Question

(2x – p + 1)^2 – (2x + p – 1)^2​

Answer 1

Step-by-step explanation:

The proof first shows that the polynomial P(x)=xp−1+2xp−2+3xp−3+⋯+(p−1)x+p only has zeros of absolute value >1. Then, it assumes it can be factored into two polynomials, namely, P(X)=Q(X)R(X), and since p=P(0)=Q(0)R(0), one of Q(0),R(0)=±1, so WLOG Q(0) has a zero of absolute value >1 which contradicts the fact that P(X) only has zeros of absolute value >1.

I don't see why Q(0)=±1 means that it has a zero with absolute value greater than one.

(The following is the proof in question.)

"We will show that all the zeros of P(x) have absolute value greater than 1. Let y be a complex zero of P(x).

Then 0=(y−1)P(y)=yp+yp−1+yp−2+⋅⋅⋅+y−p. Assuming |y|≤1, we obtain p=|yp+yp−1+yp−2+⋅⋅⋅+y|≤∑pi=1|y|i≤∑pi=11=p. This can happen only if the two inequalities are, in fact, equalities, in which case y=1. But P(1)>0, a contradiction that proves our claim. Next, let us assume that P(x)=Q(x)R(x) with Q(x) and R(x) polynomials with integer coefficients of degree at least 1. Then p=P(0)=Q(0)R(0). Since both Q(0) and R(0) are integers, either Q(0)=±1 or R(0)=±1. Without loss of generality, we may assume Q(0)=±1. This, however, is impossible, since all zeros of Q(x), which are also zeros of P(x), have absolute value greater than 1. We conclude that P(x) is irreducible."