Lagrange Four Square Theorem — Examples

Take the representations $7=2^2+1^2+1^2+1^2$ and $15=3^2+2^2+1^2+1^2$, so $x=(2,1,1,1)$ and $y=(3,2,1,1)$. The four bilinear forms of Euler's four-square product identity evaluate to

$z_1=10,z_2=1,z_3=-2,z_4=0,$

and ${10}^2+1^2+(-2{)}^2+0^2=100+1+4+0=105=7\cdot 15$. Every coordinate is computed below, the vanishing one included: $z_4=0$ is a value the formula returns and not a coordinate that has been left out.

A negative coordinate is not a defect. The third coordinate is $-2$, and only its square enters the sum, so $(10,1,-2,0)$ represents the same integer as $(10,1,2,0)$ and the two representations are equivalent up to signs. They are distinct ordered quadruples. The formulas are not arranged to produce nonnegative outputs, and no step of the identity or of the descent needs them to be.

Let $p=11$ and work with the representatives $0,1,\dots ,10$. Put

$Q=\{r:0\le r\le 10\text{ and }r\equiv y^2(\mathrm{m}\mathrm{o}\mathrm{d}11)\text{ for some }y\in \mathbb{Z}\},R=\{r:0\le r\le 10\text{ and }r\equiv -1-y^2(\mathrm{m}\mathrm{o}\mathrm{d}11)\text{ for some }y\in \mathbb{Z}\}\mathrm{.}$

Then $Q=\{0,1,3,4,5,9\}$ and $R=\{1,5,6,7,9,10\}$, each with $6=(11+1)\mathrm{/}2$ elements, and $Q\cap R=\{1,5,9\}$. Taking the common value $1$, which is $1^2$ and is also congruent to $-1-3^2$, gives

$1^2+3^2+1=11\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}11)\mathrm{.}$

This is the mechanism behind Every nonzero residue modulo an odd prime is a sum of two squares at $a=-1$, written out at $p=11$: two subsets of the $11$ residues, each of size $6$, cannot be disjoint. That proposition supplies such a common value for every odd prime, and For every prime $p$ the congruence $x^2+y^2+1\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}p)$ is solvable cites it rather than repeating the count.

Why $11$ is a good modulus for this. Here $11\equiv 3(\mathrm{m}\mathrm{o}\mathrm{d}4)$ and $-1\equiv 10$ is not in $Q$, so $-1$ is not itself a square modulo $11$ and no pair with $y=0$ works. The example is therefore not a disguised one-square case: two squares are genuinely needed, which is what the cited proposition provides.

Other solutions exist. The intersection has the further common values $5$ and $9$: the value $5$ is $4^2$ and also $-1-4^2$ modulo $11$, giving $4^2+4^2+1=33=3\cdot 11$, and the value $9$ is $3^2$ and also $-1-1^2$, giving $3^2+1^2+1=11$. So the solution pair is not unique; the cited proposition asserts that at least one common value exists and says nothing about how many do.

Write $r_m(a)$ for the least absolute remainder of $a$ modulo $m$ (The least absolute remainder modulo a positive integer). For $a=0,1,2,3,4,5,6,7$:

$r_7(a):0,1,2,3,-3,-2,-1,0,$

$r_8(a):0,1,2,3,4,-3,-2,-1.$

Modulo the odd modulus $7$ every entry satisfies $4r^2<49$ strictly, the largest value of $4r^2$ being $36$. Modulo the even modulus $8$ the entry $r_8(4)=4$ attains $2r=m$ and hence $4r^2=64=8^2$; here $4$ and $-4$ are congruent modulo $8$, and the normalisation $-m<2r\le m$ is what selects $4$ rather than $-4$.

Only an even modulus produces the tie. The equality $2r=m$ requires $m$ to be twice an integer, so for odd $m$ the bound $4r^2\le m^2$ is automatically strict, as the first table shows. This is exactly the boundary that The centred residue quadruple of $pm=a^2+b^2+c^2+d^2$ has norm $mn$ with $1\le n<m$ has to exclude by a separate argument: the norm estimate there gives only $n\le m$, and equality in all four centred coordinates at once is exactly the case $n=m$ that a separate argument must rule out.

Start from the prime $p=13$ and the multiplier $m=11$, with

$143=13\cdot 11={11}^2+3^2+3^2+2^2\mathrm{.}$

Centring the coordinates modulo $11$ gives $(0,3,3,2)$, whose norm is $22=11\cdot 2$, so the multiplier drops to $n=2$. Euler's identity applied to $(11,3,3,2)$ and $(0,3,3,2)$ gives $(22,33,33,22)$, each coordinate divisible by $11$; dividing by $11$ gives

$26=13\cdot 2=2^2+3^2+3^2+2^2\mathrm{.}$

Repeating with $m=2$: the centred quadruple is $(0,1,1,0)$, of norm $2=2\cdot 1$, so $n=1$; the identity applied to $(2,3,3,2)$ and $(0,1,1,0)$ gives $(6,4,0,0)$, and dividing by $2$ gives

$13=3^2+2^2+0^2+0^2\mathrm{.}$

The two steps use an odd modulus and then an even one, so both parities occur.

