How can I find all functions $f:\mathbb{R}\to\mathbb{R}$ such that $f(1)=1$ and $$f\left(f(x)y+\frac{x}{y}\right)=xyf\left(x^2+y^2\right)$$ for all real numbers $x$ and $y$ with $y\neq0$?

PS. This is from USA MOSP (Mathematical Olympiad Summer Program) 2007, and is one of the most difficult problems I've seen ever. This problem has been posted a couple of times on MathLinks fora, but I couldn't find a complete solution for it... And the only solution is $f(x)=\frac{1}{x}$. Please try it, thanks!

  • $\begingroup$ If the only solution is $f(x)=\frac{1}{x}$ then the function should be defined on $\Bbb{R}^*$. $\endgroup$ Jun 14, 2011 at 9:42
  • $\begingroup$ Showing $f(0)=0$ is hard, but it's true. Now I think it's better. $\endgroup$ Jun 14, 2011 at 10:24
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    $\begingroup$ Showing $f(0)=0$ is actually easy. Substitute $x=0$. Then the relation becomes $f(y f(0))=0$. If $f(0)$ were nonzero, then the argument $yf(0)$ could be any nonzero number and it would be true that $f(z)=0$ for all nonzero $z$ which contradicts $f(1)=1$. $\endgroup$ Jun 14, 2011 at 11:13
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    $\begingroup$ Is there anything to be gained by changing to polar coordinates? That would simplify the LHS. $\endgroup$
    – Simon
    Jun 16, 2011 at 16:38

4 Answers 4


Here is my approach. The equation $f(x)y+\frac{x}{y}=x^2+y^2$ for $x \neq 0$ has at least one real root, since is equivalent to a cubic equation. Denote that one root with $y_0$.

Substituting $y_0$ in the functional equation we get $f(x^2+y^2)=xy_0f(x^2+y^2)$. Notice that $x^2+y_0^2\neq 0$. If we can prove that $f(c)=0$ if and only if $c=0$ we are done, since then the relation above imples $y_0=\frac{1}{x}$ and therefore, from the relation $f(x)y_0+\frac{x}{y_0}=x^2+y_0^2$ we get $f(x)=\frac{1}{x}$.

Therefore, to prove that the only solution of the functional equation is the one mentioned above, we just need to prove the following:

$$f(c)=0 \Leftrightarrow c=0$$

I've made some small steps in proving this but the proof is not complete.

The relation $f(f(x)+x)=xf(x^2+1)$ allows us to prove that if $x \neq 0$ then $f(x)=0\Rightarrow f(x^2+1)=0$. This implies that there exists a sequence $K_n \to \infty$ such that $f(K_n)=0$.

Assume that $f$ is continuous.

Take now $x,y$ with $x^2+y^2=K_n$. The initial relation implies that $f(f(x)y+\frac{x}{y})=0,\ \forall x^2+y^2=K_n$. For $y \to 0+$ and $x>0$ we get that $f(x)y+\frac{x}{y} \to \infty$ (since $x \to \sqrt{K_n})$ This means that $f$ takes zero values in a neighborhood of $\infty$. Since $f(y+\frac{1}{y})=yf(1+y^2)$ implies $f(x)=-f(-x),\forall x$ with $|x| \geq 2$, we have the same thing in a neighborhood of $-\infty$.

Notice that for $y >1 $ we have $y^2+1>y+\frac{1}{y}$, and we can translate the "zeros" in the neighborhood of $\infty$ until we reach $2$, i.e. $f(x)=0$ on $(-\infty,-2]\cup [2,\infty)$.

I don't know where to go from here. This is not an answer, and I added an extra hypothesis towards the end, but I think this might lead to a result.

Maybe there are some more general solutions such as

$$ f(x)=\begin{cases} \frac{1}{x}, &x \in A \\ 0, & x \notin A \end{cases}$$, where $A$ is a set with some given properties.

