If polynomial $P(x)$ has integer coefficients and at least three integer roots, then $P(x)+5^m$ has no more than one integer root for $m\geq 1$

Question

I've been doing some polynomial excersises lately and in that one I got completly stuck.

Let $m \geqslant 1$ be natural number and $P(x)$ polynomial with integer coefficients which has at least three different integer roots. Prove that $P(x)+5^m$ has no more than one integer root.

At first I considered the easiest case: $(x-x_1)(x-x_2)(x-x_3)+5$, but it did not turned out in anything helpful, so I am seeking for some clues on how to crack that problem.

Also, I'd like to ask for as elementary hint/solution as possible since this question is from (inactive) high school contest.
https://om.mimuw.edu.pl/static/app_main/problems/om48_1.pdf

Answer 1

Suppose $r_1$, $r_2$, $r_3$ are three distinct integer roots of $P(x)$, and $y_1$ and $y_2$ are integer roots of $P(x) + 5^m$. Thus $P(x) = Q(x)(x-r_1)(x-r_2)(x-r_3)$ where $Q(x)$ also has integer coefficients. Now $P(y_i) = -5^m = Q(y_i) (y_i - r_1)(y_i-r_2)(y_i-r_3)$, so $y_i - r_j$ are integers that divide $5^m$, in particular they are $\pm$ powers of $5$. And $$y_1 - y_2 = (y_1 - r_1)-(y_2-r_1) = (y_1-r_2)-(y_2-r_2) = (y_1-r_3)-(y_2-r_3)$$ is written as the difference between two of those in three different ways. But that is impossible unless $y_1 - y_2 = 0$.