Refinements of Lagrange's four-square theorem

TL;DR

This paper characterizes the possible sums of the four integers in Lagrange's four-square representations of nonnegative integers, providing new insights into the structure of these representations.

Contribution

It offers a detailed characterization of the sums of the four squares in all representations of integers, extending understanding of Lagrange's theorem.

Findings

Identifies all possible sums of the four squares for each integer n

Provides explicit criteria for the sums in the representations

Enhances understanding of the structure of four-square representations

Abstract

A well-known theorem of Lagrange asserts that every nonnegative integer $n$ can be written in the form $a^2+b^2+c^2+d^2$, where $a,b,c,d \in \mathbb{Z}$. We characterize the values assumed by $a+b+c+d$ as we range over all such representations of $n$.

Tables

Table 1. Table 1. Values of T 𝑇 T satisfying ( 2 ) and ( 3 ) but not belonging to 𝒮 ​ ( n ) 𝒮 𝑛 \mathscr{S}(n) .

$n$	exceptional $T$
1	$\varnothing$
2	$\varnothing$
3	$\varnothing$
4	$\varnothing$
5	$\varnothing$
6	$\varnothing$
7	$\varnothing$
8	$\left\{\pm 2\right\}$
9	$\varnothing$
10	$\varnothing$
11	$\varnothing$
12	$\varnothing$
13	$\varnothing$
14	$\varnothing$
15	$\varnothing$
16	$\{\pm 2,\pm 6\}$

Equations

$$n=a^2+b^2+c^2+d^2.$$

$$2017=18^2+21^2+24^2+26^2.$$

$$\mathcal{S}(n)=\{a+b+c+d:a^2+b^2+c^2+d^2=n\}.$$

$${\mathcal{S}}^{+}(n):=\{a+b+c+d:a,b,c,d\in \mathbb{N}\text{ and }{a}^2+b^2+c^2+d^2=n\}.$$

$$T\equiv n\text{ (mod 2)}.$$

$$(a+b+c+d{)}^2$$

$$=4(a^2+b^2+c^2+d^2);$$

$$T^2\le 4n.$$

$$\begin{gathered} (2(a+b)-T{)}^2+(2(a+c)-T{)}^2+(2(b+c)-T{)}^2+T^2 \\ =4a^2+4b^2+4c^2+4(T-a-b-c{)}^2. \end{gathered}$$

$$(2(a+b)-T{)}^2+(2(a+c)-T{)}^2+(2(b+c)-T{)}^2=4n-T^2,$$

$$4n-T^2=A^2+B^2+C^2.$$

$$2a+2b-T=A,2a+2c-T=B,2b+2c-T=C.$$

$$a+b+c+d=T,$$

$$4(a^2+b^2+c^2+$$

$$=(2a+2b-T{)}^2+(2a+2c-T{)}^2+(2b+2c-T{)}^2$$

$$=A^2+B^2+C^2=4n-T^2,$$

$$a^2+b^2+c^2+d^2=n.$$

$$a=\frac{1}{4}(A+B-C+T),b=\frac{1}{4}(A-B+C+T),c=\frac{1}{4}(-A+B+C+T).$$

$$A+B-C+T\equiv A+B+C+T\equiv A^2+B^2+C^2+T^2\equiv 4n\equiv 0\text{ (mod 2)},$$

$$A+B-C+T\equiv 2\text{ (mod 4)}.$$

$$(A+B-$$

$$=A^2+B^2+C^2+T^2+2(AB-AC+AT-BC+BT-CT)$$

$$=4n+2(AB-AC+AT-BC+BT-CT)$$

$$\equiv 0\text{ (mod 8)};$$

$$\begin{gathered} 4n-(2^k{)}^2=4^k(32\ell +27),4n-(2^{k+1}{)}^2=4^{k+1}(8\ell +6), \\ \text{and}4n-(2^{k+2}{)}^2=4^{k+1}(8\ell +3); \end{gathered}$$

$$\mathcal{S}(n)=\{T\equiv n\text{ (mod 2)}:|T|\le 2\sqrt{n}\}.$$

$${\mathcal{S}}^{+}(n)\supseteq \{T\equiv n\text{ (mod 2)}:\sqrt{3n-2}-1\le T\le 2\sqrt{n}\}.$$

$$a^2+b^2+c^2+d^2=n$$

$$\mathcal{S}(n)=2\mathcal{S}(n/4),$$

$$\mathcal{S}(n)=2^{k+1}\mathcal{S}(n/4^{k+1}).$$

$$n\notin \bigcup _{k\ge 1}\{T^2-4^k\text{ (mod 22k+3)}\}.$$

Full text

LG is partially supported by an NSA Young Investigator grant. PP is partially supported by NSF award DMS-1402268.

