Given $p(x)$ is a polynomial with integer coefficients and that $p(a)=1$ for some integer $a$ prove that $p(x)$ has no more than two integral roots.
I've attempted a proof by contradiction assuming $p(x)$ has three or more roots, but haven't gotten anywhere on this. Help would be appreciated!
I think that the original question (where coefficients need not be integral) is false. A counter example would be
$$f(x)=(x -(a-1))(x-(a-2))(x-(a-3))\left(x-\left(a-\frac{1}{2}\right)\right)\left(x-\left(a-\frac{1}{3}\right)\right)$$
It has three distinct integral roots $a-1,a-2,a-3$, and it also satisfies $f(a)=1$.
For this to be true, we need to specify that $p$ has integer coefficients: without this assumption, $p(x) = \frac16x(x-1)(x+1)$ is a counterexample, with roots at $-1,0,$ and $1$, but $p(2)=1$.
Suppose a polynomial $p(x)$ with integer coefficients has three or more distinct integral roots. This means that $p(x) = (x-a_0)(x-a_1)(x-a_2)q(x)$, and $q(x)$ also has integer coefficients (and so takes on integer values). For $p(x)$ to equal $1$ for some $x$, we then have three cases:
At least two of $x-a_0,x-a_1,$ and $x-a_2$ are equal to $-1$, and both the third and $q(x)$ are equal to either $1$ or $-1$
$q(x)$ and one of $x-a_0,x-a_1,$ and $x-a_2$ are equal to $-1$, and the other two are equal to $1$.
All four of the factors are equal to $1$.
This is because the only ways to write $1$ as a product of four integers is as $$1 = (-1)^4 = (-1)^2 \cdot (1)^2 = (1)^4.$$
In all three of these cases, we have $x-a_i = x-a_j$ for some $i\neq j$, so $a_i = a_j$ and the roots are not distinct.
