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.

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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 $-a$.

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