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# Prime Problem

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

Note by Sudipta Biswas
3 years, 6 months ago

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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 $$504\mid p^6-1$$.

- 3 years, 6 months 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

- 3 years, 6 months ago

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

- 3 years, 6 months 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

- 3 years, 6 months ago

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 $$( 7\times 8\times 9) | p^{6}-1$$

- 3 years, 6 months ago