Prove divisibility using linear congruences I need to prove that:
$$10|(53^{53} - 33^{33})$$
I can and should only use linear congruences ($a \equiv b \mod n$) - how can I do this?
 A: $$53^{53}-33^{33}\equiv 3^{53}- 3^{33}\pmod {10}$$
Now, $$3^4=81\equiv 1\pmod 4\implies 3^{53}- 3^{33}\equiv 3^1-3^1\pmod {10}\equiv0$$ as $33\equiv1\pmod 4, 53\equiv1\pmod 4$
Alternatively, $$3^{53}- 3^{33}\equiv 3^{33}(3^{20}-1)\pmod {10}$$
Now, $3^4\equiv1\pmod 4\implies 3^{20}=(3^4)^5\equiv1^5\pmod {10}\equiv1$
A: Hint:
$53 \equiv 3 (\mod 10) \implies53^2 \equiv -1(\mod 10) \implies 53^{32} \equiv 1(\mod 10)$
$33 \equiv 3 (\mod 10) \implies33^2 \equiv-1(\mod 10) \implies 33^{32} \equiv 1\mod 10)$
$53^{33} \equiv \dots (\mod 10)$ and $33^{33} \equiv \dots(\mod 10)$ (Fill in the blanks)
A: $(53)^{4k}\equiv 1 \mod 10 \Rightarrow (53)^{53}=(53)^{52}\times53 \equiv 1\times53\equiv 3 \mod 10$
similarly $33^{4k} \equiv 1 \mod 10 \Rightarrow 33^{33} \equiv 3 \mod 10$
So the result follows.
Reason: $(10k+3)^4 \equiv 3^4\equiv  1 \mod 10\Rightarrow (10k+3)^{4l}\equiv 1 \mod 10$
Here $k,l\in \mathbb{N}$
A: $53^{53}\equiv3^{53}\equiv{(3^{4})}^{13}\cdot3\equiv3\mod 10$.
$33^{33}\equiv3^{33}\equiv{(3^{4})}^{8}\cdot3\equiv3\mod 10$.
because $a^{\varphi(n)}\equiv1\mod n$ and $\varphi(10)=4$, and thus $3^4\equiv1\mod 10$.
Therefore $53^{53}-33^{33}\equiv3-3\equiv0\mod 10$
