2004 IMO Problems/Problem 2

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Problem

Find all polynomials $f$ with real coefficients such that for all reals $a,b,c$ such that $ab + bc + ca = 0$ we have the following relations

\[f(a - b) + f(b - c) + f(c - a) = 2f(a + b + c).\]

Solution

Solution 1

From $b=c=0$, we have $f(a) + f(-a) = 2f(a) \Longrightarrow f(a) = f(-a)$, so $f$ is even, and all the degrees all of its terms are even. Let $\text{deg}\, f(x) = n$

Let $(a,b,c) = (6x, 3x, -2x)$*; then we have $f(3x) + f(5x) + f(8x) = 2f(7x)$. Comparing lead coefficients, we have $3^n + 5^n + 8^n = 2 \cdot 7^n$, which cannot be true for $n \ge 5$. Hence, we have $f(x) = c_1 x^4 + c_2 x^2$. We can easily verify by expanding that all such polynomials work.

  • The substitution arises from writing $a = \frac{-bc}{b+c}$.

Solution 2

Let $a = (1 - \sqrt {3})x$, $b = x$, and $c = (1 + \sqrt {3})x$. Then it is easy to check that $ab + bc + ca = 0$, so

\[P( - \sqrt {3}x) + P( - \sqrt {3}x) + P(2 \sqrt {3} x) = 2P(3x)\]

for all $x$. Hence, for the coefficient of $x^n$ to be nonzero, we must have $2( - \sqrt {3})^n + (2 \sqrt {3})^n = 2 \cdot 3^n$.

This does not hold for $n = 1$, and if $n$ is odd and $n \ge 3$, then the LHS is irrational and the RHS is a positive integer, so $n$ must be even.

Let $n = 2m$. Then $2 \cdot 3^m + 12^m = 2 \cdot 3^{2m}$, so $2 + 4^m = 2 \cdot 3^m$. This holds for $m = 1$ and $m = 2$, and $(4/3)^3 = 64/27 > 2$, so $2 + 4^m > 4^m > 2 \cdot 3^m$ for $m \ge 3$. Therefore, $P(x)$ must be of the form $a_2 x^2 + a_4 x^4$.

See also

  • <url>viewtopic.php?p=99448#99448 AoPS/MathLinks discussion</url>
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