Prime Problem

Prove that for every prime $p>7$ , $p^6 -1$ is divisible by $504$ .

#NumberTheory #PrimeNumbers

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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Comments

By Fermat's Little Theorem, $7\mid p^6-1$ . By Euler's theorem, $9\mid p^6-1$ and $p^4\equiv 1\pmod 8$ . Since odd squares leave residue $1$ modulo $8$ , we have $p^6=p^4\cdot p^2\equiv 1\cdot 1=1\pmod 8$ . Finally because $7,8,9$ are pairwise coprime, their product $504\mid p^6-1$ .

Jubayer Nirjhor - 7 years ago

$504=2^{3} \times 3^{2} \times 7$ and by checking we can show the 6th residues $mod 7,8,9$ are either $0,1$ but since $p$ is prime $p^{6}=1$ mod $504$ (by the Chinese remainder theorem as $7,8,9$ are coprime) and hence we are done

Daniel Remo - 7 years ago

What is Chinese remainder theorem.??? Can you give me a link to any note..??

Sudipta Biswas - 7 years ago

Upon a quick google, http://www.math.tamu.edu/~jon.pitts/courses/2005c/470/supplements/chinese.pdf looks like the best link to explain it to you

Daniel Remo - 7 years ago

@Daniel Remo – Also here I realised you don't really need it: ignoring the last step we have that $7,8,9 | p^{6}-1$ and since they are coprime we know then that $( 7\times 8\times 9) | p^{6}-1$

Daniel Remo - 7 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$