1 plus Perfect Power is not Prime Power except for 9

Theorem
The only solution to:
 * $x^m = y^n + 1$

is:
 * $\tuple {x, m, y, n} = \tuple {3, 2, 2, 3}$

for positive integers $x, y, m, n > 1$, and $x$ is a prime number.

This is a special case of Catalan's Conjecture.

Proof
It suffices to show the result for prime values of $n$.

The case $n = 2$ is covered in 1 plus Square is not Perfect Power.

So we consider the cases where $n$ is an odd prime.

where $Q$ is a polynomial in one unknown and $R$ is a degree zero polynomial, so $R$ is a constant.

We have:

Hence we have $x^m = \paren {y + 1} \paren {\map Q y \paren {y + 1} + n}$.

Since $x$ is a prime, we have:
 * $x \divides y + 1$
 * $x \divides \map Q y \paren {y + 1} + n$

Hence $x \divides n$.

Since $x > 1$ and $n$ is a prime, we must have $x = n$.

Now we write $y + 1 = x^\alpha$.

Then we have:

For $\alpha > 1$, $x \nmid \map Q y x^{\alpha - 1} + 1$.

Hence $\alpha = 1$.

This gives $y + 1 = x = n$.

The equation now simplifies to:
 * $\paren {y + 1}^m = y^n + 1$

Expanding:

hence we must have $y \divides m$.

By Absolute Value of Integer is not less than Divisors, $y \le m$.

Moreover, from $\ds \sum_{j \mathop = 1}^m \binom m j y^{j - 1} = y^{n - 1}$ we also have:
 * $y^{n - 1} > \dbinom m m y^{m - 1} = y^{m - 1}$

Therefore we also have $n > m$.

This gives $y = n - 1 \ge m$.

The two inequalities forces $y = m$.

Now our original equation is further simplified to:

From Real Sequence (1 + x over n)^n is Convergent:
 * $\paren {1 + \dfrac 1 y}^y$ is increasing and has limit $e$.

Then we have for all $y \in \N$:
 * $y + \dfrac 1 {y^y} < e < 3$

Since $\dfrac 1 {y^y} > 0$ and $y > 1$, we can only have $y = 2$.

This gives the solution $3^2 = 2^3 + 1$, and there are no others.

Also see

 * Cube which is One Less than a Square