The even step is where the tie could have bitten. At $m=2$ the centred coordinates $1$ attain $2r=m$, so the estimate $4r^2\le m^2$ is an equality in two of the four coordinates. That is admissible: what The centred residue quadruple of $pm=a^2+b^2+c^2+d^2$ has norm $mn$ with $1\le n<m$ excludes is equality in all four at once, and here two coordinates centre to $0$.

The starting representation is not the minimal one. The descent does not require the multiplier to come from Some multiple $pm$ with $1\le m<p$ is a sum of four squares; any $m$ with $1<m<p$ for which $pm$ is a sum of four squares will do, and $m=11$ was taken to make two steps rather than one.

The integer $65$ has the four-square representations

$65=1^2+8^2+0^2+0^2=4^2+7^2+0^2+0^2=2^2+5^2+6^2+0^2=2^2+3^2+4^2+6^2\mathrm{.}$

Their multisets of absolute values are $\{8,1,0,0\}$, $\{7,4,0,0\}$, $\{6,5,2,0\}$ and $\{6,4,3,2\}$, which are pairwise distinct, so no two of the four are obtained from one another by permuting coordinates or changing signs: they are essentially different in the sense of Representations as sums of four squares.

Existence and multiplicity are different questions. Lagrange's four-square theorem: every nonnegative integer is a sum of four integer squares asserts that a representation exists. How many essentially different representations a given integer has is not settled by it, and nothing above computes that number for $65$: the four displayed are exhibited, not claimed to be all.

False claim: for every prime $p$ (Prime and composite integers: $p$ is prime when $p>1$ and its only positive divisors are $1$ and $p$), any two representations of $p$ as a sum of four integer squares (Representations as sums of four squares) are equivalent up to signs and order; that is, a prime has exactly one four-square representation up to permuting the coordinates and changing their signs.

Where the claim comes from. For two squares the corresponding statement is true: A prime congruent to $1$ modulo $4$ has one two-square representation up to signs and order says that a prime congruent to $1$ modulo $4$ has one representation as a sum of two squares up to signs and order. The claim refuted above is that statement with two coordinates replaced by four.

What survives of the analogy. Nothing of the two-square proof transfers. It turns on such a prime having a primitive two-square representation whose factorisation is controlled, and with four coordinates the extra room admits genuinely different multisets of absolute values; $31$ is one witness and not a special one.

Factor $105=3\cdot 5\cdot 7$ and take the prime representations

$3=1^2+1^2+1^2+0^2,5=2^2+1^2+0^2+0^2,7=2^2+1^2+1^2+1^2\mathrm{.}$

Multiplying the first two with the identity of Euler's four-square product identity, applied to $(1,1,1,0)$ and $(2,1,0,0)$, gives $(3,-1,-2,1)$ and

$15=3^2+(-1{)}^2+(-2{)}^2+1^2\mathrm{.}$

Multiplying that result by the third, applied to $(3,-1,-2,1)$ and $(2,1,1,1)$, gives $(4,8,5,0)$ and

$105=4^2+8^2+5^2+0^2\mathrm{.}$

This is the route the proof of Lagrange's four-square theorem: every nonnegative integer is a sum of four integer squares takes: prime factors first, then closure under products.

Negative intermediate coordinates are harmless. The quadruple produced at the first multiplication has two negative entries, and they are carried into the second multiplication unchanged. Only squares are read off at the end, and the identity of Euler's four-square product identity holds for negative inputs as it does for positive ones.

The order of multiplication is a choice. Multiplying $5$ and $7$ first, or $3$ and $7$ first, produces different quadruples for the same $105$; the construction is not canonical, and Sums of four squares are closed under products claims existence rather than a preferred witness.

False claim: every nonnegative integer is a sum of three integer squares, that is, for every $n\ge 0$ there are integers $x,y,z$ with $n=x^2+y^2+z^2$.

The integers $7$ and $28$ refute it. Both are excluded by Positive integers $4^{a}m$ with $m\equiv 7(\mathrm{m}\mathrm{o}\mathrm{d}8)$ are not sums of three integer squares, as $7=4^0\cdot 7$ and $28=4^1\cdot 7$ with $7\equiv 7(\mathrm{m}\mathrm{o}\mathrm{d}8)$; and for $7$ the exclusion can also be seen by a finite search, which is carried out below so that the witness does not rest on the obstruction alone.

Two independent routes for $7$. Step 1.1 argues by residues modulo $8$ and step 1.2 by exhausting the finitely many candidates. The second uses no lemma about squares at all, so the non-existence claim does not depend on the congruence argument being right.

Why $28$ needs the halving step. The residue of $28$ modulo $8$ is $4$, which is attainable by a sum of three squares, so the congruence argument does not apply to $28$ directly. It is the divisibility by $4$, and the fact that this forces all three coordinates even, that reduces $28$ to $7$.

False claim: if two nonnegative integers are each a sum of three integer squares, then so is their product.

The pair $3$, $5$ refutes it: $3=1^2+1^2+1^2$ and $5=2^2+1^2+0^2$, while $15=3\cdot 5$ satisfies $15\equiv 7(\mathrm{m}\mathrm{o}\mathrm{d}8)$ and is therefore excluded by Positive integers $4^{a}m$ with $m\equiv 7(\mathrm{m}\mathrm{o}\mathrm{d}8)$ are not sums of three integer squares.