Our point of departure is the following signature result from a first course in number theory.

[TABLE]

For instance ($n=2017$), we have

[TABLE]

Twenty years before Lagrange’s proof, Euler had already conjectured a refinement of the four-square theorem for odd numbers $n$. The following statement can be found in a letter to Goldbach dated June 9, 1750.

Picking back up our earlier example, when $n=2017$, Euler’s conjecture is satisfied with $a=-18$, $b=21$, $c=24$, and $d=-26$. A proof of Euler’s conjecture was posted to MathOverflow by Franz Lemmermeyer in September 2010.111See https://mathoverflow.net/questions/37278/euler-and-the-four-squares-theorem.

Much more recently, Sun & Sun (apparently unaware of Euler’s conjecture) presented a number of related refinements of Lagrange’s theorem [3] (cf. [4]). One of their many results is that for every $n\in\mathbb{N}$, there is a representation (1) with $a+b+c+d$ a square, as well as one with $a+b+c+d$ a cube [3, Theorem 1.1(a)].

We can unify all the above assertions by introducing the sum spectrum

[TABLE]

Lagrange’s theorem is equivalent to $\mathscr{S}(n)\neq\emptyset$; Euler’s conjecture asserts that ${1\in\mathscr{S}(n)}$ for all odd $n\in\mathbb{N}$; and Sun & Sun’s theorem asserts that $\mathscr{S}(n)$ contains a perfect square and a perfect cube for every $n$. Our goal in this note is to completely describe the set $\mathscr{S}(n)$.

We have not found our results stated anywhere in the literature, but we do not claim they are novel. In the introduction to his resolution [1] of Fermat’s polygonal number conjecture, Cauchy poses the following problem: Décomposer un nombre entier donné en quatre quarrés dont les racines fassent une somme donnée.222Decompose a given whole number into four squares whose roots make a given sum. For Cauchy, “racines” are nonnegative; hence, he is asking for a description of

[TABLE]

Cauchy goes on to prove a partial characterization of $\mathscr{S}^{+}(n)$ (see Remark 1 below for a summary of his results), by essentially the same methods we describe below. Despite being anticipated, we believe an explicit description of $\mathscr{S}(n)$ is sufficiently interesting (and Cauchy’s work on $\mathscr{S}^{+}(n)$ sufficiently underappreciated) to warrant popularization here. Moreover, we will show how our characterization of $\mathscr{S}(n)$ immediately implies both Sun & Sun’s theorems and a generalization of Euler’s conjecture.

We begin by recording two easy observations. First, since an integer and its square have the same parity, every $T\in\mathscr{S}(n)$ satisfies

[TABLE]

Second, for any real numbers $a,b,c,d$, the Cauchy–Schwarz inequality yields

[TABLE]

it follows that every $T\in\mathscr{S}(n)$ satisfies

[TABLE]

As shown in Table 1, the necessary conditions (2) and (3) are quite often (but not always) sufficient for membership in $\mathscr{S}(n)$. The following theorem, which is our main result, tells the full story.

Note that (3) is implied by the condition on $4n-T^{2}$ and so does not need to be included explicitly as a hypothesis in Theorem 2.

To convince the reader that Theorem 2 qualifies as a complete description of $\mathscr{S}(n)$, we recall the following classical result (see the Appendix to Chapter IV of [2] for a proof).

[TABLE]

Thus if $T\in\mathscr{S}(n)$, say with $a^{2}+b^{2}+c^{2}+d^{2}=n$ and $a+b+c+d=T$, then

[TABLE]

so that $4n-T^{2}$ is a sum of three squares.

Conversely, suppose that $4n-T^{2}$ is a sum of three squares, say

[TABLE]

In view of (4), it is enough to show that—after possibly swapping the signs of $A$, $B$, $C$—there are $a,b,c\in\mathbb{Z}$ with

[TABLE]

Indeed, in that case setting $d=T-(a+b+c)$, we have

[TABLE]

and

[TABLE]

so that

[TABLE]

Thus, we focus our attention on (6).

