SHORTLISTED PROBLEMS FOR THE 73rd NMO

Let us find all functions $f : \mathbb{Q} \to \mathbb{Q}$ such that for all $x, y \in \mathbb{Q}$,
$$f(1 - xy) = (1 - f(x))(1 - f(y)).$$

Let us denote the given equation as $(*)$.

Step 1. Plug $x = 0$ into $(*)$:
$$f(1 - 0 \cdot y) = (1 - f(0))(1 - f(y)) \implies f(1) = (1 - f(0))(1 - f(y)).$$
But the left side is independent of $y$, so the right side must also be independent of $y$. Therefore, $1 - f(0) = 0$ or $1 - f(y)$ is constant.

Case 1: $1 - f(0) = 0 \implies f(0) = 1$

Then $f(1) = 0$ (since $f(1) = (1 - f(0))(1 - f(y)) = 0$ for all $y$).

Case 2: $1 - f(y)$ is constant for all $y$, so $f(y)$ is constant. Let $f(y) = c$ for all $y$.
Plug into $(*)$:
$$c = (1 - c)^2.$$
So $c = 1 - 2c + c^2 \implies c^2 - 3c + 1 = 0$.
The solutions are $c = \frac{3 \pm \sqrt{5}}{2}$, but these are not rational, so there is no constant rational solution.

So only Case 1 is possible: $f(0) = 1$, $f(1) = 0$.

Step 2. Plug $y = 1$ into $(*)$:
$$f(1 - x \cdot 1) = (1 - f(x))(1 - f(1)) \implies f(1 - x) = (1 - f(x))(1 - 0) = 1 - f(x).$$
So $f(1 - x) = 1 - f(x)$ for all $x \in \mathbb{Q}$.

Step 3. Plug $y = 0$ into $(*)$:
$$f(1 - x \cdot 0) = (1 - f(x))(1 - f(0)) \implies f(1) = (1 - f(x))(1 - 1) = 0.$$
Which is consistent with $f(1) = 0$.

Step 4. Plug $x = a$, $y = b$ into $(*)$ and use $f(1 - x) = 1 - f(x)$:
Let us try to find the general form.
Let us try $f(x) = x$.
Then $f(1 - xy) = 1 - xy$, $(1 - f(x))(1 - f(y)) = (1 - x)(1 - y) = 1 - x - y + xy$.
But $1 - xy \neq 1 - x - y + xy$ in general, so $f(x) = x$ is not a solution.

Try $f(x) = 1 - x$.
Then $f(1 - xy) = 1 - (1 - xy) = xy$, $(1 - f(x))(1 - f(y)) = x y$.
So $f(1 - xy) = (1 - f(x))(1 - f(y))$.

Therefore, $f(x) = 1 - x$ is a solution.

Let us check if there are other solutions.
Suppose $f(x) = a x + b$ for some $a, b \in \mathbb{Q}$.
From above, $f(1 - x) = 1 - f(x)$:
$$f(1 - x) = a(1 - x) + b = a - a x + b = 1 - (a x + b) = 1 - f(x).$$
So $a - a x + b = 1 - a x - b \implies a + b = 1 - b \implies a + 2b = 1$.

Also, $f(0) = b = 1$ (from above), so $b = 1$, so $a + 2 \cdot 1 = 1 \implies a = -1$.
So $f(x) = -x + 1$.

Therefore, the only solution is $f(x) = 1 - x$.

Check:
$$f(1 - xy) = 1 - (1 - xy) = xy,$$
$$(1 - f(x))(1 - f(y)) = x y.$$
So it works.

Conclusion:

The only function $f : \mathbb{Q} \to \mathbb{Q}$ satisfying $f(1 - xy) = (1 - f(x))(1 - f(y))$ for all $x, y \in \mathbb{Q}$ is
$$f(x) = 1 - x.$$