Question

Find all positive integers $m,n,$ and primes $p\ge 5$ such that $m(4m^2+m+12)=3(p^n-1)$.

Answer 1

Rewriting the given equation we have
$4m^3+m^2+12m+3=3p^n$.
L.H.S.=$(4m+1)(m^2+3)$.
Suppose that $(4m+1,m^2+3)=1$. Then $(4m+1,m^2+3)=(3p^n,1),(3,p^n),(p^n,3)$ or $(1,3p^n)$, a contradiction since $4m+1,m^2+3\ge 4$.
$\therefore (4m+1,m^2+3)>1$.
Since $4m+1$ is odd, we have $(4m+1,m^2+3)=(4m+1,16m^2+48)=(4m+1,49)=7$ or 49.
$\therefore p=7$, and $4m+1=3.7k$ or 7k for some natural number k.
If $(4m+1,49)=7$, then we have $k=1$ and $4m+1=21$ which does not lead to a solution.
$\therefore (4m+1,m^2+3)=49$.
If $7^3$ divides 4m+ 1 then it does not divide $m^2+3$, so we get $m^2+3\le 3\cdot 7^2<7^3\le 4m+1$. This implies $(m-2{)}^2<2$, so $m\le 3$, which does not lead to a solution.
$\therefore 4m+1=49$
$\therefore m=12$ and $n=4$.
Thus $(m,n,p)=(12,4,7)$ is the only solution.