Find all non-trivial triplets $(a,b,c)$ such that $ a+b^{d}\equiv0\pmod c $ for all $d>0$

Question

Let $a,b,c$ be co-prime integers $>2$ . Find all non-trivial triplets $(a,b,c)$ such that $ a+b^{d}\equiv0\pmod c $ for all $d>0$.

Answer 1

If $a,b,c$ satisfy $a+b^d \equiv 0 \pmod{c}$ for all $d>0$, then $c$ divides $a+b^2$, $a+b$, and $b^2-b=b(b-1)$. If we assume $c$ and $b$ are coprime, this implies that $c$ divides $b-1$. If $c$ divides $b-1$ then $c$ divides $a+1$, since $c$ divides their sum.

Conversely, suppose that $a \equiv -1 \pmod{c}$ and $b \equiv 1 \pmod{c}$. Then $b^d \equiv 1 \pmod{c}$ for all $d>0$, and $$a+b^d \equiv -1+1 \equiv 0 \pmod{c}.$$

Hence, for $a$, $b$, and $c$ pairwise coprime, we can conclude that $a+b^d \equiv 0 \pmod{c}$ for all $d>0$ if and only if $a \equiv -1\pmod{c}$ and $b \equiv 1\pmod{c}$.

Answer 2

$a+b^1=a+b^2\mod c$

so $b^2-b$ is a multiple of $c$.

$b$ is coprime to $c$, so $b-1$ is a multiple of $c$.

$b\equiv 1\mod c,a\equiv-1\mod c$ gives all solutions.