(P4 IMO 2019, untreated)

Find all pairs \((k, n)\) of strictly positive integers such that:

$$
k!=\left(2^{n}-1\right)\left(2^{n}-2\right)\left(2^{n}-4\right) \ldots\left(2^{n}-2^{n-1}\right)
$$
With the factor 2, we are interested in the 2-adic valuation.

On the right, it is $0+1+2+\ldots+(n-1)=\frac{n(n-1)}{2}$.

On the left, by Legendre's formula, it is $\sum_{p \geq 1}\left\lfloor\frac{k}{2^{p}}\right\rfloor \leq \sum_{p=1}^{k} \frac{k}{2^{p}}=\frac{k}{2} \times \frac{1-\left(\frac{1}{2}\right)^{k+1}}{1-\frac{1}{2}}\frac{n(n-1)}{2}$. Then, by the growth of the factorial, $\left(\frac{n(n-1)}{2}\right)!\leq\left(2^{n}-1\right)\left(2^{n}-2\right) \left(2^{n}-4\right) \ldots\left(2^{n}-2^{n-1}\right) \leq 2^{n^{2}}$. Intuitively, the factorial grows much faster than powers, so for large $n$ the inequality cannot be satisfied. Let's try to formalize this.

It is a matter of showing that from a certain rank, $2^{n^{2}}0$ such that $k!=1$: only $k=1$ works.
- For $n=2$, we must have $k!=6$ so $k=3$.
- For $n=3$, we must have $k!=168$, which is impossible.
- For $n=4$, 7 divides $(16-1)(16-2)(16-4)(16-8)$ so $k \geq 7$, in particular 9 divides $k!$, which is impossible because 9 does not divide $(16-1)(16-2)(16-4)(16-8)$.
- For $n=5: 2^{5}-1=31$ which is prime, so $k \geq 31$, hence $k!=31!\geq 29 \times\left(2^{5}-1\right)\left(2^{5}-2\right)\left(2^{5}-4\right)\left(2^{5}-8\right)\left(2^{5}-16\right)>\left(2^{5}-1\right)\left(2^{5}-2\right)\left(2^{5}-4\right)\left(2^{5}-8\right)\left(2^{5}-16\right)=k!$, absurd.

Finally, there are exactly two solution pairs: $(k, n)=(1,1)$ and $(k, n)=(3,2)$.

## 7 Functional Equations (Emilhan)

## Theoretical Reminders

- Introduction -

A functional equation is an equation where the unknowns are functions.

## Example 1.

Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying for all $x, y \in \mathbb{R}$,

$$
f(x+y)=f(x)+f(y)
$$

To solve a functional equation, one must proceed by analysis-synthesis. In the analysis phase, one gathers as much information as possible about the function using various methods that will be presented in the following. Then, it is necessary to verify that the functions found are indeed solutions.

The first method is substitution. It involves using the given equation for specific values of $x$ and $y$ to obtain information.