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Let ${\textstyle N}$ be odd

• first item
• second item

and ${\textstyle M}$ even.

${\displaystyle {\text{ R J M}}}$ ${\displaystyle {}^{\wedge }}$ ${\displaystyle {\tilde {\ }}{\tilde {\phantom {a}}}}$

${\displaystyle {\boldsymbol {\alpha \beta \gamma \delta \epsilon \zeta \eta \theta }}}$

${\displaystyle {\boldsymbol {\alpha }}}$

${\displaystyle (\nabla \times \mathbf {F} )\cdots {\mathrm {d} }\mathbf {S} }$ = ${\displaystyle \oint _{\partial S}\mathbf {F} }$ \cdot${\displaystyle {\mathrm {d} }{\boldsymbol {\ell }}}$

${\displaystyle a\operatorname {snap} b}$

${\displaystyle \displaystyle \sum _{n\in \mathbb {Z} }}$ ${\displaystyle \displaystyle \sum _{n\in \mathbb {Z} }}$

SVG: ${\displaystyle \displaystyle \sum _{n\in \mathbb {Z} }}$ MathML: ${\displaystyle {\displaystyle \sum _{n\in \mathbb{\mathbb{Z}}}}}$

\mathcal{A} ${\displaystyle {\mathcal {A}}}$ ${\displaystyle {\mathcal {A}}}$ ${\displaystyle {𝒜}}$

${\displaystyle {\mathcal {ABCDEFGHIJKLMNOPQRSTUVWXYZ}}}$ ${\displaystyle {\mathcal {abcdefghijklmnopqrstuvwxyz}}}$

${\displaystyle B^{b}|\psi \rangle }$ ${\displaystyle B^{b}|\psi \rangle }$

${\displaystyle B^{b}|\psi )}$

${\displaystyle |\psi \rangle }$

${\displaystyle |B^{b}\rangle }$

${\displaystyle B^{b}\left|\psi \right\rangle }$

${\displaystyle B^{b}\left|\psi \right\rangle }$

${\displaystyle B^{b}|\psi \rbrace }$

${\displaystyle B^{b}|\psi ]}$

${\displaystyle p(a|b)}$

${\displaystyle p(a|b)}$

${\displaystyle \textstyle \max _{B_{y}^{b}|\psi \rangle }}$

${\textstyle \max _{B_{y}^{b}|\psi \rangle }}$

${\displaystyle f:X\to Y}$ SVG ${\displaystyle f:X\to Y}$

${\displaystyle A\backslash B}$ SVG ${\displaystyle A\backslash B}$

${\displaystyle A\backslash B}$

${\displaystyle A\angle B\sphericalangle C\measuredangle D}$

${\displaystyle A\Box B\square C\blacksquare D\Diamond E\lozenge F\blacklozenge G\bigstar H}$

${\displaystyle A\triangle B\triangledown C\blacktriangle D\blacktriangledown E}$

${\displaystyle A\forall B\exists C\nexists D\And E}$

${\displaystyle A\lnot B\neg C\not \operatorname {R} D\bot E\top F}$

${\displaystyle A\S B}$

${\displaystyle A\diamondsuit B\heartsuit C\clubsuit D\spadesuit E\Game F\flat G\natural H\sharp I}$

${\displaystyle A\diagup B\diagdown C}$

PNG ${\displaystyle A\operatorname {foo} B}$ ${\displaystyle A\operatorname {foo} B}$ PNG ${\displaystyle A\operatorname {foo} B}$

${\displaystyle \operatorname {G} (V,g)}$ SVG ${\displaystyle \operatorname {G} (V,g)}$

${\displaystyle \|A\|,D(p\|q)}$ SVG ${\displaystyle \|A\|,D(p\|q)}$

${\displaystyle q(v)=\|v\|^{2}}$ SVG ${\displaystyle q(v)=\|v\|^{2}}$

${\displaystyle q(v)=|v|^{2}}$ SVG ${\displaystyle q(v)=|v|^{2}}$

<math>q(v)=|v|^2</math> MathML/MathJax ${\displaystyle q(v)=|v|^{2}}$ SVG ${\displaystyle q(v)=|v|^{2}}$

<math>q(v)=\|v\|^2</math> MathML/MathJax ${\displaystyle q(v)=\|v\|^{2}}$ SVG ${\displaystyle q(v)=\|v\|^{2}}$

<math>q(v)=\|v\|_A</math> MathML/MathJax ${\displaystyle q(v)=\|v\|_{A}}$ SVG ${\displaystyle q(v)=\|v\|_{A}}$

<math>x^2</math> MathML/MathJax ${\displaystyle x^{2}}$ SVG ${\displaystyle x^{2}}$

<math>(v)^2</math> MathML/MathJax ${\displaystyle (v)^{2}}$ SVG ${\displaystyle (v)^{2}}$

SVG: ${\displaystyle {\frac {x}{x}}\quad {\frac {x}{y}}\quad {\frac {x}{l}}\quad {\frac {x}{2}}}$ MathML: ${\displaystyle \frac{x}{x}\frac{x}{y}\frac{x}{l}\frac{x}{2}}$

SVG: ${\displaystyle {\frac {x}{x}}\quad {\frac {y}{x}}\quad {\frac {l}{x}}\quad {\frac {2}{x}}}$ MathML: ${\displaystyle \frac{x}{x}\frac{y}{x}\frac{l}{x}\frac{2}{x}}$

<math>\cancel{y}</math> MathML/MathJax ${\displaystyle {\cancel {y}}}$ SVG ${\displaystyle {\cancel {y}}}$

<math>\cancel{x}</math> MathML/MathJax ${\displaystyle {\cancel {x}}}$ SVG ${\displaystyle {\cancel {x}}}$

<math>\cancel{xyz}</math> MathML/MathJax ${\displaystyle {\cancel {xyz}}}$ SVG ${\displaystyle {\cancel {xyz}}}$

let ${\textstyle N}$ be odd

1. foo
2. bar

and ${\textstyle M}$ even.

Pick a random number ${\textstyle 1<a<N}$.|Compute ${\displaystyle K=\gcd(a,N)}$, the greatest common divisor of ${\displaystyle a}$ and ${\displaystyle N}$.|If ${\displaystyle K\neq 1}$, then ${\displaystyle K}$ is a nontrivial factor of ${\displaystyle N}$, with the other factor being ${\textstyle {\frac {N}{K}}}$ and we are done.|Otherwise, use the quantum subroutine to find the order ${\displaystyle r}$ of ${\displaystyle a}$.|If ${\displaystyle r}$ is odd, then go back to step 1.|Compute ${\displaystyle g=\gcd(N,a^{r/2}+1)}$. If ${\displaystyle g}$ is nontrivial, the other factor is ${\textstyle {\frac {N}{g}}}$, and we're done. Otherwise, go back to step 1. }}It has been shown that this will be likely to succeed after a few runs. In practice, a single call to the quantum order-finding subroutine is enough to completely factor ${\displaystyle N}$ with very high probability of success if one uses a more advanced reduction.

1. Cite error: The named reference siam was invoked but never defined (see the help page).
2. Ekerå, Martin (June 2021). "On completely factoring any integer efficiently in a single run of an order-finding algorithm". Quantum Information Processing. 20 (6): 205. arXiv:2007.10044. Bibcode:2021QuIP...20..205E. doi:10.1007/s11128-021-03069-1.