erkamseker wrote: $a_{1}=5$ and $a_{n+1}=a_{n}^{3}-2a_{n}^{2}+2$ for all $n\geq1$. $p$ is a prime such that $p=3(mod 4)$ and $p|a_{2011}+1$. Show that $p=3$. Lemma: if (a;b)=1 and p is prime p=4k-1 k-Natural a^2+b^2 is not multiple p we have a{n+1}=a{n}^3-2*a{n}^2+2 n -Natural => (a{n+1}-2)/(a{n}-2)=a{n}^2 (a{n}-2)/(a{n-1}-2)=a{n-1}^2 ..... .... .... (a{2+1}-2)/(a{2}-2)=a{2}^2 (a{2}-2)/(a{1}-2)=a{1}^2 => (a{n+1}-2)/(a{1}-2)=(a{n}a{n-1}....a{2}a{1})^2 a{n}a{n-1}....a{2}a{1}=A => a{n+1}-2=(a{1}-2)* A^2=3*A^2 => a{n+1}+1=3*(1+A^2) we have p|a{n+1}+1 => if p >3 => p|1+A^2 but for Lemma => p=3 sorry me because my englisg bad

javlik wrote: erkamseker wrote: $a_{1}=5$ and $a_{n+1}=a_{n}^{3}-2a_{n}^{2}+2$ for all $n\geq1$. $p$ is a prime such that $p=3(mod 4)$ and $p|a_{2011}+1$. Show that $p=3$. Lemma: if (a;b)=1 and p is prime p=4k-1 k-Natural a^2+b^2 is not multiple p we have a{n+1}=a{n}^3-2*a{n}^2+2 n -Natural => (a{n+1}-2)/(a{n}-2)=a{n}^2 (a{n}-2)/(a{n-1}-2)=a{n-1}^2 ..... .... .... (a{2+1}-2)/(a{2}-2)=a{2}^2 (a{2}-2)/(a{1}-2)=a{1}^2 => (a{n+1}-2)/(a{1}-2)=(a{n}a{n-1}....a{2}a{1})^2 a{n}a{n-1}....a{2}a{1}=A => a{n+1}-2=(a{1}-2)* A^2=3*A^2 => a{n+1}+1=3*(1+A^2) we have p|a{n+1}+1 => if p >3 => p|1+A^2 but for Lemma => p=3 sorry me because my englisg bad could you rewrite it in LaTeX?

LaTeXed and reworded version of javlik's solution: =========================================================== It is well known that if $p \equiv 3 \mod 4$ for some prime $p$, then if $\gcd(a, b)=1$ then $a^2+b^2 \not\equiv 0 \mod p$. Note note that from the recursion we have the following equivalences: \begin{align*} \frac{a_{n+1}-2}{a_n-2} &= a_n^2\\ \frac{a_n-2}{a_{n-1}-2} & = a_{n-1}^2\\ &\vdots\\ \frac{a_3-2}{a_2-2}&= a_2^2\\ \frac{a_2-2}{a_1-2} &= a_1^2\\ \end{align*} Multiplying the relations gives a telescoping product, which yields \[\frac{a_{n+1}-2}{a_1-2} = \left(\prod_{i=1}^{n} a_i\right)^2=A^2\] Since $a_1=5$ we get that $a_{n+1} +1 = 3\left(1+A^2\right)$. So then for some prime $p$, $p | a_{n+1}+1 = 3\left(1+A^2\right)$. If $p\neq 3$ then $p | 1+A^2$, but from the lemma this is impossible. So then $p=3$ as desired.

Click to reveal hidden text

we get $ a_{2011}+1=3(a^2+1) $ if $ a^2+1\div p $ then $ (a+i)(a-i)\vdots p $ since $ p=4k+3 $ then $ p $ is prime in complex number then $ 1\vdots p $ contradiction then $ p=3 $