## Task A-3.2.

Determine all triples of natural numbers ( $p, m, n$ ) such that $p$ is a prime number and

$$
2^{m} p^{2}+1=n^{5}
$$
## Solution.

If we write the equation in the form $2^{m} p^{2}=n^{5}-1$ and factorize the right side using the formula for the difference of fifth powers, we get

$$
2^{m} p^{2}=(n-1)\left(n^{4}+n^{3}+n^{2}+n+1\right)
$$

The factor $n^{4}+n^{3}+n^{2}+n+1$ is an odd number, so $n-1$ is divisible by $2^{m}$.

It also follows that $p$ is an odd number.

On the other hand, since $n$ is a natural number, it is clear that $n^{4}+n^{3}+n^{2}+n+1 > n-1$. Therefore, $p$ cannot divide $n-1$, because then $n-1$ would be at least $2^{m} p$, and $n^{4}+n^{3}+n^{2}+n+1$ would be at most $p$, which is less than $2^{m} p$. Thus, we have

$$
2^{m}+1=n, \quad p^{2}=n^{4}+n^{3}+n^{2}+n+1
$$

Notice that the second equation is equivalent to the following:

$$
\begin{aligned}
& n^{4}+n^{3}+n^{2}+n=p^{2}-1 \\
& n(n+1)\left(n^{2}+1\right)=(p-1)(p+1)
\end{aligned}
$$

Substituting $n=2^{m}+1$ into the last equation, we get

$$
\left(2^{m}+1\right)\left(2^{2 m}+2^{m+1}+2\right)\left(2^{m}+2\right)=(p-1)(p+1)
$$

or

$$
4\left(2^{m}+1\right)\left(2^{2 m-1}+2^{m}+1\right)\left(2^{m-1}+1\right)=(p-1)(p+1)
$$

Since $p$ is odd, the right side is the product of two consecutive even numbers and is divisible by 8. The left side is not divisible by 8, unless $m=1$. The only solution is therefore $m=1, n=3, p=11$.