2025/015) Prove $4^{2n}+10n \equiv 1 \pmod {25}$

 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  

   $4^{2n}+10n \equiv 1 \pmod {25}$