2004 IMO Problems/Problem 2

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 $a_2 x^2 + a_4 x^4$ .

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2004 IMO Problems/Problem 2

Problem

Contents

Solution

Solution 1

Solution 2

See also