n^5+n^4

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n^5+n^4

Post Number:#1  Unread postby Katy729 » Sun Jun 18, 2017 10:48 pm

Determine all pairs of positive integers $(m,n)$ such that
$n^5+n^4=7^m-1$
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Re: n^5+n^4

Post Number:#2  Unread postby 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.
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