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 $y^y$ |
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.
$\blacksquare$
Also see
Sources
- 1870: G.C. Gerono: Note 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)