Lagmental Vicfred

The Supplementary Laws Handle Minus One and Two by Vicfred

tags: mathematics, quadratic residues, quadratic reciprocity, sums of squares

Two compact parity formulas complete the practical reciprocity algorithm. I will separate the object being defined from the consequence being claimed.

The data

For an odd prime \(p\), the Legendre symbol \(\left(\frac ap\right)\) records whether \(a\) is a nonzero square modulo \(p\). Reciprocity exchanges numerator and denominator up to a sign.

$$ \left(\frac{-1}{p}\right)=(-1)^{(p-1)/2} $$

There are two layers here: the object \(\mathsf D\) and the law \(\mathsf C\). Writing them separately makes the direction of \(\Longrightarrow\) visible and keeps an accidental converse from slipping in.

$$ \left(\frac2p\right)=(-1)^{(p^2-1)/8} $$

Derivation

The middle display is intentionally dense: it is where signs, bounds, multiplicities, or normalising factors are most likely to be lost.

$$ \begin{array}{c|cccc}p\bmod8&1&3&5&7\\\hline\left(\frac2p\right)&1&-1&-1&1\end{array} $$

Invariant content

The two-row display is also a debugging tool: if the conclusion changes when only notation changes, some hidden choice has entered the argument.

$$ \begin{aligned} \mathsf{D}\;&:\quad \left(\frac{-1}{p}\right)=(-1)^{(p-1)/2},\\[5pt] \mathsf{C}\;&:\quad \left(\frac2p\right)=(-1)^{(p^2-1)/8}. \end{aligned} $$

Scope

The symbol is defined modulo an odd prime and is not ordinary division. Composite odd denominators require the Jacobi symbol, which can equal one without certifying a square.

$$ \boxed{\begin{gathered} \text{compact conclusion}\\[-2pt] \left(\frac2p\right)=(-1)^{(p^2-1)/8} \end{gathered}} $$

The notation is dense, but it is doing honest work: every delimiter records scope and every index records dependence. Removing one should require a mathematical reason.

This article was posted on Tue 04 April 2023. Facts and circumstances may have changed since publication.
Please contact me before jumping to conclusions if something seems wrong or unclear.