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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 ?

amWhy
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Anubhab Giri
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  • You are correct that $N^6 = 13k \pm 1$ but that really is too weak a result. You are using too weak a consequence of FLT. Use FLT directly $N^6 = N^{7-1} \equiv 1,0 \mod 7$. Much more applicable. – fleablood Sep 27 '18 at 15:51
  • Your first line shows that if an integer is a sixth power then it's a cube and a square. But if you are going to apply Fermat's Theorem to 6th powers to address the problem in your title, you need to show the converse, that is, if $x = y^2 = z^3$, then $x$ is a sixth power. – Adam Bailey Sep 27 '18 at 17:56
  • A proof that if an integer is both a square and a cube then it is a sixth power of an integer may be found here. – Adam Bailey Sep 29 '18 at 17:32

4 Answers4

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

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

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

lhf
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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 $p\nmid n$

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

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

1123581321
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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.

fleablood
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