numtheory/sq2factor(deprecated) - Help

numtheory

 sq2factor
 integer factorization in Z(sqrt(2))

 Calling Sequence sq2factor(z)

Parameters

 z - integer, list or set of integers in $Z\left(\sqrt{2}\right)$

Description

 • Important: The numtheory[sq2factor] command has been deprecated.  Use the superseding command NumberTheory[FactorNormEuclidean] instead.
 • The sq2factor function returns the integer factorization of z.
 • All integers of $Z\left(\sqrt{2}\right)$ have the form $a+b\sqrt{2}$, where a and b are rational integers.
 • The answer is in the form: $±1u{\left({f}_{1}\right)}^{{e}_{1}}\mathrm{...}{\left({f}_{n}\right)}^{{e}_{n}}$ such that $z=±1\cdot u{f}_{1}^{{e}_{1}}\dots {f}_{n}^{{e}_{n}}$ where ${f}_{1},\dots ,{f}_{n}$ are distinct prime factors of z, ${e}_{1},\dots ,{e}_{n}$ are non-negative integer numbers, $u$ is a unit in $Z\left(\sqrt{2}\right)$ and is represented under the form ${w}^{n}$ or ${\stackrel{&conjugate0;}{w}}^{n}$ or ${\left(-w\right)}^{n}$ or ${\left(-\stackrel{&conjugate0;}{w}\right)}^{n}$ where $w$ is the fundamental unit (i.e, $w=1+\sqrt{2}$), and $n$ is a non-negative integer.
 • The expand function may be applied to cause the factors to be multiplied together again.
 • The command with(numtheory,sq2factor) allows the use of the abbreviated form of this command.

Examples

 > $\mathrm{with}\left(\mathrm{numtheory}\right):$
 > $\mathrm{sq2factor}\left({\left(1-\sqrt{2}\right)}^{-4}\right)$
 ${\left({1}{+}\sqrt{{2}}\right)}^{{4}}$ (1)
 > $\mathrm{sq2factor}\left(83424959\right)$
 ${}\left({9503}{+}{1855}{}\sqrt{{2}}\right){}{}\left({9503}{-}{1855}{}\sqrt{{2}}\right)$ (2)
 > $\mathrm{expand}\left(\right)$
 ${83424959}$ (3)
 > $\mathrm{sq2factor}\left(9232-932\sqrt{2}\right)$
 ${{}\left(\sqrt{{2}}\right)}^{{5}}{}\left({-}{1}{+}\sqrt{{2}}\right){}{}\left({1}{+}{3}{}\sqrt{{2}}\right){}{}\left({5}{+}\sqrt{{2}}\right){}{}\left({17}{+}{59}{}\sqrt{{2}}\right)$ (4)
 > $\mathrm{expand}\left(\right)$
 ${9232}{-}{932}{}\sqrt{{2}}$ (5)