Exercise 6. Find all strictly positive integers $p, q$ such that

$$
\mathrm{p} 2^{\mathrm{q}}=\mathrm{q} 2^{\mathrm{p}}
$$
Solution to Exercise 6

First case: if $\mathrm{p}=\mathrm{q}$, then the equality is true.

Second case: if $p \neq q$, we can assume without loss of generality that $p>q$, the case $q>p$ being treated similarly. We note that any odd divisor of $p$ is an odd divisor of $q$, and vice versa. Thus, if we write $p=a 2^{b}$ and $q=c 2^{d}$ with $a, c$ odd and $b, d \in \mathbb{N}$ (and it is always possible to do so according to the fundamental theorem of arithmetic), then $a=c$. The ratio $\mathrm{p} / \mathrm{q}$ is therefore a power of 2 (different from 1 since $p \neq q$). Let $e \in \mathbb{N}^{*}$ such that $p=2^{e} \mathbf{q}$. We then have

$$
2^{e+q}=2^{2^{e} q}
$$

so $\mathrm{e}+\mathrm{q}=2^{e} \mathrm{q}$ by identifying the exponents, or $\mathrm{e}=\mathrm{q}\left(2^{e}-1\right)$. A quick induction shows that $2^{n}-1>n$ for all $n \geqslant 2$: this is indeed true for $n=2$, and if $2^{n}>1+n$ for some integer $n$, then $2^{n+1}=2 \times 2^{n}>2(1+n)>2 n+2 \geqslant(n+1)+1$. Therefore, if $e>2$, $q\left(2^{e}-1\right)>q e \geqslant e$, a contradiction. Thus, $\mathrm{e}=1$, so $1=\mathrm{q}\left(2^{1}-1\right)$, or $\mathrm{q}=1$. We conclude that the only solution is $\mathrm{p}=2$ and $\mathrm{q}=1$.

Conclusion: the solutions to the problem are the pairs $(p, p)$ for $p \in \mathbb{N}^{*}$ as well as the pairs $(2,1)$ and $(1,2)$.

## Exercises of Group A