We shall prove if by binomial expansion
we have $4^{2n}$
$= (5-1)^{2n}$
$ = \sum_{k=0}^{2n}{2n \choose k}5^{2n-2}(-1)^{k}$
$ = \sum_{k=0}^{2n-2}{2n \choose k}5^{2n-2}(-1)^k - {2n \choose 2n-1}*5 + 1$ separating last 2 terms
1st sum each term is divisible by $5^2$ that is 25
so we are left with
$4^{2n} \equiv - 10n +1 \pmod {25}$
or
$4^{2n}+10n \equiv 1 \pmod {25}$
