Prove by induction that for all $n$, $8$ is a factor of $7^{2n+1} +1$ I want to prove by induction that for all $n$, 8 is a factor of $$7^{2n+1}+1$$
I have proved it true for the base case and assumed it true for $n=k$, but when I cannot figure out when to go towards the end of proving it true for $n=k+1$ assuming it is true for $n=k$.
I let $$7^{2k+1}+1 = 8m$$
Then I work with $$7^{2(k+1)+1}+1$$ to get eventually $$7^2(8m)$$
I am not sure if this is correct or if it is I am not sure how to prove that this too is dividable by $8$.
I would appreciate any help.
Thanks

use modules...
$7 \equiv -1 \pmod 8$, so $$7^{2n + 1} + 1 \equiv (-1)^{2n + 1} + 1 \equiv -1 + 1 \equiv 0 \pmod 8$$ so it is divisble by $8$
(Note that $2n + 1$ is odd, so $(-1)^{2n+1} = -1$)