Solving for $a,b,c$ in terms of $A,B,C$ gives

[TABLE]

We claim that $A,B,C$, and $T$ must all have the same parity. To see this, note that (5) gives {A^{2}+B^{2}+C^{2}\equiv-T^{2}{\text{\rm\ (mod~{}4)}}}. If $T$ is odd then {A^{2}+B^{2}+C^{2}\equiv 3{\text{\rm\ (mod~{}4)}}}, and a moment’s thought shows that all of $A,B$, and $C$ must be odd. Similarly, if $T$ is even, then A^{2}+B^{2}+C^{2}\equiv 0{\text{\rm\ (mod~{}4)}}, and this forces $A,B$, and $C$ to all be even. In either case, the difference between any pair of $A$, $B$, and $C$ is even, so the difference between any pair of $a$, $b$, and $c$ is an integer. It follows that if any of $a,b,c\in\mathbb{Z}$, then all three are in $\mathbb{Z}$. Moreover,

[TABLE]

and so the only way we can fail to have $a\in\mathbb{Z}$ (and hence all of $a,b,c\in\mathbb{Z}$) is if

[TABLE]

If $T$ is odd, then $A$ is odd, and so if necessary we can replace $A$ with $-A$ to avoid (7). If $T$ is even, we will show that (7) cannot occur. Indeed, (7) implies that ${8\nmid(A+B-C+T)^{2}}$. But

[TABLE]

here we used that n\equiv T\equiv 0{\text{\rm\ (mod~{}2)}} and that all of $A,B,C$, and $T$ are even. ∎

Let us see how Theorem 2 makes quick work of both the conjecture of Euler and the theorems of Sun & Sun. We begin with the latter. If $4n$ itself is a sum of three squares, then Theorem 2 shows that $T=0\in\mathscr{S}(n)$, and [math] is both a square and a cube. Otherwise, by the Legendre–Gauss theorem, $4n=4^{k+1}(8\ell+7)$, where $k$ and $\ell$ are nonnegative integers. Then

[TABLE]

invoking the Legendre–Gauss theorem once more, we see that all three of these numbers are sums of three squares. By Theorem 2, all of $2^{k},2^{k+1},2^{k+2}$ must belong to $\mathscr{S}(n)$. Clearly, the set $\{2^{k},2^{k+1},2^{k+2}\}$ contains both a square and a cube.

As for Euler’s conjecture, we prove the following generalization (which, for most $n$, gives a very simple description of $\mathscr{S}(n)$):

[TABLE]

[TABLE]

See [1, Corollary I of Theorem I, Theorem IV, and Corollary II of Theorem III].

We conclude this note with a few remarks about the structure of $\mathscr{S}(n)$ for general $n$. When $8\mid n$, it is easy to see that any integer solution to

[TABLE]

has all of $a,b,c,d$ even. Thus, there is a bijection $(a,b,c,d)\leftrightarrow(a/2,b/2,c/2,d/2)$ between representations of $n$ as a sum of four squares and representations of $n/4$. Consequently,

[TABLE]

where the notation on the right-hand side means dilation by a factor of $2$. Iterating, if $k$ is the largest nonnegative integer for which $2^{2k+3}\mid n$, we find that

[TABLE]

We have from our choice of $k$ that $2\mid n/4^{k+1}$ while $8\nmid n/4^{k+1}$.

The observations of the last paragraph show that to describe $\mathscr{S}(n)$, it is enough to consider those cases where $8\nmid n$. When $4\nmid n$, Proposition 3 tells us the answer. However, when $4\mid n$, it does not seem that there is much to be said beyond what follows immediately from Theorem 2 and the Legendre–Gauss theorem.

The situation becomes both clearer and a bit cleaner if one is willing to shift perspective. Rather than first picking $n$ and asking for a description of the elements of $\mathscr{S}(n)$, we may pick $T$ and ask for which $n$ we have $T\in\mathscr{S}(n)$.

1. (1)

If $T$ is odd, then $T\in\mathscr{S}(n)$. 2. (2)

If $T$ is twice an odd integer, then $T\in\mathscr{S}(n)$ if and only if n\not\equiv 0{\text{\rm\ (mod~{}8)}}. 3. (3)

Suppose that $4\mid T$. Then $T\in\mathscr{S}(n)$ if and only if

[TABLE]

Here the right-hand side is an infinite union of disjoint residue classes modulo $2^{2k+3}$, over positive integers $k$.

We leave the (routine) proof of this proposition to the interested reader.