# 1 plus Square is not Perfect Power

## Contents

## Theorem

The equation:

- $x^p = y^2 + 1$

has no solution in the integers for $x, y, p > 1$.

## Proof

Suppose $p$ is even.

Write $p = 2 k$.

Then:

\(\displaystyle 1\) | \(=\) | \(\displaystyle y^2 - x^{2 k}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {y - x^k} \paren {y + x^k}\) | Difference of Two Squares |

Since both $y - x^k$ and $y + x^k$ are integers, they must be equal to $\pm 1$.

Summing them up, we have $2 y$ is one of $-2, 0, 2$.

Thus $y$ is one of $-1, 0, 1$, and we ignore these solutions due to our condition $y > 1$.

Now suppose $p$ is odd.

Suppose $y$ is odd.

Then $x^p = y^2 + 1$ is even.

Hence $x$ is even.

Then:

\(\displaystyle 0\) | \(\equiv\) | \(\displaystyle x^p\) | \(\displaystyle \pmod 8\) | as $p \ge 3$ | |||||||||

\(\displaystyle \) | \(\equiv\) | \(\displaystyle y^2 + 1\) | \(\displaystyle \pmod 8\) | ||||||||||

\(\displaystyle \) | \(\equiv\) | \(\displaystyle 1 + 1\) | \(\displaystyle \pmod 8\) | Odd Square Modulo 8 | |||||||||

\(\displaystyle \) | \(\equiv\) | \(\displaystyle 2\) | \(\displaystyle \pmod 8\) |

which is a contradiction.

Hence $y$ must be even, and $x$ must be odd.

From Gaussian Integers form Euclidean Domain, we can define greatest common divisors on $\Z \sqbrk i$, and it admits unique factorization.

We factorize $y^2 + 1$:

- $x^p = y^2 + 1 = \paren {1 + i y} \paren {1 - i y}$

The greatest common divisors of $1 + i y$ and $1 - i y$ must divide their sum and product.

Their sum is $2$ while their product is $y^2 + 1$, which is odd.

Therefore we see that $1 + i y$ and $1 - i y$ are coprime.

From unique factorization we must have that both $1 + i y$ and $1 - i y$ is a product of a unit and a $p$th power.

By Units of Gaussian Integers, the units are $\pm 1$ and $\pm i$.

Hence

- $\exists u \in \set {\pm 1, \pm i}: \exists \alpha \in \Z \sqbrk i: 1 + i y = u \alpha^p, 1 - i y = \bar u \bar \alpha^p$

Since $p$ is odd:

- $1^p = 1$
- $\paren {-1}^p = -1$
- $i^p = \pm i$
- $\paren {-i}^p = -i^p = \mp i$

therefore there is some unit $u' \in \set {\pm 1, \pm i}$ such that $u'^p = u$.

By writing $\beta = u' \alpha$:

- $1 + i y = u'^p \alpha^p = \beta^p, 1 - i y = \bar \beta^p$

Write $\beta = a + i b$, where $a, b \in \Z$.

- $2 a = \beta + \bar \beta \divides \beta^p + \bar \beta^p = 2$

this gives $a = \pm 1$.

We also have:

\(\displaystyle 1 + y^2\) | \(=\) | \(\displaystyle \beta^p \bar \beta^p\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {\beta \bar \beta}^p\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {a^2 + b^2}^p\) | Product of Complex Number with Conjugate | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {1 + b^2}^p\) |

since $1 + y^2$ is odd, $b$ must be even.

Hence:

\(\displaystyle 1 + i y\) | \(=\) | \(\displaystyle \beta^p\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {a + i b}^p\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{k \mathop = 0}^p \binom p k a^{p - k} \paren {i b}^k\) | Binomial Theorem | ||||||||||

\(\displaystyle \) | \(\equiv\) | \(\displaystyle a^p + p a^{p - 1} i b\) | \(\displaystyle \pmod 4\) | $k \ge 2$ vanish as all terms containing $b^2$ is divisible by $4$ |

In particular, comparing real parts gives $1 \equiv a^p \pmod 4$.

