# Lagrange's four-square theorem

Lagrange's four-square theorem, also known as Bachet's conjecture, states that any natural number can be represented as the sum of four integer squares

$p = a_0^2 + a_1^2 + a_2^2 + a_3^2\$

where the four numbers a0, a1, a2, a3 are integers. For illustration, 3, 31 and 310 can be represented as the sum of four squares as follows:

3 = 12 + 12 + 12 + 02
31 = 52 + 22 + 12 + 12
310 = 172 + 42 + 22 + 12.

This theorem was proven by Joseph Louis Lagrange in 1770, and corresponds to Fermat's theorem on sums of two squares.

## Historical development

The theorem appears in the Arithmetica of Diophantus, translated into Latin by Bachet in 1621.

Adrien-Marie Legendre improved on the theorem in 1798 by stating that a positive integer can be expressed as the sum of three squares if and only if it is not of the form 4k(8m + 7). His proof was incomplete, leaving a gap which was later filled by Carl Friedrich Gauss.

## Proof using the Hurwitz integers

One of the ways to prove the theorem relies on Hurwitz quaternions, which are the analog of integers for quaternions.[1] The Hurwitz quaternions consist of all quaternions with integer components and all quaternions with half-integer components. These two sets can be combined into a single formula

$\alpha = \frac{1}{2} E_0 (1 + \mathbf{i} + \mathbf{j} + \mathbf{k}) +E_1\mathbf{i} +E_2\mathbf{j} +E_3\mathbf{k} = a_0 +a_1\mathbf{i} +a_2\mathbf{j} +a_3\mathbf{k}$

where E0, E1, E2, E3 are integers. Thus, the quaternion components a0, a1, a2, a3 are either all integers or all half-integers, depending on whether E0 is even or odd, respectively. The set of Hurwitz quaternions forms a ring; that is to say, the sum or product of any two Hurwitz quaternions is likewise a Hurwitz quaternion.

The generalized Euclidean algorithm identifies the greatest common right or left divisor of two Hurwitz quaternions, where the "size" of the remainder ρ is measured by the norm. The norm N(α) of a quaternion α is the nonnegative real number

$\mathrm{N}(\alpha) = \alpha\bar\alpha = a_0^2+a_1^2+a_2^2+a_3^2$,

where $\bar\alpha=a_0 -a_1\mathbf{i} -a_2\mathbf{j} -a_3\mathbf{k}$ is the conjugate of α. Note that the norm is always an integer. (If the coefficients are half-integer, then their squares are a number, equivalent to 1 modulo 4, divided by 4, so the sum is again an integer.)

Since quaternion multiplication is associative, and real numbers commute with other quaternions, the norm of a product of quaternions equals the product of the norms:

$\mathrm{N}(\alpha\beta)=\alpha\beta\overline{(\alpha\beta)}=\alpha\beta\overline{\beta}\overline{\alpha}=\alpha \mathrm{N}(\beta)\bar\alpha=\alpha\bar\alpha \mathrm{N}(\beta)= \mathrm{N}(\alpha) \mathrm{N}(\beta).$

The ring of Hurwitz quaternions is Euclidean, since for any quaternion $\alpha=a_0+a_1\mathbf{i}+a_2\mathbf{j}+a_3\mathbf{k}$ with rational coefficients we can choose a Hurwitz quaternion $\beta=b_0+b_1\mathbf{i}+b_2\mathbf{j}+b_3\mathbf{k}$ so that $\mathrm{N}(\alpha-\beta)<1$ by first choosing $b_0$ so that $|a_0-b_0| \leq 1/4$ and then $b_1, b_2, b_3$ so that $|a_i-b_i| \leq 1/2$ for $i=$ 1, 2, 3. Then we obtain

$\mathrm{N}(\alpha-\beta)=(a_0-b_0)^2+(a_1-b_1)^2+(a_2-b_2)^2+(a_3-b_3)^2 \leq \left(\frac{1}{4} \right)^2+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2=\frac{13}{16}<1.$

Therefore, the ring of Hurwitz quaternions is Euclidean and, as a consequence, also a unique factorization domain.

