1 plus Perfect Power is not Prime Power except for 9

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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.


\(\displaystyle x^m\) \(=\) \(\displaystyle y^n + 1\)
\(\displaystyle \) \(=\) \(\displaystyle y^n - \paren {-1}^n\) as $n$ is odd
\(\displaystyle \) \(=\) \(\displaystyle \paren {y - \paren {-1} } \sum_{j \mathop = 0}^{n - 1} y^{n - j - 1} \paren {-1}^j\) Difference of Two Powers
\(\displaystyle \) \(=\) \(\displaystyle \paren {y + 1} \paren {\map Q y \paren {y + 1} + R}\) 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:

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

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:

\(\displaystyle x^m\) \(=\) \(\displaystyle \paren {y + 1} \paren {\map Q y \paren {y + 1} + n}\)
\(\displaystyle \) \(=\) \(\displaystyle x^\alpha \paren {\map Q y x^\alpha + x}\)
\(\displaystyle \) \(=\) \(\displaystyle x^{\alpha + 1} \paren {\map Q y x^{\alpha - 1} + 1}\)

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:

\(\displaystyle \paren {y + 1}^m\) \(=\) \(\displaystyle \sum_{j \mathop = 0}^m \binom m j y^j 1^{m - j}\) Binomial Theorem
\(\displaystyle \) \(=\) \(\displaystyle 1 + \sum_{j \mathop = 1}^m \binom m j y^j\)
\(\displaystyle \) \(=\) \(\displaystyle y^n + 1\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sum_{j \mathop = 1}^m \binom m j y^{j - 1}\) \(=\) \(\displaystyle y^{n - 1}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \binom m 1 y^0\) \(=\) \(\displaystyle 0\) \(\displaystyle \pmod y\) as $y > 1$
\(\displaystyle \leadsto \ \ \) \(\displaystyle m\) \(=\) \(\displaystyle 0\) \(\displaystyle \pmod y\) 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 $\displaystyle \sum_{j \mathop = 1}^m \binom m j y^{j - 1} = y^{n - 1}$ 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:

\(\displaystyle \paren {y + 1}^y\) \(=\) \(\displaystyle y^{y + 1} + 1\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {1 + \frac 1 y}^y\) \(=\) \(\displaystyle y + \frac 1 {y^y}\) 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. 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)