Let $b_n={a_n}^2+na_n$ $a_{n+2}-a_{n+1}=b_{n+1}-b{n}$ $\sum_{i=1}^{n-1}a_{n+2}-a_{n+1}=\sum_{i=1}^{n-1}b_{n+1}-b{n}$ $a_{n+1}=b_n={a_n}^2+na_n$ Let $\mathbb{P}$ be set of primes that divide at least one term of the sequence. Assume that $\mathbb{P}$ is finite and $\mathbb{P}=\{p_1, p_2, \cdots p_k\}$. We know that $a_n|a_{n+1}$ so there exist a $N$ such that for all $n>N$ and for all $i\in \{1,2, \cdots, k\}$, $p_i|a_n$. There exist a sufficient large $t$ such that $tp_1p_2\cdots p_k+1>N$. For $n= tp_1p_2\cdots p_k+1$, $a_n+n$ is relatively prime with all $p_i$ so there is a prime that not in $\mathbb{P}$ and divide $a_{n+1}$. Contradiction. $\mathbb{P}$ is not finite.

For b) one can show that the primes $3,5,19$ work by computing the first terms and noticing that the sequence will be periodic.

pablock wrote: For b) one can show that the primes $3,5,17$ work by computing the first terms and noticing that the sequence will be periodic. No, actually it is not true for 17. But 19 satisfies the condition. We use this fact for prove this: If $a_n=a_m(modk)$ and $n=m(modk)$ then the sequence will be $(m-n)$ $periodic$ in module $k$

How to find the third prime without guessing?

ISE wrote: pablock wrote: For b) one can show that the primes $3,5,17$ work by computing the first terms and noticing that the sequence will be periodic. No, actually it is not true for 17. But 19 satisfies the condition. We use this fact for provide this: If $a_n=a_m(modk)$ and $n=m(modk)$ then the sequence will be $(m-n)$ $periodic$ in module $k$ Sorry; corrected

rmtf1111 wrote: How to find the third prime without guessing? I have tried it until I can find. Otherwise, if there is an easier way, it would not be 2nd question though, in my opinion. If there is an easier solution, I don't know and could not find. Also the term divisible by $p$ is not so far for $p=7,11,13,17$ and this makes it triable enough. (last edit for @grupyorum)

any solution?