42nd Balkan Mathematical Olympiad (BMO 2025 )

Problem 3.

Fnd all functions $f\colon \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x,y \in \mathbb{R}$ ,
$f(x+yf(x))+y = xy + f(x+y).$

Solution 1

We claim that the only solutions are $\boxed{f(x)=x}$ and $\boxed{f(x)=2-x}$ . These clearly work.

First note that if there exists some $u$ such that $f(u)=0$ , then we can immediately conclude that $f$ is linear by placing $x=u$ , which leads to the two above solutions. We will now assume for the sake of contradiction that there does not exist such a $u$ :

By letting $y=a-x$ , we obtain that $f(…)+(a-x)=x(a-x)+f(a)$ . Since $f(…)\ne 0$ by the above assumption, we see that there does not exist an $x$ such that $x(a-x)-(a-x)+f(a)=0$ .

Therefore, given any $a$ , we have that $f(a)$ is not in the range of the quadratic $(x-1)(x-a)=x^2-(a+1)x+a$ , which means that

$f(a)<\left(\frac{a+1}{2}\right)^2-\frac{(a+1)^2}{2}+a=-\frac{(a-1)^2}{4}$
With this bound, we can now control the size of $f$ quite well:
Let $y=-\frac{x}{f(x)}$ so that $f(0)-\frac{x}{f(x)}=-\frac{x^2}{f(x)}+f\left(x-\frac{x}{f(x)}\right)$ . Thus:

$f(0)+\frac{x^2-x}{f(x)}=f\left(x-\frac{x}{f(x)}\right) \qquad (*)$
Now if we let $x$ tend to infinity in $(*)$ we can see that $\left|\frac{x^2-x}{f(x)}\right|<\left|\frac{x^2-x}{(x-1)^2/4}\right|\mapsto 4$ . This leads us to a contradiction, since then as $x$ approaches infinity, the LHS in $(*)$ is bounded between $f(0)-4$ and $f(0)+4$ (which are constants), while the RHS approaches negative infinity!

Solution 2

Answers are $f(x)=x$ and $f(x)=2-x$ which indeed fit. Let $P(x,y)$ be the assertion.
Claim: $f(1)=1$ .
Proof: $P(1,y)$ gives $f(y+1)=f(yf(1)+1)$ . We have
$f(1+x+yf(1+x))=xy+f(1+x+y)$

Plug $P(xf(1),yf(1))$ to get $xy=xyf(1)^2$ . If $f(1)=-1$ , then $f(x+1)=f(1-x)$ or $f(x)=f(2-x)$ . Since $f\not \equiv -1$ , pick $f(t)\neq -1$ and $t\neq 1$ . Plugging $P(t,\frac{2-2t}{1+f(t)})$ yields $(\frac{2-2t}{f(t)+1})(t-1)=0$ which is impossible.

$P(0,1)$ yields $f(f(0))=0$ and $P(f(0),y-f(0))$ implies $y-f(0)=(y-f(0))f(0)+f(y)$ hence $f$ is linear. All linear solutions are $f(x)=x$ and $f(x)=2-x$ as desired. $\blacksquare$

Solution 3

The only solutions are $f(x) = x$ and $f(x) = 2 - x$ , which clearly work. Now we prove they are the only solutions. Suppose there existed another solution. Let $P(x,y)$ be the given assertion.

Claim: $f$ is not linear.
Proof: Suppose $f$ was linear (and keep in mind we are asssuming $f$ is not the identity or $2 - x$ ). Since $f$ is clearly not constant, it must be injective as well. Let $f(x) = ax + b$ .

$P(1,y)$ gives $f(1 + yf(1)) = f(1 + y)$ , so since injective, $f(1) = 1\implies a + b = 1$ .

$P(0,y): f(yf(0)) + y = f(y)$ , so setting $y = 1$ gives $f(f(0)) = 0$ , so $f(b) = 0\implies ab + b = 0$ , so $a = -1$ or $b = 0$ . But these give $b = 2$ and $a =1$ respectively, so $f(x) = x \forall x$ or $f(x) = 2 - x \forall x$ , absurd. $\square$

Claim: $0$ is not in the image of $f$ .
Proof: If $0$ was in the image of $f$ , fix $u$ with $f(u) = 0$ . Setting $x = u$ gives $y = uy + f(u + y)$ , so $f$ is clearly linear. $\square$

Claim: If $a$ is in the image of $f$ , then anything less than $a$ is in the image of $f$ as well.
Proof: Let $a = f(c)$ . $P(x, c - x)$ gives that $f(c) + x(c-x) - (c - x) = f(c) + (-x^2 + (c+1) x - c)$ is in the image of $f$ . Let $P(x) = -x^2 + (c+1) x - c$ . Since $P$ gets arbitrarily small (negative leading coefficient) and $c$ is a root of $P$ , by IVT, $P$ is surjective over negative real numbers. $\square$

The two claims combined implies that $f(x)$ is negative for all $x$ .

Setting $x = 0$ gives $f(yf(0)) = f(y) - y$ , so $f(y) < y$ must hold. Now let $g(x) = x - f(x) > 0$ .

We have

$xy + f(x+y) = f(x + yf(x)) + y < x + y + y f(x) \implies g(x + y) > y g(x)$

Setting $y = c - x$ gives that $g(c) > (c - x) g(x)$ for all reals $x,c$ , so for all $x < c$ ,

$g(x) < \frac{g(c)}{c - x}$

Thus (as $g$ is always positive), $\lim_{x \to -\infty} g(x) = 0$ . From $P(0,y)$ , we have $f(yf(0)) = -g(y)$ . Taking $y$ to negative infinity gives that

$\lim_{x \to -\infty} f(x f(0)) = 0 \implies \lim_{x \to \infty} f(x) = 0$

because $f(0)$ is negative.

$P(x,1): f(x + f(x)) + 1 = x + f(x+1)$ . Setting $x \to \infty$ gives that the RHS approaches infinity, but the LHS is always less than $1$ , a contradiction.