Since $p$ is odd, we have $a = 1$.

Now we have:

\(\displaystyle 1 + i y\) | \(=\) | \(\displaystyle \paren {1 + i b}^p\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{k \mathop = 0}^p \binom p k 1^{p - k} \paren {i b}^k\) | Binomial Theorem | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle 1\) | \(=\) | \(\displaystyle \sum_{k \mathop = 0}^{\paren {p - 1} / 2} \binom p {2 k} b^{2 k} \paren {-1}^k\) | Comparing Real Parts; only even $k$ remain | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 1 - \binom p 2 b^2 + \sum_{k \mathop = 2}^{\paren {p - 1} / 2} \binom p {2 k} b^{2 k} \paren {-1}^k\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 1 - \binom p 2 b^2 + \sum_{k \mathop = 2}^{\paren {p - 1} / 2} \paren {\frac {p \paren {p - 1} } {2 k \paren {2 k - 1} } } \binom {p - 2} {2 k - 2} b^{2 k} \paren {-1}^k\) | Factors of Binomial Coefficient | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 1 - \binom p 2 b^2 + \binom p 2 b^2 \sum_{k \mathop = 2}^{\paren {p - 1} / 2} \paren {\frac 1 {k \paren {2 k - 1} } } \binom {p - 2} {2 k - 2} b^{2 k - 2} \paren {-1}^k\) | Binomial Coefficient with Two | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \binom p 2 b^2\) | \(=\) | \(\displaystyle \binom p 2 b^2 \sum_{k \mathop = 2}^{\paren {p - 1} / 2} \paren {\frac 1 {k \paren {2 k - 1} } } \binom {p - 2} {2 k - 2} b^{2 k - 2} \paren {-1}^k\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle 1\) | \(=\) | \(\displaystyle \sum_{k \mathop = 2}^{\paren {p - 1} / 2} \paren {\frac 1 {k \paren {2 k - 1} } } \binom {p - 2} {2 k - 2} b^{2 k - 2} \paren {-1}^k\) |

The summands on the right hand side may not be an integer, but if we can show:

- In canonical form, the numerator of each summand is even

then the equation is never satisfied.

This is because the sum of all the terms will be a rational number with even numerator and odd denominator, which cannot equal to $1$.

Since $2 k + 1$ is always odd and $\paren {-1}^k \dbinom {p - 2} {2 k - 2}$ is always an integer, we only need to check $\dfrac {b^{2 k - 2} } k$.

Since $b$ is even:

- $2^{2 k - 2} \divides b^{2 k - 2}$

But we have:

\(\displaystyle 2^{2 k - 2}\) | \(\ge\) | \(\displaystyle 2^k\) | as $k \ge 2$ | ||||||||||

\(\displaystyle \) | \(>\) | \(\displaystyle k\) | N less than M to the N |

Hence the largest power of $2$ that divides $k$ is less than $2^{2 k - 2}$.

Therefore the numerator of $\dfrac {b^{2 k - 2} } k$ is even.

And thus all the equations above are never satisfied.

So our original equation:

- $x^p = y^2 + 1$

has no solution in the integers for $x, y, p > 1$.

$\blacksquare$

## Historical Note

This is a special case of Catalan's Conjecture, and is due to Victor-Amédée Lebesgue.

(Not to be confused with Henri Léon Lebesgue.)

## Sources

- 1850: Victor-Amédée Lebesgue:
*Sur l'impossibilité en nombres entiers de l'équation $x^m = y^2 + 1$*(*Nouv. Ann. Math.***Ser. 1****Vol. 9**: pp. 178 – 181)

- 2014: Yuri F. Bilu, Yann Bugeaud and Maurice Mignotte:
*The Problem of Catalan*: Chapter $2$: Even Exponents