The recurrence relation can be rewritten as $2a_{n+1}=(a_{n}-4)(a_{n}+5)$ If $p=2$:Αssume all $a_{n}$ are odd. We are going to prove the following claim: Claim:For all $k\in N$ there exist a unique $c$ such that $a_{n}\equiv c \pmod 2^{k}$ for all n. Proof: We must have $a_{1}\equiv 1 \pmod 4$. Assume now the claim holds up to $k$. We must have: $\frac{(a_{n}-4)(a_{n}+5)}{2}\equiv c \pmod 2^{k}$ or $a_{n}^{2}+a_{n}\equiv 20+2c \pmod 2^{k+1}$(1) for all $n$. Assume there are $n,m$ such that $a_{n}\not\equiv a_{m} \pmod 2^{k+1}$. By (1) we deduce that $2^{k+1}\mid (a_{n}-a_{m})(a_{n}+a_{m}+1)$ . But $a_{n}+a_{m}+1$ is odd by the parity of $a_{i}$ thus $2^{k+1}\mid a_{n}-a_{m}$ a contradiction. Thus the claim is proven. Since $a_{2}\equiv a_{1}\pmod 2^{N}$ for all $N$, we deduce that $a_{1}=a_{2}$ which is a contradiction by the recurrence relation. If $p=5$ choose $a_{1}=6$ to do the work. Assume for contradiction that there is an odd prime $p$ not equal to 5, which divides some term of the sequence no matter which $a_{1}$ we selected beforehand. Take $a_{1}=p+5$. ClaimIf $a_{k} \equiv 5 \pmod p$ for some k then $a_{k+1} \equiv 5 \pmod p$. Proof:Just notice that $2a_{k+1} \equiv (a_{k}-4)(a_{k}+5) \equiv 10 \pmod p$ implying that $a_{k+1} \equiv 5 \pmod p$. Now since we chosed the first term equivalent to 5 $\pmod p$, each term must be equivalent to 5 $\pmod p$ according to the previous claim. Hence there is no term divisible by p. Thus the final answer to this problem is only $p=2$.

The recurrence relation can be rewritten as $2a_{n+1}=(a_{n}-4)(a_{n}+5)$ If $p=2$:Fix $a_{1}$ and assume all $a_{n}$ are odd. We are going to prove the following claim: Claim:For all $k\in N$ there exist a unique $c$ such that $a_{n}\equiv c \pmod 2^{k}$. Proof: We must have $a_{1}\equiv 1 \pmod 4$. Assume now the claim holds up to $k$. We must have: $\frac{(a_{n}-4)(a_{n}+5)}{2}\equiv c \pmod 2^{k}$ or $a_{n}^{2}+a_{n}\equiv 20+2c \pmod 2^{k+1}$(1) for all $n$. Assume there are $n,m$ such that $a_{n}\not\equiv a_{m} \pmod 2^{k+1}$. By (1) we deduce that $2^{k+1}\mid (a_{n}-a_{m})(a_{n}+a_{m}+1)$ . But $a_{n}+a_{m}+1$ is odd by the parity of $a_{i}$ thus $2^{k+1}\mid a_{n}-a_{m}$ a contradiction. Thus the claim is proven. Since $a_{2}\equiv a_{1}\pmod 2^{N}$ for all $N$, we deduce that $a_{1}=a_{2}$ which is a contradiction by the recurrence relation. If $p=5$ choose $a_{1}=6$ to do the work. Assume for contradiction that there is an odd prime $p$ not equal to 5, which divides some term of the sequence no matter which $a_{1}$ we selected beforehand. Take $a_{1}=p+5$. ClaimIf $a_{k} \equiv 5 \pmod p$ for some k then $a_{k+1} \equiv 5 \pmod p$. Proof:Just notice that $2a_{k+1} \equiv (a_{k}-4)(a_{k}+5) \equiv 10 \pmod p$ implying that $a_{k+1} \equiv 5 \pmod p$. Now since we chosed the first term equivalent to 5 $\pmod p$, each term must be equivalent to 5 $\pmod p$ according to the previous claim. Hence there is no term divisible by p. Thus the final answer to this problem is only $p=2$.