If (a,b,c)=(x,y,1)->x,y could be any 2 natural numbers,c>1 gives no natural solution. hence these are the only solutions.

And here is why: for $c>1$, we have \[ a^{b+20} = \frac{c^{b+21}-1}{c-1} = c^{b+20}+c^{b+19}\cdots+1\in\left(c^{b+20},(c+1)^{b+20}\right)\cap \mathbb{Z}, \]forcing $c<a<c+1$, which is absurd.

Very easy problem for p3. $$ANSWER:\boxed{(a,b,1)}$$