Important: The numtheory package 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: $\pm 1u{\left(f_1\right)}^{e_1}\mathrm{...}{\left(f_{\mathrm{n}}\right)}^{e_{\mathrm{n}}}$ such that $z=\pm 1\cdot u{f}_1^{e_1}\dots f_n^{e_n}$ 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.

$\mathrm{with}\left(\mathrm{numtheory}\right)\:$

$\mathrm{sq2factor}\left({\left(1-\mathrm{sqrt}\left(2\right)\right)}^{-4}\right)$

${\left(1+\sqrt{2}\right)}^4$

$\mathrm{sq2factor}\left(83424959\right)$

$\left(9503+1855\sqrt{2}\right)\left(9503-1855\sqrt{2}\right)$

$\mathrm{expand}\left(\right)$

$83424959$

$\mathrm{sq2factor}\left(9232-932\mathrm{sqrt}\left(2\right)\right)$

${\left(\sqrt{2}\right)}^5\left(\sqrt{2}-1\right)\left(1+3\sqrt{2}\right)\left(5+\sqrt{2}\right)\left(17+59\sqrt{2}\right)$

$\mathrm{expand}\left(\right)$

$9232-932\sqrt{2}$