Let $f\in \mathbb{Q} [x]$ be a polynomial, and $a\_0 = a$ be an arbitrary integer. Let us define a sequence $\{a\_n \} $ by the recurrence relationship : $$a\_n = f(a\_{n-1} ). $$ I want to show that $a\_n $ cannot always be a prime number, with $\{a\_n \}$ being pairwise distinct. I am pretty sure that this is a very well known fact, but I cannot easily find this.


Note : I posted this question also in SE Math. 


<https://math.stackexchange.com/questions/1568513/non-existence-of-a-prime-generating-polynomial-recurrence-relation>



Non-existence of a prime generating polynomial recurrence relation
Let $a\_0 = 2^{2^k} + 1$ for $k$ sufficiently large, and let


$$a\_n = (a\_{n-1} - 1)^2 + 1.$$


Then $a\_n = 2^{2^{k+n}} + 1$, so this sequence can't always be prime regardless of the value of $k$ iff there are infinitely many composite [Fermat numbers](https://en.wikipedia.org/wiki/Fermat_number), and as far as I know this is wide open. 


