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_2$$x^2$ +.....+ $a_n$$x^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_1$$a_0$ + $a_2$$a_0$$^2$ ....+ $a_n$$a_0$$^n$ = $a_0$(1 + $a_1$ + $a_2$$a_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


1 Answer 1


It's almost right.

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.


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