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Writeup: IMO 2004 Shortlist A4

Problem Statement:

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 relation:

f(ab)+f(bc)+f(ca)=2f(a+b+c).

My Solution

First, note that the the set S of all polynomials satisfying the problem statement is closed under addition.

Now, note that S must be even. This comes from the fact that if we set b=c, then a=12b so we must have

f(32b)+f(32b)=2f(32b)$so$f(32b)=f(32b).

Now, note that if we set c=12b, then we must have a=\fracb3 and we have

f(43b)+f(12b)+f(56b)=2f(7/6b).

Let the degree of f be d. Then note that we must have

(43)d+(12)d+(56)d2(76)d=0.

A quick proof of this fact involves using limits. We know that if

f(43b)+f(12b)+f(56b)

and

2f(7/6b)

are the same polynomial P(b), then limxP(b)/xd will be the same for both polynomials (and this limit exists and is definite).

But if we don't have

(43)d+(12)d+(56)d2(76)d=0

then the limit will not hold true so we have a contradiction.

Finally, note that d=2 and d=4 both satisfy this equation. Note, also, that 4376. So if we have

(43)n2(76)n

then we will have

(43)n+1>(43)n·762(76)n+1.

Note that

(43)6+(12)6+(56)6>(43)6>2(76)6

so the only possible polynomials we can consider are linear combinations of x2 and x4, because we can't have odd polynomials and we can't have even polynomials above degree 4.

Note that

(ab)2+(bc)2+(ac)22(a+b+c)2=4(ab+bc+ac)=0

and

(ab)4+(bc)4+(ac)42(a+b+c)4=6(ab+bc+ac)(2a2+2b2+2c2+ab+bc+ac)=0

so the polynomials f(x)=x2 and f(x)=x4 do work.

Therefore the only polynomials satisfying the problem statement are of the form

f(x)=ax2+bx4.