# Proving if $f(x) \in \mathbb{Z}[x]$ and $f(a) \cdot f(b) = -(a-b)^{2}\neq 0$ then $f(a)+f(b)=0$

I am having trouble solving this problem.

• Suppose $f(x) \in \mathbb{Z}[x]$, and $f(a) \cdot f(b) = -(a-b)^{2}$ for distinct $a,b \in \mathbb{Z}$, then how do we prove that $f(a)+f(b)=0$?

Don't know how to proceed.

-
What are $a$ and $b$? Maybe I am missing something, but, put $x=y$ and get $(f(x))^2 = 0$. –  Aryabhata Jan 26 '11 at 4:28
You did not use $a$ and $b$ anywhere... –  Mariano Suárez-Alvarez Jan 26 '11 at 4:28
What is the source of the problem? (I think a=x, b=y.) –  Jonas Meyer Jan 26 '11 at 4:30
@Moron: Yes, i did give a try, in your manner. –  anonymous Jan 26 '11 at 4:38
@Chandru1: You realize that you are using $x$ for two entirely distinct purposes? Very bad form. –  Arturo Magidin Jan 26 '11 at 5:01

By translation we can assume that $b = 0$. Let $f = x \mapsto P(x)x + C$. We are given that $(P(a)a + C)C = -a^2$. I claim that $a|C$. Indeed, suppose that $(a,C) = g$. Then also $(a,P(a)a+C) = g$, so that $g = a$ (otherwise the product won't be divisible by $-a^2$). Since $a$ divides both factors, they must both be equal (up to sign) to $a$. So one is $a$ and the other $-a$.

-

The problem appears as one of the "Quickies" in the problems section of Mathematics Magazine Vol. 66, No. 1, February 1993. It was proposed by Bjorn Poonen. The answers to the Quickies appear in the same issue on page 64.

From page 57:

Suppose $P(x)\in\mathbf{Z}[x]$ and $P(a)P(b)=-(a-b)^2$ for some distinct $a,b\in\mathbf{Z}$. Prove $P(a)+P(b)=0$.

From page 64:

Since $a-b$ divides $P(a)-P(b)$, the roots $P(a)/(a-b)$ and $-P(b)/(a-b)$ of $x^2-[(P(a)-P(b))/(a-b)]x+1$ are integers with product $1$, by the rational root theorem. Thus $P(a)/(a-b)=-P(b)/(a-b)=\pm1$, so $P(a)+P(b)=0$.

-