Proving $24\mid5^{2n}-1$ using modular arithmetic So, in general I am aware of how to use modular arithmetic to prove a divisibility. But I have the following problem:

Prove that $24\mid5^{2n}-1$ for all $n\in\mathbb Z$. 

I know that theoretically, I could show 23 different cases that the expression is congruent to $0\bmod24$, but that seems like it might be excessive to me.
Is there a faster way to show this?
 A: You want to prove what $5^{2n}$ is congruent to in modulo 24
Notice that $5^{2n} = (5^2)^n$
$5^2=25 \equiv 1 \pmod {24}$
So $5^{2n} \equiv 1^n \equiv 1 \pmod {24} $
A: You could use $5^{2n}-1=(5^2-1)(5^{2n-2}+5^{2n-4}+\ldots +1)$ where you are summing the geometric series.
A: Use induction. It is true for $n=1$. Assuming it is true for $n$, show it is true for $n+1$:
$$5^{2(n+1)}-1=25(5^{2n}-1)+24 = 0 (mod \ 24).$$
A: 1)
Prove $8|5^{2n} - 1$ and $3|5^{2n} - 1$
a) $5^{2n} - 1 = (5^n -1)(5^n + 1)$.  $5^n \pm 1 \equiv 1^n\pm 1 \equiv 0\mod 2$ so $2|5^2 - 1$ and $2|5^2 + 1$ and so $4|5^{2n}-1$.  If $5^n - 1 = 2k$ then $5^{2n} + 1= 2k + 2 = 2(k+1)$.  Either $k$ is even or $k+1$ is even so either $2k$ is even and $2(k+1)$ is divisible by $4$, or $2k$ is divisible by $4$ and $2(k+1)$ is even.  Either way.  $2k*2(k+1)= 5^{2n} -1$ is divisible by $8$.
b)$5^{2n}-1\equiv (-1)^{2n}-1 \equiv 1^n - 1 \equiv 0 \mod 3$.
So $3|5^{2n}-1$.
2)
Or notice $5^{2n} -1 = 25^n -1 \equiv 1^n -1 \equiv 0 \mod 24$.
(I will admit, I did not see the obvious right away.)
