1 plus Square is not Perfect Power

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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:

\(\ds 1\) \(=\) \(\ds y^2 - x^{2 k}\)
\(\ds \) \(=\) \(\ds \paren {y - x^k} \paren {y + x^k}\) 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:

\(\ds 0\) \(\equiv\) \(\ds x^p\) \(\ds \pmod 8\) as $p \ge 3$
\(\ds \) \(\equiv\) \(\ds y^2 + 1\) \(\ds \pmod 8\)
\(\ds \) \(\equiv\) \(\ds 1 + 1\) \(\ds \pmod 8\) Odd Square Modulo 8
\(\ds \) \(\equiv\) \(\ds 2\) \(\ds \pmod 8\)

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 $a, b \in \Z$.

By Sum of Two Odd Powers:

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

this gives $a = \pm 1$.

We also have:

\(\ds 1 + y^2\) \(=\) \(\ds \beta^p \bar \beta^p\)
\(\ds \) \(=\) \(\ds \paren {\beta \bar \beta}^p\)
\(\ds \) \(=\) \(\ds \paren {a^2 + b^2}^p\) Product of Complex Number with Conjugate
\(\ds \) \(=\) \(\ds \paren {1 + b^2}^p\)

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


Hence:

\(\ds 1 + i y\) \(=\) \(\ds \beta^p\)
\(\ds \) \(=\) \(\ds \paren {a + i b}^p\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^p \binom p k a^{p - k} \paren {i b}^k\) Binomial Theorem
\(\ds \) \(\equiv\) \(\ds a^p + p a^{p - 1} i b\) \(\ds \pmod 4\) $k \ge 2$ 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:

\(\ds 1 + i y\) \(=\) \(\ds \paren {1 + i b}^p\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^p \binom p k 1^{p - k} \paren {i b}^k\) Binomial Theorem
\(\ds \leadsto \ \ \) \(\ds 1\) \(=\) \(\ds \sum_{k \mathop = 0}^{\paren {p - 1} / 2} \binom p {2 k} b^{2 k} \paren {-1}^k\) Comparing Real Parts; only even $k$ remain
\(\ds \) \(=\) \(\ds 1 - \binom p 2 b^2 + \sum_{k \mathop = 2}^{\paren {p - 1} / 2} \binom p {2 k} b^{2 k} \paren {-1}^k\)
\(\ds \) \(=\) \(\ds 1 - \binom p 2 b^2 + \sum_{k \mathop = 2}^{\paren {p - 1} / 2} \paren {\frac {p \paren {p - 1} } {2 k \paren {2 k - 1} } } \binom {p - 2} {2 k - 2} b^{2 k} \paren {-1}^k\) Factors of Binomial Coefficient
\(\ds \) \(=\) \(\ds 1 - \binom p 2 b^2 + \binom p 2 b^2 \sum_{k \mathop = 2}^{\paren {p - 1} / 2} \paren {\frac 1 {k \paren {2 k - 1} } } \binom {p - 2} {2 k - 2} b^{2 k - 2} \paren {-1}^k\) Binomial Coefficient with Two
\(\ds \leadsto \ \ \) \(\ds \binom p 2 b^2\) \(=\) \(\ds \binom p 2 b^2 \sum_{k \mathop = 2}^{\paren {p - 1} / 2} \paren {\frac 1 {k \paren {2 k - 1} } } \binom {p - 2} {2 k - 2} b^{2 k - 2} \paren {-1}^k\)
\(\ds \leadsto \ \ \) \(\ds 1\) \(=\) \(\ds \sum_{k \mathop = 2}^{\paren {p - 1} / 2} \paren {\frac 1 {k \paren {2 k - 1} } } \binom {p - 2} {2 k - 2} b^{2 k - 2} \paren {-1}^k\)

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:

\(\ds 2^{2 k - 2}\) \(\ge\) \(\ds 2^k\) as $k \ge 2$
\(\ds \) \(>\) \(\ds k\) 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

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