Difference between revisions of "Catalan's theorem"
Borishaase (talk | contribs) (Created page with "For <math>1 + x^m = y^n</math> where <math>m, n, x, y \in {}^{\omega}\mathbb{N}_{\ge 2}</math>, only the solution <math>m_0 = 3</math>, <math>n_0 = 2</math>, <math>x_0 = 2</ma...") |
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For <math>1 + x^m = y^n</math> where <math>m, n, x, y \in {}^{\omega}\mathbb{N}_{\ge 2}</math>, only the solution <math>m_0 = 3</math>, <math>n_0 = 2</math>, <math>x_0 = 2</math> and <math>y_0 = 3</math> exists. | For <math>1 + x^m = y^n</math> where <math>m, n, x, y \in {}^{\omega}\mathbb{N}_{\ge 2}</math>, only the solution <math>m_0 = 3</math>, <math>n_0 = 2</math>, <math>x_0 = 2</math> and <math>y_0 = 3</math> exists. | ||
| − | '''Indirect proof:''' Assuming <math>\acute{x}, y \in {}^{\omega}2\mathbb{N}</math> yields the contradiction <math>1 + x^m \equiv 2 \equiv y^n \equiv 0</math> resp. <math>4</math> mod <math>8</math>. For <math>p \mid y,\ x = \hat{r},\ y = \hat{s} + 1</math> and <math>2 < m \ne n > 2</math>, subsequently <math>(\hat{s} + 1)^n - 1 \equiv \hat{r}^m \equiv 0</math> mod <math>p \in {}^{\omega}\mathbb{P}_{\ge 3}</math> is assumed. Then also <math>(\hat{s} + 1)^{\hat{n}} - 1 = (\hat{s} + 2)^n\ \hat{s}^n \equiv \hat{r}^m \equiv 0</math> mod <math>p</math> and <math>2^{kn}(t^2 + t)^n\ 2^{-gn} \equiv 2^{lm}\hat{r}^m \equiv 0</math> mod <math>2^{\text{max}(kn,\ lm)}</math> hold where <math>t = \check{s}</math> and maximal <math>g \in \mathbb{N}^{*}</math>, from which <math>kn = lm</math> due to <math>(t^2 + t)2^{-gn}, \hat{r}^m \in \mathbb{N} \setminus 2\mathbb{N}</math> and even <math>m = n</math> is implied contradicting the premise. Both <math>(\hat{s} + 1)^3 - 1 = \hat{r}^2</math> are left, which can be refuted by the [[w:Integer factorization|<span class="wikipedia"> | + | '''Indirect proof:''' Assuming <math>\acute{x}, y \in {}^{\omega}2\mathbb{N}</math> yields the contradiction <math>1 + x^m \equiv 2 \equiv y^n \equiv 0</math> resp. <math>4</math> mod <math>8</math>. For <math>p \mid y,\ x = \hat{r},\ y = \hat{s} + 1</math> and <math>2 < m \ne n > 2</math>, subsequently <math>(\hat{s} + 1)^n - 1 \equiv \hat{r}^m \equiv 0</math> mod <math>p \in {}^{\omega}\mathbb{P}_{\ge 3}</math> is assumed. Then also <math>(\hat{s} + 1)^{\hat{n}} - 1 = (\hat{s} + 2)^n\ \hat{s}^n \equiv \hat{r}^m \equiv 0</math> mod <math>p</math> and <math>2^{kn}(t^2 + t)^n\ 2^{-gn} \equiv 2^{lm}\hat{r}^m \equiv 0</math> mod <math>2^{\text{max}(kn,\ lm)}</math> hold where <math>t = \check{s}</math> and maximal <math>g \in \mathbb{N}^{*}</math>, from which <math>kn = lm</math> due to <math>(t^2 + t)2^{-gn}, \hat{r}^m \in \mathbb{N} \setminus 2\mathbb{N}</math> and even <math>m = n</math> is implied contradicting the premise. Both <math>(\hat{s} + 1)^3 - 1 = \hat{r}^2</math> are left, which can be refuted by the [[w:Integer factorization|<span class="wikipedia">integer factorisation</span>]] of <math>r</math> and <math>s</math>, and <math>(\hat{s} + 1)^2 - 1 = 4s(s + 1) = \hat{r}^3</math>, which results in <math>r = s = 1</math> and thus leads to the solution specified.<math>\square</math> |
== See also == | == See also == | ||
Revision as of 08:30, 10 August 2024
For [math]\displaystyle{ 1 + x^m = y^n }[/math] where [math]\displaystyle{ m, n, x, y \in {}^{\omega}\mathbb{N}_{\ge 2} }[/math], only the solution [math]\displaystyle{ m_0 = 3 }[/math], [math]\displaystyle{ n_0 = 2 }[/math], [math]\displaystyle{ x_0 = 2 }[/math] and [math]\displaystyle{ y_0 = 3 }[/math] exists.
Indirect proof: Assuming [math]\displaystyle{ \acute{x}, y \in {}^{\omega}2\mathbb{N} }[/math] yields the contradiction [math]\displaystyle{ 1 + x^m \equiv 2 \equiv y^n \equiv 0 }[/math] resp. [math]\displaystyle{ 4 }[/math] mod [math]\displaystyle{ 8 }[/math]. For [math]\displaystyle{ p \mid y,\ x = \hat{r},\ y = \hat{s} + 1 }[/math] and [math]\displaystyle{ 2 \lt m \ne n \gt 2 }[/math], subsequently [math]\displaystyle{ (\hat{s} + 1)^n - 1 \equiv \hat{r}^m \equiv 0 }[/math] mod [math]\displaystyle{ p \in {}^{\omega}\mathbb{P}_{\ge 3} }[/math] is assumed. Then also [math]\displaystyle{ (\hat{s} + 1)^{\hat{n}} - 1 = (\hat{s} + 2)^n\ \hat{s}^n \equiv \hat{r}^m \equiv 0 }[/math] mod [math]\displaystyle{ p }[/math] and [math]\displaystyle{ 2^{kn}(t^2 + t)^n\ 2^{-gn} \equiv 2^{lm}\hat{r}^m \equiv 0 }[/math] mod [math]\displaystyle{ 2^{\text{max}(kn,\ lm)} }[/math] hold where [math]\displaystyle{ t = \check{s} }[/math] and maximal [math]\displaystyle{ g \in \mathbb{N}^{*} }[/math], from which [math]\displaystyle{ kn = lm }[/math] due to [math]\displaystyle{ (t^2 + t)2^{-gn}, \hat{r}^m \in \mathbb{N} \setminus 2\mathbb{N} }[/math] and even [math]\displaystyle{ m = n }[/math] is implied contradicting the premise. Both [math]\displaystyle{ (\hat{s} + 1)^3 - 1 = \hat{r}^2 }[/math] are left, which can be refuted by the integer factorisation of [math]\displaystyle{ r }[/math] and [math]\displaystyle{ s }[/math], and [math]\displaystyle{ (\hat{s} + 1)^2 - 1 = 4s(s + 1) = \hat{r}^3 }[/math], which results in [math]\displaystyle{ r = s = 1 }[/math] and thus leads to the solution specified.[math]\displaystyle{ \square }[/math]