How many triples $\left(m,n,p\right)$ where $p$ is a prime satisfy $2^mp^2+1=n^5$?
Found one of the solutions for $\left(m,n,p\right)$ to be $\left(1,3,11\right)$ by analysis without using modular arithmetic. How should I attempt this problem? Thanks
Note that if $(m,n,p)$ is a solution, then $$2^mp^2=n^5-1=(n-1)(n^4+n^3+n^2+n+1).$$ In particular $n-1$ divides $2^mp^2$, so $n=2^ap^b+1$ for some $a\leq m$ and $b\leq2$. Plugging this in yields $$2^mp^2+1=n^5=\left(2^ap^b\right)^5+5\left(2^ap^b\right)^4+10\left(2^ap^b\right)^3+10\left(2^ap^b\right)^2+5\left(2^ap^b\right)+1,$$ and a bit of rearranging shows that $$2^{m-a}p^{2-b}=\left(2^ap^b\right)^4+5\left(2^ap^b\right)^3+10\left(2^ap^b\right)^2+10\left(2^ap^b\right)+5.$$ The right hand side is odd so $m=a$ and $p>2$, and we are left with $$p^{2-b}=2^{4m}p^{4b}+5\cdot2^{3m}p^{3b}+5\cdot2^{2m+1}p^{2b}+5\cdot2^{m+1}p^b+5.$$ From this it is not hard to see that $b=0$, so $n=2^m+1$. This leaves you with solving $$p^2=\frac{(2^m+1)^5-1}{2^m}=16^m+5\cdot8^m+10\cdot4^m+10\cdot2^m+5.$$ For $m>1$ the right hand side is congruent to $5$ mod $8$ and hence not a square, so the only solution is $(m,n,p)=(1,3,11)$.