Proof for any natural n that: $8|5^n+2*3^{n-1}+1$

Question

I used this method for proving this statement but I came up with a problem.

$ 5^n+2*3^{n-1}+1 \equiv 1 + 25^{n/2} + 2 * 81^{(n-1)/4} \equiv 4 \pmod{8}$

What is the problem with my solution?

Answer 1

I'm assuming you are using the fact that:

$25\equiv1\pmod{8}$ and $81\equiv1\pmod{8}$ to get $2+1+1=4$

However $25^{n/2}$ will only have a modulo of $1$ if $n/2$ is a whole number, i.e. $n$ is even. Similarly $81^{(n-1)/4}$ will only have a module of $1$ if $(n-1)/4$ is a whole number, i.e. $n=4k+1$ for integer $k$. However this isn't a even number so you have a contradiction and your calculation will never work for any values of $n$.

Answer 2

Ian has pointed out why your method wouldn't work. I am just proposing a way to prove it.

If $n$ is even, say $n=2k$, then $$5^n+2(3^{n-1})+1 \equiv 5^{2k}+2(3^{2k-1})+1 \equiv (25^k)+2(3^{-1})(9^k)+1 \equiv 1+2(3)(1)+1 \equiv 0 \mod 8$$

If $n$ is odd, say $n=2k+1$, then $$5^n+2(3^{n-1})+1 \equiv 5^{2k+1}+2(3^{2k})+1 \equiv (25^k)(5)+2(9^k)+1 \equiv 5+2(1)+1 \equiv 0 \mod 8$$