TITLE: Showing $64$ divides $5^n-8n^2+4n-1$ without induction
QUESTION [9 upvotes]: I want to show that for all positive integer values of $n$, the number $5^n-8n^2+4n-1$ is divisible by $64$. Of course, I can easily do it by induction, but are there any number theoretic ways I can utilise to prove the divisibility? 
Thanks in advance for any help.

REPLY [3 votes]: Write $a_n=5^n-8n^2+4n-1 = 5^n+(-8n^2+4n-1)1^n$.
Then $a_n$ satisfies a linear recurrence implied by $(x-5)(x-1)^3$:
$$
a_{n+4} = 8a_{n+3}- 18a_{n+2} + 16a_{n+1} - 5a_{n}
$$
The particular expression for the recurrence is not important, except that it has integer coefficients.
Bottom line: It suffices to prove that $64$ divides $a_n$ for $n=0,1,2,3$. This is immediate because $a_0=a_1=a_2=0$ and $a_3=64$.