<s>[INST] Under what conditions to have $ (b+1)^n \bmod 2b = (b+1)$ Under What conditions, for any integer $n \geq 1$, we will have:
$$ (b+1)^n \bmod 2b = (b+1)$$
I tested it with a couple of numbers, and it is true for $3, 5, 7$, etc. 
Please prove your theory.
It is based on a comment of this question
 [/INST] For $b$ odd,  $b=2k+1$ and $$b^2=b(2k+1)=2b\cdot k+b \equiv b \bmod (2b)$$
Now $$(b+1)^2 = b^2+2b+1 \equiv b^2+1 \equiv (b+1) \bmod (2b)$$ as required.
Then given $(b+1)^k \equiv (b+1)\bmod(2k)$,
 clearly $(b+1)^{k+1} = (b+1)^k(b+1) \equiv (b+1)(b+1) \equiv (b+1) \bmod (2b)$,
 and the result holds for all $n$.
For $b$ even, $b=2k$, $b^2 \equiv 0 \bmod (2b)$ and $(b+1)^2 \equiv 1 \bmod (2b)$,
so the original requirement does not hold.

Note that with even $b$ and odd $n$, the original condition will hold, since all even powers of $(b+1)$ will reduce to $1 \bmod (2b)$ So we could say: 
$$ (b+1)^n \equiv (b+1) \bmod (2b) \qquad \text{whenever }b\text{ or }n\text{ odd}$$

We can also make some observations on $(b-1)^n \bmod (2b)$. For odd $b$, powers of $(b-1)$ alternate between $(b+1)$ and $(b-1) \bmod (2b)$, and for even $b$, powers of $(b-1)$ alternate between $1$ and $(b-1) \bmod (2b)$. So it turns out that 
$$(b-1)^n \equiv (b-1) \bmod (2b) \qquad \text{for }n\text{ odd}$$
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