  • $\begingroup$ Very nice, Beni! Thanks for this nice approach. I'll try to show that $f(c)=0 \iff c=0$. $\endgroup$ Jun 15, 2011 at 3:25
  • $\begingroup$ Nice, thinking :) $\endgroup$
    – user9413
    Jun 16, 2011 at 15:35
  • $\begingroup$ Thank you. Too bad I cannot find the end of the proof... $\endgroup$ Jun 16, 2011 at 15:39

This is a partial answer but too long for comment.

First plug in $x = 1$ to get $$f(y + {1 \over y}) = yf(1 + y^2)$$ and $y = 1$ to get $$f(f(x) + x) = xf(x^2 + 1)$$ Observe that the RHS of both expressions has the same form and therefore $$f(y + {1 \over y}) = f(f(y) + y)$$

Assuming $f$ is injective we immediately obtain $f(y) = {1 \over y}$. I am not sure how hard it is to obtain injectivity of $f$. Perhaps as hard as solving the problem altogether, so this answer of mine might not help much.

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    $\begingroup$ You can get the same result by substituting $x=\frac{1}{y}$, yielding $f(f(\frac{1}{y})y + \frac{1}{y^2}) = f(\frac{1}{y^2} + y^2)$. This yields that $f(\frac{1}{y})y = y^2$ if $f$ is 1-1. $\endgroup$ Jun 14, 2011 at 16:24
  • $\begingroup$ @Thomas: ah, that is indeed neater. $\endgroup$
    – Marek
    Jun 14, 2011 at 17:43

Let $K = \{x\in \mathbb R : f(x) = 0, x \ne 0 \}$ and suppose $x\in K$.
Beni Bogosel showed that under those conditions $x^2 + 1 \in K$ and there is a sequence $(x_n)_{n\in \mathbb N}\subseteq K$ such that $\lim_n x_n = +\infty$.
$f(x + 1/x) = xf(1 + x^2)$ implies that $\frac {x^2 + 1} x \in K$ too.
From the invariance of $xyf(x^2 + y^2)$ with respect to the transformation $x \to y, y\to x$ we deduce $$f(f(x)y + x/y) = f(f(y)x + y/x)$$ If $k_1, k_2\in K$, the previous equality becomes $$f(k_1/k_2) = f(k_2/k_1)$$
Applying the above equality to $x^2 + 1$ and $\frac {x^2 + 1} x$ we obtain $$f(1/x) = f(x) = 0$$ even $1/x$ belongs to $K$.
Now, we can conclude there exists a sequence $(y_n)_{n\in \mathbb N}\subseteq K$ such that $\lim_n y_n = 0$.
If $k\in K$ and $0 < k < 1$ we have $k^2 < k$ and $$f\left(k/\sqrt{k - k^2}\right) = k\sqrt{k - k^2}f(k) = 0$$ so $\sqrt{\frac k {1 - k}}\in K$ and the same is for $$\left(\sqrt{\frac k {1-k}}\right)^2 + 1 = \frac 1 {1 - k}$$ Applying the previous result to the sequence $(y_n)_{n\in \mathbb N}$ we can construct a third sequence $(z_n)_{n\in \mathbb N}\subseteq K$ such that $\lim_n z_n = 1$. Assuming $f$ is continuous we can write: $$0 = \lim_n f(z_n) = f\left(\lim_n z_n\right) = f(1) = 1$$ Therefore we must conclude $K = \emptyset$.
For $x, y \ne 0$ we have $$f(f(y)x + y/x) = yxf(y^2 + x^2) = xy f(x^2 + y^2) = f(f(x)y + x/y)$$ if $f(x) = f(y) = 0$ the previous equality becomes $$f(y/x) = f(x/y)$$

  • $\begingroup$ +1 for the work. I would like to see a more detailed proof that if $k_1,k_2 \in K$ then $f(k_1/k_2)=f(k_2/k_1)$. If you could prove this more clearly, then the problem is finalized, with your arguments. $\endgroup$ Jun 16, 2011 at 15:42
  • $\begingroup$ I had reached the same conclusion (if f is continuous at 1, then there's only one solution). The case where we don't assume continuity is still avoiding me, though. You can construct a lot of points $x_0$ where $f(x_0) = 0$. I have been able to prove that if $f(x_0) = 0$, $x_0 \neq 0$, then $f(2x_0^2) = \frac{1}{x_0^2}$, so one way to reach a contradiction would be to show that $2K^2$ and $K$ intersect. Oh and by the way, if there is $x_0 \neq 0$ such that $f(x_0) = 0$, then $f(x) = -f(-x)$ for all $x \in \mathbb{R}$. $\endgroup$
    – J. J.
    Jun 17, 2011 at 17:32

If anyone is interested, here is a proof that $f$ vanishes only at $0$, copied from my post on AoPS.

