Lagrange Four Square Theorem

Let $n$ be a nonnegative integer. A representation of $n$ as a sum of four squares is an ordered quadruple $(a,b,c,d)\in {\mathbb{Z}}^4$ with $n=a^2+b^2+c^2+d^2$. The integer $n$ is a sum of four integer squares when at least one such quadruple exists.

Two representations $(a,b,c,d)$ and $(a^{\mathrm{\prime }},b^{\mathrm{\prime }},c^{\mathrm{\prime }},d^{\mathrm{\prime }})$ of the same $n$ are equivalent up to signs and order when one is obtained from the other by permuting the four coordinates and independently changing the sign of any of them, and essentially different when they are not so equivalent. Since changing a sign does not change an absolute value and permuting the coordinates permutes them, two representations are equivalent exactly when the multisets $\{\mathrm{\mid }a\mathrm{\mid },\mathrm{\mid }b\mathrm{\mid },\mathrm{\mid }c\mathrm{\mid },\mathrm{\mid }d\mathrm{\mid }\}$ and $\{\mathrm{\mid }{a}^{\mathrm{\prime }}\mathrm{\mid },\mathrm{\mid }{b}^{\mathrm{\prime }}\mathrm{\mid },\mathrm{\mid }{c}^{\mathrm{\prime }}\mathrm{\mid },\mathrm{\mid }{d}^{\mathrm{\prime }}\mathrm{\mid }\}$ coincide.

The coordinates range over all of $\mathbb{Z}$. They may be zero, they may be negative, and they may repeat. So an integer written as a sum of one, two or three squares is a sum of four squares as well, its remaining coordinates being $0$; and $-1$ is an admissible coordinate wherever $1$ is, with the same square. This matters where the identity of Euler's four-square product identity is applied, since its output coordinates are frequently negative or zero even when its inputs are not.

Only nonnegative $n$ is defined. A square is nonnegative and a sum of four nonnegative integers is nonnegative, so no negative integer admits a quadruple and the definition would be vacuous there.

The value $0$. A sum of four squares vanishes exactly when each square vanishes, so $0=0^2+0^2+0^2+0^2$ and $(0,0,0,0)$ is the only representation of $0$. Every representation of a positive integer therefore has at least one nonzero coordinate.

Agreement with the two-square convention. The convention here is the one Representations and primitive representations as sums of two squares uses for pairs: a representation is an ordered tuple, so order and sign are recorded, and equivalence up to signs and order is what quotients them out. The only change is the number of coordinates. That every nonnegative integer has at least one representation is Lagrange's four-square theorem: every nonnegative integer is a sum of four integer squares; the definition itself asserts nothing about existence.

Let $x_1,x_2,x_3,x_4$ and $y_1,y_2,y_3,y_4$ be integers, and set

$\begin{array}{rlrl}{\displaystyle z_1}& {\displaystyle =x_1{y}_1+x_2{y}_2+x_3{y}_3+x_4{y}_4,}& {\displaystyle z_2}& {\displaystyle =x_1{y}_2-x_2{y}_1-x_3{y}_4+x_4{y}_3,}\\ {\displaystyle z_3}& {\displaystyle =x_1{y}_3+x_2{y}_4-x_3{y}_1-x_4{y}_2,}& {\displaystyle z_4}& {\displaystyle =x_1{y}_4-x_2{y}_3+x_3{y}_2-x_4{y}_1\mathrm{.}}\end{array}$

Then

$(x_1^2+x_2^2+x_3^2+x_4^2)(y_1^2+y_2^2+y_3^2+y_4^2)=z_1^2+z_2^2+z_3^2+z_4^2\mathrm{.}$

Let $m$ and $n$ be nonnegative integers. If each of $m$ and $n$ is a sum of four integer squares (Representations as sums of four squares), then $mn$ is a sum of four integer squares.

What the identity does and does not give. The four coordinates $z_i$ are determined by the two chosen quadruples, so a different choice of representation of $m$ or of $n$ generally produces a different representation of $mn$. The statement asserts existence only; it makes no claim about how many representations $mn$ has, nor that every representation of $mn$ arises this way.

Why the closure is needed. Reducing Lagrange's four-square theorem: every nonnegative integer is a sum of four integer squares to Every prime is a sum of four integer squares requires exactly this: a factorisation of $n$ into primes is useless unless the property being proved is inherited by products.

Let $p$ be a prime (Prime and composite integers: $p$ is prime when $p>1$ and its only positive divisors are $1$ and $p$). Then there are integers $x,y$ with $x^2+y^2+1\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}p)$, the congruence being that of Congruence modulo an integer: $a\equiv b(\mathrm{m}\mathrm{o}\mathrm{d}n)$ when $n\mid (a-b)$, including the moduli $0$ and $1$.

