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QUESTION

Prove that $a^{49}-a$ is divisible by 70 for every integer $a$.



ANSWER


70=2.7.5. By the corollary to Fermat's little theorem (Corollary
4.3), $a^7\equiv a$ mod 7, $a^5\equiv a$ mod 5 and $a^2\equiv a$
mod 2 for all $a$. Hence $a^{49}\equiv (a^7)^7\equiv a^7\equiv a$
mod 7, $a^{49}\equiv(a^5)^9a^4\equiv a^9.a^4\equiv
a^13\equiv(a^5)^2a^3\equiv a^2a^3\equiv a^5\equiv a$ mod 5, and
$a^{49}\equiv(a^2)^24a\equiv a^24.a\equiv(a^2)^{12}a\equiv
a^12.a\equiv(a^2)^6a\equiv a^6a\equiv(a^2)^3a\equiv a^3.a\equiv
a^4\equiv (a^2)^2\equiv a^2\equiv a$ mod 2.

Thus $a^47-a$ is divisible by 7,5 and 2, and hence (cor.1.7) by
their product 70.



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