3.2. If we write down the given equation in the form $2^m p^2 = n^5 - 1$ and factorise the right-hand side, we get
$$2^m p^2 = (n-1)(n^4 + n^3 + n^2 + n + 1).$$
Factor $n^4 + n^3 + n^2 + n + 1$ is odd, so $n-1$ is divisible by $2^m$.
We immediately see that $p$ is odd.
On the other hand, since $n$ is positive, we clearly have $n^4 + n^3 + n^2 + n + 1 > n - 1$. Hence $p$ cannot divide $n-1$, because otherwise $n-1$ 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 $2^m p$. Hence, we have
$$2^m + 1 = n, \quad p^2 = n^4 + n^3 + n^2 + n + 1.$$
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## Croatia2016_booklet — Page 18
FINAL ROUND – NATIONAL COMPETITION
Let us notice that the second equation is equivalent to the following equations:
$$n^4 + n^3 + n^2 + n = p^2 - 1,$$
$$n(n + 1)(n^2 + 1) = (p - 1)(p + 1).$$
By plugging $n = 2^m + 1$ into the last equation we get
$$(2^m + 1)(2^m + 2)(2^{2m} + 2^{m+1} + 2) = (p - 1)(p + 1),$$
which leads us to