Where the odd case comes from. For odd $p$ the work is done by Every nonzero residue modulo an odd prime is a sum of two squares at $a=-1$: the set $Q$ of square classes modulo $p$, the zero class included, and the set of classes $-1-z$ for $z\in Q$ are two subsets of $\mathbb{Z}\mathrm{/}p$ with $(p+1)\mathrm{/}2$ elements each, so they meet, and a common value gives $x^2\equiv -1-y^2$. The hypothesis $p\nmid a$ of that proposition is what step 1.3 discharges, by an argument that does not use oddness; oddness is needed only to make the proposition applicable at all.

Why $p=2$ is separate. The cited proposition is stated for odd primes, so it says nothing at $p=2$; the pair $(1,0)$ settles that case by computation rather than by weakening the proposition's hypothesis.

Let $m$ be an integer with $m\ge 1$ and let $a\in \mathbb{Z}$. Then there is exactly one integer $r$ with

$a\equiv r(\mathrm{m}\mathrm{o}\mathrm{d}m)\text{and}-m<2r\le m,$

that is, exactly one integer $r$ satisfying $a\equiv r(\mathrm{m}\mathrm{o}\mathrm{d}m)$ and $-m<2r\le m$, and consequently $4r^2\le m^2$. Call this $r$ the least absolute remainder of $a$ modulo $m$.

Where the tie falls. The normalisation is deliberately half-open on the right: $2r=m$ is permitted and $2r=-m$ is not. The case $\mathrm{\mid }2r\mathrm{\mid }=m$ can arise only for even $m$, where $r_0$ may equal the integer $t$ with $m=2t$; then $t$ and $t-m$ are congruent modulo $m$ with $\mathrm{\mid }t\mathrm{\mid }=\mathrm{\mid }t-m\mathrm{\mid }$, and the convention keeps $t$. Without the half-open choice both values would satisfy $\mathrm{\mid }2r\mathrm{\mid }\le m$ and the uniqueness clause would be false as stated.

The bound is stated in integers. Writing $4r^2\le m^2$ rather than $\mathrm{\mid }r\mathrm{\mid }\le m\mathrm{/}2$ avoids introducing a quotient that need not be an integer, and it is the form the estimates in Some multiple $pm$ with $1\le m<p$ is a sum of four squares and The centred residue quadruple of $pm=a^2+b^2+c^2+d^2$ has norm $mn$ with $1\le n<m$ use. The inequality $4r^2\le m^2$ is an equality exactly when $2r=m$.

Let $p$ be a prime (Prime and composite integers: $p$ is prime when $p>1$ and its only positive divisors are $1$ and $p$). Then there is an integer $m$ with $1\le m<p$ for which $pm$ is a sum of four integer squares (Representations as sums of four squares).

Why the centring is needed. The pair $(u,v)$ produced by [L1] is subject to no size condition, so $u^2+v^2+1$ can be an arbitrarily large multiple of $p$. Replacing $u,v$ by their least absolute remainders leaves the congruence class untouched and buys the two bounds $4x^2\le p^2$ and $4y^2\le p^2$, which is what forces the multiplier below $p$.

The bound is not tight, and does not need to be. What the coordinate bounds give at step 5.1 is $4pm\le 2p^2+4$, and only the weaker $4pm<4p^2$ is used. The comparison $2p^2+4<4p^2$ needs just $p^2>2$, which every prime satisfies, so $p=2$ needs no separate treatment even though the coordinate bound $4x^2\le p^2$ can be attained there, at $x=\pm 1$.

The case $m=1$. Nothing excludes it, and it is the case $p=x^2+y^2+1$ in which the prime is already a sum of four squares.

Let $p$ be a prime (Prime and composite integers: $p$ is prime when $p>1$ and its only positive divisors are $1$ and $p$), let $m$ be an integer with $1<m<p$, and let $a,b,c,d$ be integers with $pm=a^2+b^2+c^2+d^2$. Write $a^{\mathrm{\prime }},b^{\mathrm{\prime }},c^{\mathrm{\prime }},d^{\mathrm{\prime }}$ for the least absolute remainders of $a,b,c,d$ modulo $m$ (The least absolute remainder modulo a positive integer). Then there is an integer $n$ with $1\le n<m$ and

$a^{\mathrm{\prime }2}+b^{\mathrm{\prime }2}+c^{\mathrm{\prime }2}+d^{\mathrm{\prime }2}=mn\mathrm{.}$