# Proving that a polynomial cannot output a prime for all natural numbers

Id seen this problem in various math books and contest math forums. It states:

Prove that there is no polynomial p(x) = $$a_0$$ + $$a_1$$x + $$a_2x^2$$ +.....+ $$a_nx^n$$ of n degree with integer coefficients such that p(0), p(1), p(2)... are all primes.

I got around to solving this today and was surprised when I came across, literally, a two line solution:

1.) putting x = 0, we get p(0) = $$a_0$$, which would then, according to the question be prime, and hence a positive integer, going by the standard definition of a prime number

2.) putting x = $$a_0$$, we get p($$a_0$$) = prime = $$a_0$$ + $$a_1a_0$$ + $$a_2a_0^2$$ ....+ $$a_na_0^n$$ = $$a_0$$(1 + $$a_1$$ + $$a_2a_0$$....). Obviously, however, since the coefficients are integers, both these terms are also integers, which would imply that p($$a_0$$) is not prime, which is a contradiction.

i have a feeling theres an obvious error with this, but cant quite point it out. If there is an error, please do tell me the correct solution

The only problem is that it could be that $$p(a_0) = a_0$$, which would happen if $$1 + a_1 + a_2 a_0 + \ldots + a_n a_0^{n-1} = 1$$. But then you can look at $$p(k a_0)$$ for positive integers $$k$$: these would also be divisible by $$a_0$$, by a similar argument, so they would have to be $$a_0$$. But a polynomial of degree $$n > 0$$ (you did assume $$n > 0$$, right?) can only take a given value at most $$n$$ times.