(Corée 2012)

Find all triplets of strictly positive integers $(m, n, p)$ with $p$ prime, such that $2^{m} p^{2} +$ $1=n^{5}$.
We write $2^{m} p^{2}=n^{5}-1=(n-1)\left(n^{4}+n^{3}+n^{2}+n+1\right)$. Since $n^{4}+n^{3}+n^{2}+n+1$ is odd, $2^{m} \mid n-1$.

Thus, $p^{2}=\frac{n-1}{2^{m}}\left(n^{4}+n^{3}+n^{2}+n+1\right)$. Since $p^{2}$ can only be written as a product of two terms in the form $p \cdot p$ or $1 \cdot p^{2}$, and $\frac{n-1}{2^{m}}<n^{4}+n^{3}+n^{2}+n+1$, it follows that $\frac{n-1}{2^{m}}=1$ and $p^{2}=n^{4}+n^{3}+n^{2}+n+1$. Therefore, $n=2^{m}+1$ and $p^{2}=\left(2^{m}+1\right)^{4}+\left(2^{m}+1\right)^{3}+\left(2^{m}+1\right)^{2}+\left(2^{m}+1\right)+1$. If $m \geqslant 2$, then $p^{2} \equiv 5[8]$, but 5 is not a quadratic residue modulo 8, which is absurd. Thus, $m=1$. The only solution is therefore $(1,3,11)$.