Prove that if $n^2 - 1$ is divisible by a prime number $p$ such that $n - 1$ is not divisible by $p$, then $n + 1$ is also divisible by $p$.

If this proposition is false, please give at least $3$ counter-examples, and try to modify the proposition so that it becomes true.

If the proposition is true, please try to prove this even more general proposition:

If $B^n - 1$ is divisible by a prime number $p$ such that $B^m - 1$ is not divisible by $p$ for all $m < n$, then $B^{n-1} + 1$ is also divisible by $p$.

.

Note: This question is actually related to numeral bases.

If $B + 1$ is divisible by a prime $p$ then the reciprocal of $p$ in base $B$ has a repeating mantissa of 2 digits.

If $B^2 - 1$ is divisible by a prime $p$ (and $B - 1$ is not) then the reciprocal of $p$ in base $B$ has a repeating mantissa of 2 digits.

I'm trying to show that these two propositions are kind of equivalent.

More generally: If $B^n - 1$ is divisible by a prime $p$ (and $B^m - 1$ is not, for all $m < n$) then the reciprocal of $p$ in base $B$ has a repeating mantissa of $n$ digits.

And we have: $n = \frac{p-1}{k}$ for a positive integer $k$.

(I've found these things myself so maybe I've made some mistakes.)

.

Edit: Then what about this generalization?

If $B^{2n} - 1$ is divisible by a prime number $p$ such that $B^m - 1$ is not divisible by $p$ for all $m < 2n$, then $B^n + 1$ is also divisible by $p$.