Let $S=\{x\ne0: f(x)=0\}$, and suppose $S$ is nonempty for the sake of contradiction. (We assume $f(0)=0$ and the reduction posted earlier.)

For nonzero $x,y$, let $P(x,y)$ denote the assertion $f(f(x)y+x/y) = xyf(x^2+y^2) = f(f(y)x+y/x)$.

(1) $P(x,1)\implies f(f(x)+x) = xf(x^2+1) = f(x+1/x)$, so $s\in S\implies s^2+1,s+s^{-1}\in S$.

(2) For $s\in S$, $$ P(s,y)\implies f(s/y) = sy f(s^2+y^2) = f(f(y)s+y/s). $$ In particular, $P(s,-y)$ reveals that $f$ is odd. Also, if $t\in S$, $$ P(s,t)\implies f(s/t) = stf(s^2+t^2) = f(t/s), $$ so $P(s+s^{-1},s^2+1) \implies f(s^{-1}) = f(s) = 0\implies s^{-1}\in S$ (according to (1)) and $P(s,s)\implies s^2f(2s^2) = f(1) = 1\implies 2s^2\notin S$.

(3) If $s,t\in S$ with $t>s^2$, then $$ P(s,\sqrt{t-s^2}) \implies \frac{s}{\sqrt{t-s^2}}\in S\implies \frac{\sqrt{t-s^2}}{s}\in S. $$ Combined with (1), this gives $$ \left(\frac{\sqrt{t-s^2}}{s}\right)^2+1 = \frac{t}{s^2}\in S \implies \frac{t-s^2}{s} = \frac{s}{(s/\sqrt{t-s^2})^2}\in S $$ as well. Now WLOG fix $s\in(0,1)$ (if necessary, we can replace $s$ with one of $\pm s,\pm s^{-1}$, since $\pm1\notin S$). Then $t=s>s^2$ yields $\sqrt{s/(1-s)},\sqrt{(1-s)/s}\in S$ and $1-s = (s-s^2)/s \in S$ (from the first and second implications, respectively). But then (1) forces $(1-s)^2+1\in S$, so if we take $t=(1-s)^2+1>1>s^2$ instead, we get $$ 2\frac{1-s}{s} = \frac{(1-s)^2+1-s^2}{s} \in S $$ from the second implication. Finally, (twice) using the fact that $uv^2\in S$ whenever $u,v\in S$ and $uv^2\ne0,\pm1$ (*) twice, we first obtain $$ 2 = \left(2\frac{1-s}{s}\right)\left(\sqrt{\frac{s}{1-s}}\right)^2 \in S, $$ and then $2(s^{-1})^2>1$ must lie in $S$, contradicting the end of (2).

Edit (thanks to tenniskidperson3 for pointing out). (*) was originally incorrectly stated and missing proof. Here's a proof: By the second implication in (3), we know that if $u,v\in S$ and $uv^2>1$, then since $1/v\in S$, $u>(1/v)^2$ gives $u/(1/(1/v))^2 = uv^2$ in $S$. If $u,v\in S$ and $0<uv^2<1$ instead, then since $1/u,1/v\in S$, we have $(1/u)(1/v)^2>1$ so $(1/u)(1/v)^2\in S$ by the previous sentence, and then the inverse $uv^2$ must lie in $S$. If instead $u,v\in S$ and $uv^2<0$ but $uv^2\ne-1$, then negating $u$ shows from the previous two sentences that $uv^2\in S$.


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