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

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.

-

## closed as off-topic by Jonas Meyer, MagicMan, John, LutzL, Claude LeiboviciMar 25 at 6:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jonas Meyer, MagicMan, John, Claude Leibovici
If this question can be reworded to fit the rules in the help center, please edit the question.

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 $-a$.