Exercise 2. Determine all pairs of positive integers $(n, p)$ where $p$ is a prime number and such that $n+p$ is a power of $n$.

A power of $n$ is of the form $n^{k}$ for $k$ a natural number.
Solution to Exercise 2 Let $(\mathrm{p}, \mathrm{n})$ be a potential solution pair. We have an integer $k$ such that

$$
\mathrm{n}+\mathrm{p}=\mathrm{n}^{\mathrm{k}}
$$

First, if $k=0$, then $n+p=1$. Since $p$ is prime, $p>1$ and $n \geqslant 0$, so $n+p=1<p \leqslant p+n$ which is absurd. Therefore, $k \geqslant 1$.

We can then rewrite the equation as follows:

$$
p=n\left(n^{k-1}-1\right)
$$

We obtain that $n$ divides $p$. Since $p$ is prime, it must be that $n=1$ or $n=p$.

If $n=1$, then $1+p=1$ which is absurd since $p \neq 0$.

Therefore, $n=p$. Then $2p=p^k$ so 2 divides $p$. Thus, $p=2$ and the equation becomes $2+2=2^k$ so $k=2$. Conversely, the pair $(2,2)$ satisfies $2+2=2^2$ and thus $2+2$ is a power of 2.

The only solution pair is therefore $(2,2)$.

Grader's Comment The exercise is well done. Several approaches could have worked. However, it was quite easy to forget some special cases.