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 \(504\mid p^6-1\).

\(504=2^{3} \times 3^{2} \times 7\) 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
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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 \(( 7\times 8\times 9) | p^{6}-1\)

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TopNewestBy 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 \(504\mid p^6-1\).

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\(504=2^{3} \times 3^{2} \times 7\) 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

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

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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

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