<s>[INST] Divisibility of sum of exponents Consider the sequence
$$r, \ ra, \ ra^2, \ ra^3, ... \ , ra^n  \mod M  $$
such that:
$$ ra^{n+1} \equiv  r   \mod M$$
and $a  \ne 1$ and $a,r$ are both coprime to $M$
Is it always true then that:
$$ r + ra + \ ra^2 + \ ra^3 ... + \ ra^n \equiv 0 \mod M$$
Consider the case of $M = 7$ just to get a better feel for the problem:
For $ r = 2 \ a = 3 $
$$ 2 + 2*3 + 2*3^2 ... 2*3^5 = 2 \frac{1 - 3^6}{1 -3} \equiv 0  \mod 7  $$
We can also look at
$r = 2 \ a = 4 $
$$ 2 + 2*2 + 2*4  \equiv 2 + 4 + 1 \equiv 0 \mod 7 $$
Why does this appear to be the case? Is there counter-example
 [/INST] Your example of $M = 7$ does not make sense unless you mean $ra^{n+1} \equiv r$. Then, since $\gcd(M,r) = 1$, we have $a^{n+1} \equiv 1 \pmod M$. You also probably mean $a$ is not $\equiv 1$.
Since $a$ is not $\equiv 1$, the sum is equal to $r \cdot \frac{a^{n+1}-1}{a-1} \equiv 0 \pmod M$. 
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