Prove that $\forall$ n $\in$ $\mathbb{N}$ $7^n\equiv 1 \bmod{20}$ iff $n=4k$ Prove that $\forall$ n $\in$ $\mathbb{N}$ 
$7^n\equiv 1 \bmod{20}$ iff  $n=4k$
 A: Hint:
$$7\equiv7, \ 7^2\equiv9, \ 7^3\equiv3, \ 7^4\equiv1, 7^5\equiv7 \ldots \pmod{20}$$
A: Hint for the other direction: $n=4k$ implies $7^n=7^{4k}=(7^4)^k$
A: Let's prove the converse of the statement: If $7^{n} \equiv 1 \bmod 20$*, then* $n = 4k$.
Assume $7^{n} \equiv 1 \bmod 20$.  With P..'s response, you should see that for every power n divisible by 4, $7^{n} \equiv 1 \bmod 20$.  We can use Euler's Theorem and Chinese Remainder Theorem to verify this.  Note that 20 is the composite number, which is factored to $4 \times 5$, and $gcd(4,5) = 1$.  By Euler's theorem, we can write the following equivalences:
$7^{2} \equiv 1 \bmod 4$, which is equivalent to $7^{4} \equiv 1 \bmod 4$ by squaring both sides.
$7^{4} \equiv 1 \bmod 5$
So by Chinese Remainder Theorem $7^{4} \equiv 1 \bmod 20$.  But $7$ to any exponent divisible by $4$ gives $1 \bmod 20$.  Thus $n = 4k$.
Proving the statement another way is straightforward.  Observe that if $n = 4k$:
$7^{n} = 7^{4k} = (7^{4})^{k} = 1^{k} = 1$ for any positive integer $k$.
Thus, $7^{n} \equiv 1 \bmod 20$.
