Prove (or disprove) that $3^{2n+3}+40n-27$ is divisible by $64$ for any $n$ integer

Question

So I want to prove that is in title. I'm working with modular arithmetic, so if any of you can show me a hint, it will be perfect.

Answer 1

We find that $$ \frac{3^{2n+3}+40n-27}{8}=\frac{27(9^n-1)+40n}{8}=27(9^{n-1}+9^{n-2}+\cdots+1)+5n. $$ And also $$ 27(9^{n-1}+9^{n-2}+\cdots+1)+5n\equiv 27(1+1+\cdots +1)+5n\equiv32n\equiv 0\ \;\text{(mod}\;8). $$ Therefore $64 \;|\; 3^{2n+3}+40n-27$ for all $n\ge 1$.

Answer 2

Use induction and modular arithmetic: suppose that, for some $n$, $3^{2n+3}+40n-27\equiv 0\mod64$. We have to show that $$3^{2n+5}+40(n+1)-27=9\cdot3^{2n+3}+40n+13\equiv 0\mod64.$$ The inductive hypothesis can be rewritten as $$3^{2n+3}\equiv -40n +27\mod64,$$ so that $$9\cdot3^{2n+3}+40n+13\equiv-360n+243+40m+13=-320n+256\equiv 0\mod 64. $$