Prove that the number $7^n+1$ is divisible by $8$ if $n$ is odd. In the case where $n$ is even, give the remainder of the division of $7^n+1$ by $8$. 
Prove that the number $7^n+1$ is divisible by $8$ if $n$ is odd. In the case where $n$ is even, give the remainder of the division of $7^n+1$ by $8$.



*

*If $n$ is odd


$7 \equiv -1 \mod 8$
$7^n \equiv (-1)^n \mod 8$
$7^n \equiv -1 \mod 8$
$7^n +1 \equiv 0 \mod 8$
Therefore, $7^n+1$ is divisible by $8$ if $n$ is odd.


*

*If $n$ is even


$7 \equiv -1 \mod 8$
$7^n \equiv (-1)^n \mod 8$
$7^n \equiv 1 \mod 8$
$7^n +1 \equiv 2 \mod 8$
Therefore, the remainder of the division of $7^n+1$ is $2$.
Is that true, please?

Yes, you are right. Another way to present it would be like so:
For $n$ odd,
$$7^n+1 \equiv (-1)^n+1 \equiv -1+1 \equiv 0 \pmod 8$$
and for $n$ even,
$$7^n+1 \equiv (-1)^n+1 \equiv 1+1 \equiv 2 \pmod 8$$
