IMO 2004 · Problem 2

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

Official text of the problem. Official problem paper (PDF)

Answer

\[ \boxed{f(x)=Ax^2+Bx^4,\qquad A,B\in\mathbb R.} \]

Solution

Taking \(a=b=c=0\) gives \(3f(0)=2f(0)\), so \(f(0)=0\). Next, taking \((a,b,c)=(t,0,0)\) gives \[ f(t)+f(-t)=2f(t). \] Thus \(f\) is even.

For any real \(t\), choose \[ a=(1+\sqrt3)t,\qquad b=(1-\sqrt3)t,\qquad c=t. \] These satisfy \[ ab+bc+ca=\bigl(-2+1-\sqrt3+1+\sqrt3\bigr)t^2=0. \] The required equation, together with evenness, therefore yields \[ f(2\sqrt3\,t)+2f(\sqrt3\,t)=2f(3t). \] Replacing \(\sqrt3\,t\) by \(x\), we obtain \[ f(2x)+2f(x)=2f(\sqrt3\,x) \qquad\text{for every real }x. \]

Suppose \(f\) is nonzero. Since it is even and \(f(0)=0\), its degree is \(2n\) for some integer \(n\ge1\). Comparing leading coefficients in the last identity gives \[ 4^n+2=2\cdot3^n. \] If \(n\ge3\), then \[ \frac{4^n}{3^n}\ge\left(\frac43\right)^3 =\frac{64}{27}>2, \] contradicting this equality. Hence \(n\le2\), so necessarily \[ f(x)=Ax^2+Bx^4. \] The zero polynomial also has this form.

It remains to verify that every such polynomial works. Let \(ab+bc+ca=0\), and put \[ x=a-b,\qquad y=b-c,\qquad z=c-a,\qquad s=a+b+c. \] Then \(x+y+z=0\), and \[ x^2+y^2+z^2 =2(a^2+b^2+c^2-ab-bc-ca) =2s^2. \] Consequently, \[ xy+yz+zx=-\frac{x^2+y^2+z^2}{2}=-s^2. \] Using \(x+y+z=0\) again gives \[ x^2y^2+y^2z^2+z^2x^2 =(xy+yz+zx)^2-2xyz(x+y+z)=s^4. \] Therefore, \[ x^4+y^4+z^4 =(x^2+y^2+z^2)^2 -2(x^2y^2+y^2z^2+z^2x^2) =2s^4. \] Thus, for arbitrary real \(A,B\), \[ f(a-b)+f(b-c)+f(c-a) =2As^2+2Bs^4 =2f(a+b+c). \] This proves both necessity and sufficiency.

The solution was checked against Evan Chen's published IMO solution notes.