Congruence

**terencet** · September 30th 2008, 08:55 AM

Prove that for any integer a, a^4 ≡ 0 or 1 (mod 5)

**Moo** · September 30th 2008, 09:11 AM

Hello,

Originally Posted by terencet

Prove that for any integer a, a^4 ≡ 0 or 1 (mod 5)

Test this out

If $a \equiv 0 (\bmod 5)$ then $a^4 \equiv ?? (\bmod 5)$ ?
If $a \equiv 1 (\bmod 5)$ then $a^4 \equiv ?? (\bmod 5)$ ?
If $a \equiv 2 (\bmod 5)$ then $a^4 \equiv ?? (\bmod 5)$ ?
If $a \equiv 3 (\bmod 5)$ then $a^4 \equiv ?? (\bmod 5)$ ?
If $a \equiv 4 (\bmod 5)$ then $a^4 \equiv ?? (\bmod 5)$ ?

Otherwise, you can use Fermat's little theorem, which says : for p prime, if a is not a multiple of p, then $a^{p-1} \equiv 1 (\bmod p)$

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