IMO 2019/4 – Mattia Giuri's bizarre blog

Find all couples of integers $(k,n)$ which satisfy

$k!=(2^n-1)(2^n-2)(2^n-4)...(2^n-2^{n-1})$

The Idea

We notice that $(1,1) , (3,2)$ work and we claim that other solutions don’t exist, so we want to get an upper and lower bound for $n$ .

Proof

Now let $v_p(x)$ be the greatest exponent of prime $p$ in the scomposition of $x$ , we notice that $v_2(RHS)=\frac{n(n-1)}{2}$ and $v_2(RHS)=\sum_{i=1}^{\infty}\lfloor{\frac{k}{2^i}}\rfloor < \sum_{i=1}^{\infty}\frac{k}{2^i}=k$ .

Hence, $k>\frac{n(n-1)}{2}$

Now we see that $v_3(LHS)=\sum_{i=1}^{\infty}\lfloor{\frac{k}{3^i}}\rfloor>\lfloor\frac{k}{3} \geq \frac{k}{3}-1$ while for the right hand side, we see that all the factors divisible by 3 are the ones which (once all factor 2 are removed) can be written in the form $4^h-1$ . So (by the Lifting The Exponent Lemma) $v_3(RHS)= \sum_{h=1}^{\lfloor\frac{n}{2}\rfloor} v_3(h)+1=v_3(\lfloor\frac{n}{2}\rfloor!)+\frac{n}{2}$ . Now similarly as before $v_3(\lfloor\frac{n}{2}\rfloor!) = \sum_{i=1}^{\infty}\lfloor\frac{\lfloor\frac{n}{2}\rfloor}{3^i}\rfloor <\sum_{n=1}^{\infty}\frac{\frac{n}{2}}{3^i}= \frac{n}{4}$ . So $v_3(RHS)=\frac{3n}{4}$ , hence $\frac{3n}{4}>\frac{k}{3}-1$ or $\frac{9n}{4}+3>k$ .