# 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.

 $\ds x^m$ $=$ $\ds y^n + 1$ $\ds$ $=$ $\ds y^n - \paren {-1}^n$ as $n$ is odd $\ds$ $=$ $\ds \paren {y - \paren {-1} } \sum_{j \mathop = 0}^{n - 1} y^{n - j - 1} \paren {-1}^j$ Difference of Two Powers $\ds$ $=$ $\ds \paren {y + 1} \paren {\map Q y \paren {y + 1} + R}$ Division Theorem for Polynomial Forms over Field

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

We have:

 $\ds R$ $=$ $\ds \map Q {-1} \paren {-1 + 1} + R$ $\ds$ $=$ $\ds \sum_{j \mathop = 0}^{n - 1} \paren {-1}^{n - j - 1} \paren {-1}^j$ $\ds$ $=$ $\ds \sum_{j \mathop = 0}^{n - 1} \paren {-1}^{n - 1}$ $\ds$ $=$ $\ds \sum_{j \mathop = 0}^{n - 1} 1$ as $n$ is odd $\ds$ $=$ $\ds n$

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:

 $\ds x^m$ $=$ $\ds \paren {y + 1} \paren {\map Q y \paren {y + 1} + n}$ $\ds$ $=$ $\ds x^\alpha \paren {\map Q y x^\alpha + x}$ $\ds$ $=$ $\ds x^{\alpha + 1} \paren {\map Q y x^{\alpha - 1} + 1}$

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:

 $\ds \paren {y + 1}^m$ $=$ $\ds \sum_{j \mathop = 0}^m \binom m j y^j 1^{m - j}$ Binomial Theorem $\ds$ $=$ $\ds 1 + \sum_{j \mathop = 1}^m \binom m j y^j$ $\ds$ $=$ $\ds y^n + 1$ $\ds \leadsto \ \$ $\ds \sum_{j \mathop = 1}^m \binom m j y^{j - 1}$ $=$ $\ds y^{n - 1}$ $\ds \leadsto \ \$ $\ds \binom m 1 y^0$ $=$ $\ds 0$ $\ds \pmod y$ as $y > 1$ $\ds \leadsto \ \$ $\ds m$ $=$ $\ds 0$ $\ds \pmod y$ Binomial Coefficient with One

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:

 $\ds \paren {y + 1}^y$ $=$ $\ds y^{y + 1} + 1$ $\ds \leadsto \ \$ $\ds \paren {1 + \frac 1 y}^y$ $=$ $\ds y + \frac 1 {y^y}$ Dividing both sides by $y^y$
$\paren {1 + \dfrac 1 y}^y$ 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.

$\blacksquare$

## Sources

• 1870: G.C. GeronoNote sur la résolution en nombres entiers et positifs de l’équation $x^m = y^n + 1$ (Nouv. Ann. Math. Ser. 2 Vol. 9: pp. 469 – 471)