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Question: Prove that there is no non-constant polynomial $P(x)$ with integer coefficients such that $P(n)$ is a prime number for all positive integers $n$.

My approach: For the sake of contradiction let us assume that $\forall n\in\mathbb{N}$, $P(n)$ is prime, where $P$ is a non-constant polynomial.

Now, observe that $P(2)=2k+a_0=p$ for some integer $k$ and some prime $p$. This implies that $p=a_0+2k$. Now if $a_0$ even, then $2|p$, which implies that $p=2$.

This implies that $\forall k\in\mathbb{N},$ $P(2k)$ is even and since it is equal to a prime, implies that $P(2k)=2,$ $\forall k\in\mathbb{N}$.

Now let $G(x)=P(x)-2,$ $\forall x$, then for any $k\in\mathbb{N}$, we have $G(2k)=0$, which implies that $G$ has an infinite number of roots, which in turn implies that $G$ is the zero polynomial. Thus $P(x)-2=0, \forall x\implies P(x)\equiv 2$. But, $P$ is a non-constant polynomial, which implies that $P(x)\equiv 2,$ results in a contradiction.

Thus $a_0$ is not even and it is odd.

This implies that $\forall $ even $x$, $P(x)$ is an odd prime.

After this I went on to prove that $a_0=-1$ or $a_0=1$.

How to proceed after this? Please be rigorous.

Bill Dubuque
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Sanket Biswas
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1 Answers1

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I'm going to give a hint based on one of the most famous polynomials in this branch of mathematics.

The polynomial $n^2+n+41$ evaluates to $43$ when $n=1$. By modular arithmetic congruences and the signs of the coefficients it will then be a positive multiple of $43$ for $n=43k+1$ and every natural number $k$. For such $n$ this multiple of $43$ can be prime only if it is $43$ itself, but $n^2+n+41=43$ can have only a finite number of roots for $n$.

Do you see how to generalize this argument?

Oscar Lanzi
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