In order to make sense, I think that the correct statement is : "Prove that p divides a_(n+1) - 1." And, in fact, i'm quite sure of that, because it's true . Let's prove it : Let U_n = a_(n-1)*a_(n+1) - (a_n) ² . We have U_1 = 5, and for each n > 0 : U_(n+1) = a_n*a_(n+2) - (a_(n+1)) ² = a_n*a_(n+1) + (a_n) ² - (a_(n+1)) ² = a_(n+1)(a_n - a_(n+1)) + (a_n) ² = -U_n. It follows that U_n = 5*(-1) ^n-1 for each n > 0, that is a_(n-1)*a_(n+1) = (a_n) ² + 5*(-1) ^n-1 . (1) Let p be a prime number which divides a_(2k) - 2. Thus a_(2k) = 2 mod[p]. From (1), it follows that (a_(2k)) ² - 5 = (a_(2k+1) - a_(2k))*a_(2k+1) that gives -1 = (a_(2k+1)) ² - 2*a_(2k+1) mod[p] thus p divides (a_(2k+1) - 1) ² and since p is prime we are done. Pierre.

Omid Hatami wrote: $a_{0}=2,a_{1}=1$ and for $n\geq 1$ we know that : $a_{n+1}=a_{n}+a_{n-1}$ $m$ is an even number and $p$ is prime number such that $p$ divides $a_{m}-2$. Prove that $p$ divides $a_{n+1}-1$. Sorry, what is "n", in the last part?

Let $b_n=a_{2n}$ We can see that $b_{n+2}=3b_{n+1}-b_n$ and $b_0=2$ , $b_1=3$ By induction $b_{n+1}^2-3b_{n+1}b_{n}+b_n^2=-5$ $m=2k$ if $p|a_{2k}-2$ then $b_k \equiv 2 \pmod p$ $\rightarrow$ $b_{k+1}^2-6b_{k+1}+9 \equiv 0 \pmod p $ $\rightarrow$ $b_{k+1} \equiv 3 \pmod p$ $a_{m+1}=a_{2k+1}=a_{2k+2}-a_{2k}=b_{k+1}-b_k\equiv 3 -2=1 \pmod p$ $\blacksquare$