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

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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

1 Answer 1

up vote 7 down vote accepted

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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