Prove that $4^{2n} + 10n -1$ is a multiple of 25 Prove that if $n$ is a positive integer then $4^{2n} + 10n - 1$  is a multiple of $25$
I see that proof by induction would be the logical thing here so I start with trying $n=1$ and it is fine. Then assume statement is true and substitute $n$ by $n+1$ so I have the following:
$4^{2(n+1)} + 10(n+1) - 1$
And I have to prove that the above is a multiple of 25. I tried simplifying it but I can't seem to get it right. Any ideas? Thanks.

Another solution, via congruences mod. $25$. First note $16\bmod 25$ generates a cyclic group of order $5$:
$$16^2\equiv 6,\quad16^3\equiv 6\cdot 16\equiv-4,\quad16^4\equiv6^2\equiv 11, \quad 16^5\equiv6\cdot-4\equiv1\mod25.$$
So let's examine each case:


*

*If $n\equiv 0\mod 5$, $\;16^n+10n-1\equiv1+0-1=0$.

*If $n\equiv 1\mod 5$, $\;16^n+10n-1\equiv16+10-1=25\equiv 0$.

*If $n\equiv 2\mod 5$, $\;16^n+10n-1\equiv 6+20-1=25\equiv 0$.

*If $n\equiv 3\mod 5$, $\;16^n+10n-1\equiv -4+30-1=25\equiv 0$.

*If $n\equiv 4\mod 5$, $\;16^n+10n-1\equiv 11+40-1=50\equiv 0$.


Thus in each case,  $\;16^n+10n-1$ is divisible by $25$.
