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.