Take a look at this problem: Polynomial $p(a) = 1$, why does it have at most 2 integer roots?
" Suppose is (x) is a polynomial with integer coefficients. Show that if p(a)=1 for some integer a then p(x) has at most two integer roots. "
There is an explained solution, but I don't get it. The idea is, you write: $p(x) = (x-x_1)(x-x_2)(x-x_3)q(x)$, in order to try and prove by contradiction.
If we write $p(x) = 1$, we get $1=(x-x_1)(x-x_2)(x-x_3)q(x)$. Since we can only factor 1 using -1 and 1, we know that these 4 factors have to be some form of 1 and -1.
So far, so good. The key part of the proof is that apparently these factors need to be distinct, and thus we cannot have 3 such roots. This is what I don't get. Why do they need to be distinct? Why can't we just have (-1)(1)(1)(-1)?
