1. ## Modulo proof

Let n E Z. and supposed that 5 does not divide n. Prove that n^4 is congruent to 1 mod 5.

2. Originally Posted by kel1487
Let n E Z. and supposed that 5 does not divide n. Prove that n^4 is congruent to 1 mod 5.
Nothing fancy here. For example,
$n \equiv 1~\text{ (mod 5)}$
so
$n^4 \equiv (1)^4 \equiv 1~\text{ (mod 5)}$

Similarly for $n \equiv 2~\text{ (mod 5)}$
$n^4 \equiv (2)^4 \equiv 16 \equiv 1~\text{ (mod 5)}$

etc.

-Dan

3. Thanks!! I was thinking too much into it!!

4. If n is not divisible by 5, then one of $n+1$, $n-1$, $n+2$, $n-2$ must be divisible by 5.

Hence the product $(n+1)(n-1)(n+2)(n-2)=(n^2-1)(n^2-4)$ must be divisible by 5.

Note however that $n^2-4\equiv n^2+1\pmod{5}$.

Hence $(n^2-1)(n^2+1)=n^4-1$ is divisible by 5; in other words $n^4\equiv1\pmod{5}$.

In general, Fermat's little theorem states that if p is prime and p does not divide n, then $n^{p-1}\equiv1\pmod{p}$.