# Let $x$, $y$ and $z$ be prime numbers. Find the number of ordered triplets such that $x^y + 1 = z$

One of the possible ordered triplets is (2,2,5).

However I am unable to disprove that there will never be another triplet of prime numbers satisfying that condition. something like (x,y,z) = (2,y,z). In this case, we will have the final result of the expressions as odd on both the sides of the equality.

When $$z$$ is even, the only solution is when $$z=2$$, or $$x^y=1$$. This can only be possible if $$x=1$$ or $$y=0$$, so there is no solution.
When $$z$$ is odd, $$x^y$$ must be even, so $$x$$ must be even, which means $$x=2$$. However by this post, if $$2^y+1$$ is prime, $$y$$ must be a power of $$2$$. The only $$y$$ that satisfies this is $$2$$, so $$(2,2,5)$$ is the only solution.
• A parallel observation (implicit in your reference to another post, but more generally true). If $y\gt 1$ is odd then the left-hand side has a factor $x+1$ and cannot be prime, so we must have $y=2$. Commented Sep 28, 2018 at 9:30