We has $ a_{2n + 1}\equiv 0(\mod 2)$and $ a_{2n + 1} > 2,\forall n\in N$ So $ a_{2n + 1}$ is composite.

Actually, only if $ n\equiv0,3\pmod4$, not for odd $ n$ (for example, $ a_5$ is odd - and even prime). But the problem is still way too easy. I don't think this is the correct statement.

The following is the correct statement: Prove that the sequence $ a_n=1+2^2+3^3+\ldots+n^n$ contains infinitely many odd composite numbers. It was also given at All-Soviet MO in 1988, as 5th (or 1st in the second day) problem for Grade 9.

freemind wrote: The following is the correct statement: Prove that the sequence $ a_n = 1 + 2^2 + 3^3 + \ldots + n^n$ contains infinitely many odd composite numbers. That looks a bit better indeed. However don't we have the same kind of trivial solution with mod 3 then? freemind wrote: It was also given at All-Soviet MO in 1988, as 5th (or 1st in the second day) problem for Grade 9. Interesting, do you have a reference for that?

Let $ p, q$ be coprime integers. Prove that the sum get's divisible by $ p$ and coprime to $ q$ (both at the same time) infinitely often.

ZetaX wrote: Let $ p, q$ be coprime integers. Prove that the sum get's divisible by $ p$ and coprime to $ q$ (both at the same time) infinitely often. Exactly my thoughts. Maybe I should put the extended version in PEN. Do you have a nice and elementary solution, ZeTaX?

$ (k + mp(p - 1))^{k + mp(p - 1)}\equiv k^k\mod p$, therefore $ p|a_{mp^2(p - 1)} \ \forall m\in N$. Exactly, if $ a_n = \sum_{k = 1}^n f(k)$ and $ f(k)\mod p$ had period $ T_p$, then $ p|a_{mpT_p} = \sum_{k = 1}^{T_p}\sum_{l = 0}^{mp - 1} f(k + lT_p)\equiv \sum_{k = 1}^{T_p}f(k)mp\equiv 0\mod p$ for any $ m\in N$.

I think problem very easy obviously $a_{4k+3}\equiv 0(mod 2)$ Then there are infinitely many $n$ such that $a_n$ is composite.

you should read again the problem

very easy problem.