$a_1=-5$, $a_2=-6$ and for all $n \geq 2$ the ${(a_n)^\infty}_{n=1}$ sequence defined as, \[a_{n+1}=a_n+(a_1+1)(2a_2+1)(3a_3+1)\cdots((n-1)a_{n-1}+1)((n^2+n)a_n+2n+1)).\] If a prime $p$ divides $na_n+1$ for a natural number n, prove that there is a integer $m$ such that $m^2\equiv5(modp)$
Problem
Source: Turkey TST 2014 Day 3 Problem 8
Tags: number theory proposed, number theory
14.03.2014 12:06
I think you have mistakes. Please, see the condition \[ a_{n+1}=a_n + ( a_1+1 ) ( 2 a_2 + 1) ( 3a_3+1)\cdots (n - 1)a_{n - 1}+1)((n^2 + n)a_n+2n+1)) .\]
14.03.2014 14:27
Yeah, there is a parenthesis before the $a_{n-1}$'s coefficient. I've just edited it.
16.03.2014 12:05
something is still wrong with the last term...
16.03.2014 16:46
Yeah, although it looks like wrong, this is the problem written on the paper.
02.07.2014 22:13
Does anybody has a solution to this strange problem?
14.07.2014 15:30
Any body solution?
14.07.2014 17:50
It's enough to prove that $b_n=na_n+5n^2+1$ is always a perfect square. For $n=1,2,3,4,5$ it equals $1^2,3^2,20^2,501^2,313026^2$, respectively. So i think it's true
27.07.2014 12:46
It surely does!!! But I don't know how to solve it.
02.08.2014 03:02
Let $b_n=na_n+1$ and $B_n=\prod_{i=1}^nb_i$. Obviously $2|B_1|B_n$. Also $2|a_n,n>1$ because $a_{n+1}=a_n\mod2,n>1$, The equation becomes $a_{n+1}=a_n+B_{n-1}((n+1)b_n+n)=a_n+(n+1)B_n+nB_{n-1}$. Multiply by $n+1$ and plus $1$, $b_{n+1}=(n+1)a_n+1+(n+1)^2B_n+(n^2+n)B_{n-1}$ Multiply by $B_n$, $B_{n+1}=(n+1)^2B_n^2+(n^2+n)B_nB_{n-1}+(n+1)a_nB_n+B_n \\=[(n+1)B_n+\frac n2B_{n-1}+\frac{a_n}2]^2+B_n-\frac14(nB_{n-1}+a_n)^2$. Hence, $B_{n+1}-\frac14((n+1)B_n+a_{n+1})^2 \\=B_{n+1}-\frac14(2(n+1)B_n+a_n+nB_{n-1})^2 \\=B_{n+1}-[(n+1)B_n+\frac n2B_{n-1}+\frac{a_n}2]^2 \\=B_n-\frac14(nB_{n-1}+a_n)^2$ Now $B_n-\frac14(nB_{n-1}+a_n)^2=\cdots=B_2-\frac14(2B_1+a_2)^2=-5,\forall n>1$ In other words, we can express $B_n$ as $x^2-5$. Since $na_n+1=b_n|B_n$, we are done!
02.08.2014 17:56
WOW it is FANTASTIC!!