Show that a cube number which is also a square is of the form 7n, 7n+1

Question

I cannot reach at the answer but I have done the following: $N^6= (N^2)^3= (N^3)^2$
Using a derivation of Fermat's theorem $(N^\frac{p-1}{2} +1)(N^\frac{p-1}{2} -1)=kp$
Taking p=13, I got $N^6=13k+1, 13k-1$
How to reach the asked conclusion ?

Answer 1

Take $p=7$ which by Fermat's little theorem, divides $n^p-n=n(n^{p-1}-1)$ for all integer $n$

Now consider the two cases $p| n$

and $p\nmid n$

Answer 2

If $7\not| N$ then $gcd(7,N)=1$, so $ N^{7-1}\equiv 1(mod\ 7)\Rightarrow N^6=7n+1$

If $7|N \Rightarrow 7|N^6\Rightarrow N^6=7n$

Answer 3

The squares mod $7$ are $0,1,2,4$.

The cubes mod $7$ are $0,1,6$.

The intersection of these two sets is $0,1$.

Answer 4

Why use the prime $13$ and the derivation of FLT when you can use the prime $7$ and FLT itself?

$N^{7-1}\equiv 1 \mod 7$ if $7\not \mid a$. (i.e. $N^6$ is of the form $N^6 = 7n + 1$

... and...

$N^{7-1} \equiv 0\mod 7$ if $7\mid N$. (i.e. $N^6$ is of the form $N^6 = 7n$).

So any number that is both a square and a cube is a sixth power and the result follows.