.


Given revenue and cost functions R = 30x - .4x^2 and C = 8x + 9 , where x is the daily production, find the rate of change of profit with respect to time when 10 units are produced and the rate of change of production is 8 units per day.

the profit function $ P(x)$ will be::
$ R\minus{}C\equal{}P(x)\equal{}\minus{}4x^2\plus{}22x\minus{}9$
$ \frac{dP}{dt}\equal{}\minus{}8x\frac{dx}{dt}\plus{}22\frac{dx}{dt}$

so,,when $ \frac{dx}{dt}\equal{}8$ and $ x\equal{}10$ we have:
$ \frac{dP}{dt}\equal{}\minus{}8(10)(8)\plus{}22(8)\equal{}\minus{}464$

sorry,,I'm not so good in economics.. :blush:


conics
hyperbola
Asymptote
special factorizations
calculus
calculus computations

find the center of the conic section by completing the square and writing in the proper form. Graph the curve.
9x^2-y^2-18x-2y=1

I think the completing the square part is (3x-9)^2-(y-1)^2=83  but I'm not sure what to do next.

In your case, upon completing the squares and rearranging, you should have:-
$ \frac{(x\minus{}1)^{2}}{1^{2}}\minus{}\frac{(y\plus{}1)^{2}}{3^{2}}\equal{}1$

I have generalised as follows:-

For a hyperbola $ \frac{(x\minus{}a)^{2}}{\alpha ^{2}}\minus{}\frac{(y\minus{}b)^{2}}{\beta ^{2}}\equal{}1$

The foci are given by $ \left( \pm \sqrt{\alpha ^{2}\plus{}\beta ^{2}}\plus{}a,\,b\right)$

The asymptotes are given by $ y\equal{}\pm \frac{\beta }{\alpha }\left( x\minus{}a\right) \plus{}b$

Hope this helps.

thanks that does help alot.
I am still a little confused on what the graph would look like.


There aren't any threads on it.

Are we not allowed to discuss it?

Most of you always get 6's on it?
Never mind then, I guess I'm just bringing up a joke for you guys.

I usually just get 5's...

Eh no I get like 5s due to ABSOLUTELY DUMB 5+3=7 MISTAKES

Ahh, k.

That sometimes happens to me too, except it sometimes results in me getting a 4.  :blush:

We always get solutions right after, so I guess that's why there's not a lot of discussion about it.  Or maybe because usually, there's not really much to discuss.

[quote="tjhance"]hm i should have a perfect this year so far

except like it is not on the site for some reason

oh well[/quote]

lol fail on round 4:

So on problem 6 teh dude was like about some messenger and stuffs

So I gots Rt 2 since I didn't read th eproblem and thought he asked for teh time it took for the dudes to meet >.>

this is why I failed mathcounts