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Solution for February 2013

The Problem:

Find all real-valued functions $f(x)$ such that
$$f\left(x^3+y^3\right) = x^2f(x) + yf(y^2) $$ for all real numbers $x$ and $y$.

This was Problem 6 on the 46th Ukrainian Mathematical Olympiad final round (2006), proposed by T.M. Mitelman. Correct solutions were submitted by

Lamis Alsheikh (Syria)

Luigi Bernardini (Italy)

Aleksandar Blazhevski (Macedonia)

Radouan Boukharfane (Québec)

Lou Cairoli (USA)

Ioan Viorel Codreanu (Romania

Bernard Collignon (France)

Hubert Desprez (France)

Mei-Hui Fang (Austria)

Philippe Fondanaiche(France)

Jan Fricke (Germany)

Georges Ghosn (Québec)

Pierre Gobin (France) Matthew Lim (USA)
Patrick J. LoPresti (USA) Ángel Plaza
Heri Setiyawan (Indonesia) Albert Stadler (Switzerland)
Hakan Summakoğlu (Turkey) Bruno Tisserand (France)
Daniel Văcaru (Romania) Arthur Vause (UK)

There was one incorrect submission.

The solution:

It is easy to see that for any real number $m$, the function $\boldsymbol{ f(x)=mx}$ satisfies our functional equation:
On the left we have
$$f(x^3+y^3) = m(x^3+y^3),$$
while on the right,
$$x^2f(x) = x^2\cdot mx\quad \mbox{ and } \quad yf(y^2) = y\cdot my^2.$$
We must now show that no other function satisfies the given identity. To that end, suppose that $f(x)$ is a function that satisfies

\[ f\left(x^3+y^3\right) = x^2f(x) + yf(y^2).\] (1)

Then $f(x^3) = f(x^3+0) =_{(1)} x^2f(x)$, and $f(x^3) = f(0+x^3) =_{(1)} xf(x^2)$. That is,

\[ f(x^3) = x^2f(x) = xf(x^2).\] (2)

When $x=0$ equation (2) tells us that $f(0)=0$; when $x \ne 0$ we can divide through by $x$ to get $xf(x) = f(x^2)$. Thus, for all real $x$,

\[ f(x^2)=xf(x).\] (3)

Furthermore, equation (2), when put back into (1), tells us that
$$f\left(x^3+y^3\right) = x^2f(x) + yf(y^2) =_{(2)} f(x^3)+f(y^3)$$
for all real numbers $x$ and $y$. It follows that for any reals $a$ and $b$ we can set $x=a^{1/3}$ and $b=y^{1/3}$, and deduce that

\[ f(a+b) = f(a) + f(b).\] (4)

We recognize (4) to be Cauchy's functional equation, which has infinitely many other solutions in addition to the linear solutions $f(x)=mx$. Several of our correspondents felt obliged to make further assumptions (such as continuity) to eliminate the unwanted nonlinear solutions. The majority of our solvers saw, however, that equation (3) already does the job without the help of further assumptions! Note on the one hand that for all $x$,
$$f\left((x+1)^2\right) =_{(3)} (x+1)f(x+1) =_{(4)} (x+1)(f(x)+f(1)) = xf(x)+xf(1)+f(x)+f(1).$$
On the other hand,
f\left((x+1)^2\right) = f\left((x^2+x)+(x+1)\right) &=_{(4)}& f\left((x^2+x)\right) + f(x+1)\\
&=_{(4)}& f(x^2) + f(x) + f(x) + f(1)\\
& =_{(3)}& xf(x) + f(x) + f(x) + f(1).
It follows that
$$xf(x)+xf(1)+f(x)+f(1) = xf(x) + f(x) + f(x) + f(1),$$
whence, $f(x) = f(1)\cdot x$. Since $f(1)$ is an arbitrary real number we have proved that if $f(x)$ satisfies equation (1) for all real numbers $x$, then there exists a real number $m$ for which $f(x) = mx$.



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