This came up an a training piece for the Putnam Competition and also in Ireland and Rosen.
The question posed was basically:
Let $p(x)$ be a polynomial with integer coefficients satisfying that $p(0)$ and $p(1)$ are odd. Show that $p$ has no integer zeros.
I&R give an example:
$p(x) = x^2 – 117x + 31$ and show (no problem) that for any $n$ whether even or odd, $p(n)$ will be odd. And claim that this shows $p(n)$ will never be $0$.
I can see, e.g., that $x^2 + 2x + 1$ will be odd substituting an even $n$ and even for an odd $n$.
But would appreciate help in understanding the underlying math and what is happening here.
Also, as a second part, can a general statement about the existence of an integer solution be made if $n$, even and odd, generates an even and an odd as in the last example.
I can see that if you look at these equations (mod $2$), you can distinguish whether there is an integer solution. And I would guess this is intimately connected with the question.
Thanks as always.
Best Answer
We know
$$f(0) \equiv 1 \pmod{2}$$
This means that the constant term of $f$ is an odd integer.
We also know
$$f(1) \equiv 1 \pmod{2}$$
This means that the sum of the coefficients of $f$ is odd.
Consider $f(x)$ for a general $x \in \mathbb{Z}$.
Case 1: $x$ is even.
If $x$ is even, all of the powers of $x$ must also be even, and since the product of even numbers with any integer coefficients is also even, everything but the constant term of $f$ is even. Thus, $f(x)$ must be odd.
Case 2: $x$ is odd.
If $x$ is odd, all of the powers of $x$ must also be odd. We have a bunch of odd numbers $x^i$ (including $x^0)$, and we are adding multiples of them together. We know that the sum of the coefficients is odd; this means we have, in total, an odd number of $x^i$s. The sum of an odd number of odd numbers must be odd, and so $f(x)$ is odd.
We've proven that $f(x)$ is odd for integer $x$. It follows, a fortiori, that $f(x)$ is nonzero.