Since any natural number can be factored into powers of primes, the four-square theorem holds for all natural numbers if it is true for all prime numbers. It is true for 2 = 02 + 02 + 12 + 12. To show this for an odd prime integer p, represent it as a quaternion (p, 0, 0, 0) and assume for now (as we shall show later) that it is not a Hurwitz prime; that is, it can be factored into two non-unit Hurwitz quaternions

p = αβ.

The norms of p, α, β are nonnegative integers such that

$\mathrm N(p)=p^2=\mathrm N(\alpha\beta)=\mathrm N(\alpha)\mathrm N(\beta)$

This shows that both $\mathrm N(\alpha)$ and $\mathrm N(\beta)$ are equal to p (since they are integers), and p is the sum of four squares

p = N(α) = a02 + a12 + a22 + a32.

If it happens that the α chosen has half-integer coefficients, it can be replaced by another Hurwitz quaternion. Choose ω=(±1 ± i ± j ± k)/2 in such a way that γ ≡ ω + α has even integer coefficients. Then

$p=(\bar\gamma-\bar\omega)\omega\bar\omega(\gamma-\omega)=(\bar\gamma\omega-1)(\bar\omega\gamma-1)$

Since γ has even integer coefficients, $(\bar\omega\gamma-1)$ will have integer coefficients and can be used instead of the original α to give a representation of p as the sum of four squares.

As for showing that p is not a Hurwitz prime, Lagrange proved that any odd prime p divides at least one number of the form 1 + l2 + m2, where l and m are integers.[1] The latter number can be factored in Hurwitz quaternions:

1 + l2 + m2 = (1 + l i + m j) (1 − l im j).

If p could not be factored in Hurwitz quaternions, it would be a Hurwitz prime number by definition. Then, by unique factorization, p would have to divide either 1 + l i + m j or 1 − l im j to form another Hurwitz quaternion. But for p > 2, the number

1/p ± l/p i ± m/p j

is not a Hurwitz integer. Therefore, every p > 2 (as well as 2 itself) can be factored in Hurwitz quaternions, and the four-square theorem holds.

## Generalizations

Lagrange's four-square theorem is a special case of the Fermat polygonal number theorem and Waring's problem. Another possible generalisation is the following problem: Given natural numbers a, b, c, d, can we solve

n = ax12 + bx22 + cx32 + dx42

for all positive integers n in integers x1, x2, x3, x4? The case a = b = c = d = 1 is answered in the positive by Lagrange's four-square theorem. The general solution was given by Ramanujan. He proved that if we assume, without loss of generality, that abcd then there are exactly 54 possible choices for a, b, c, d such that the problem is solvable in integers x1, x2, x3, x4 for all n. (Ramanujan listed a 55th possibility a = 1, b = 2, c = 5, d = 5, but in this case the problem is not solvable if n = 15.[2])

## Algorithms

Michael O. Rabin and Jeffrey Shallit[3] have found randomized polynomial-time algorithms for computing a representation n = x12 + x22 + x32 + x42 for a given integer n, in expected running time O((logn)2).

## Uniqueness

The sequence of positive integers whose representation as a sum of four squares is unique (up to order) is:

1, 2, 3, 5, 6, 7, 8, 11, 14, 15, 23, 24, 32, 56, 96, 128, 224, 384, 512, 896 ... (sequence A006431 in OEIS).

These integers consist of the seven odd numbers 1, 3, 5, 7, 11, 15, 23 and all numbers of the form 2 × 4k, 6 × 4k or 14 × 4k.

The sequence of positive integers which cannot be represented as a sum of four non-zero squares is:

1, 2, 3, 5, 6, 8, 9, 11, 14, 17, 24, 29, 32, 41, 56, 96, 128, 224, 384, 512, 896 ... (sequence A000534 in OEIS).

These integers consist of the eight odd numbers 1, 3, 5, 9, 11, 17, 29, 41 and all numbers of the form 2 × 4n, 6 × 4n or 14 × 4n.