To solve $1+2^mp^2=q^5$

Question

How do we find all posible solutions of $1+2^mp^2=q^5$ for positive integer $m$ and primes $p,q$ ? $m=1,q=3,p=11$ is a solution , is there any other solution ?

Answer 1

Following Ivan Loh, $2^mp^2=(q-1)(q^4+q^3+q^2+q+1)$ implies $(q^4+q^3+q^2+q+1)$ equals either $p$ or $p^2$ (because it's odd). In either case $p\gt q^2$, which implies $p$ cannot divide $q-1$, hence we must have $q^4+q^3+q^2+q+1=p^2$ and thus $q-1=2^m$. Writing now $q=2^m+1$, we wind up with

$$2^{4m}+5\cdot2^{3m}+5\cdot2^{2m+1}+5\cdot2^{m+1}+5=p^2$$

Because $p^2\equiv1$ mod $8$ for any odd number, we must have $m\lt2$. This leaves only $m=1$, $q=3$, $p=11$.

Answer 2

I don't have time to flesh out the details, so here's a quick proof sketch.

Writing $2^mp^2=(q-1)(q^4+q^3+q^2+q+1)$, note that $(q^4+q^3+q^2+q+1)$ is odd so $2^m \mid q-1$. Then if $p \mid q-1$ we get $q-1>\sqrt{2^mp^2}>(q^4+q^3+q^2+q+1)$, a contradiction. Thus $p^2=q^4+q^3+q^2+q+1$ so $4p^2=4q^4+4q^3+4q^2+4q+4$ but we can bound the RHS of this strictly between consecutive perfect squares for sufficiently large $q$, leaving us with finitely many small cases to check.