Discrete math - Mod() : $x^n - y^n$ for $n ≥ 0$

Question

I have been working with modulo for a while now. But I just saw an equation that I have not been able to solve.

What are the last digits in $4^n-6^n$ for $n ≥ 0$.

Hint: This can be solved in many ways, to associate it with the theme, you should try to find a method that uses mod(10).

Is someone having a nice article or video lecture about this or perhaps you know how to explain it?

Solve

Thanks to @Henrik, I started to think of it as smaller pieces.

$4^2 mod(10),6^2 mod(10) => R=6 $ - Not valid

$4^3 mod(10) => R=4 $

$6^2 mod(10) => R=6 $

$n = 3$

Regards, Petter

Answer 1

Powers of $4$ end at $$4,6,4,6,4,6,....$$ and powers of $6$ end at $$6,6,6,6,6,.....$$

Thus $$4^n - 6^n $$ end with $$ 2,0,2,0,2,0,2,0,.......$$

Of course we have a sequence of negative numbers ending at these digits.

$$ -2,-20, -152,-1040,.....$$