Post Number:#2 by ahmedittihad » Thu Jun 22, 2017 6:20 am
We rearrange the equation as, $n^5+n^4+1= 7^m $.
Observe that $n^5+n^4+1=(n^2+n+1)(n^3-n+1)$.
So, each of $n^2+n+1$ and $n^3-n+1$ is $7^x $ for some integer $x \geq 0$.
But using the euclidean algorithm, we find that $gcd(n^2+n+1, n^3-n+1)$ divides $7$.
So we consider the $4$ equations
$n^2+n+1=1$
$n^3-n+1=1$
$n^2+n+1=7$
$n^3-n+1=7$
We get that the only solution is $n=2$.
Frankly, my dear, I don't give a damn.