Prove using mathematical induction: for $n \ge 1, 5^{2n} - 4^{2n}$ is divisible by $9$ I have to prove the following statement using mathematical induction.
For all integers, $n \ge 1, 5^{2n} - 4^{2n}$ is divisible by 9. 
I got the base case which is if $n = 1$ and when you plug it in to the equation above you get 9 and 9 is divisible by 9. 
Now the inductive step is where I'm stuck.
I got the inductive hypothesis which is $ 5^{2k} - 4^{2k}$
Now if P(k) is true than P(k+1) must be true. $ 5^{2(k+1)} - 4^{2(k+1)}$
These are the step I gotten so far until I get stuck:
$$ 5^{2k+2} - 4^{2k+2} $$
$$ = 5^{2k}\cdot 5^{2} - 4^{2k} \cdot 4{^2} $$
$$ = 5^{2k}\cdot 25 - 4^{2k} \cdot 16 $$
Now after this I have no idea what to do. Any help is appreciated. 

The induction hypothesis can be written
$$
5^{2k}-4^{2k}=9m
$$
for some integer $m$. Therefore $5^{2k}=4^{2k}+9m$ and so
$$
5^2\cdot5^{2k}-4^2\cdot4^{2k}=
25(9m+4^{2k})-16\cdot4^{2k}=
9\cdot 25m+(25-16)\cdot 4^{2k}=
9\cdot 25m-9\cdot 4^{2k}
$$

Alternatively, $4\equiv -5\pmod{9}$, so
$$
5^{2k}-4^{2k}\equiv 5^{2k}-(-5)^{2k}\equiv 5^{2k}-5^{2k}\pmod{9}
$$
