$P(x)$ has integer coefficients and admits $4$ integer roots. Prove that $P(x) = 2$ does not admit integer roots.

Question

I was reading a book (21 Aulas de Matemática Olímpica, C. Shine) and I came across the following question:

$P(x)$ has integer coefficients and admits $4$ integer roots. Prove that $P(x) = 2$ does not admit integer roots.

I have no idea how to solve it, but because of the topic that was being addressed, I suppose the solution needs the polynomial remainder theorem. Can anyone give any help?

Answer 1

If one allows multiple roots, the statement is not quite true: consider $$P(x) = (x+1)^3 (x+2)$$ or $$P(x) = (x+1)^2 (x-1)(x-2)$$ for example.

If the four integer roots are distinct, then $$P(x) = (x-a)(x-b)(x-c)(x-d)g(x)$$ for distinct integers $a, b, c, d$. Then $P(t) = 2$ for an integer $t$ would mean that an integer divisor of $2$ (positive or negative) can be written as a product of four different integers - that is not possible.