12
$\begingroup$

This is kind of a follow-up to the question I posted here about expressing integers as the sum of two squares. Is there a similar general method for expressing integers as the sum of four squares? I believe the Lagrange's Four-Square Theorem states that all positive integers are expressible as the sum of four squares of integers, but how do you find these numbers. As an example consider the value $1638$. How can we find the four squares?

$\endgroup$
13
$\begingroup$

Similar to the Brahmagupta-Fibonacci two-square identity. Euler has a four square identity which involves the sum of 4 squares:

$$(a_1^2+a_2^2+a_3^2+a_4^2)(b_1^2+b_2^2+b_3^2+b_4^2) =\\ \quad(a_1b_1 - a_2b_2 - a_3b_3 - a_4b_4)^2 + (a_1b_2+a_2b_1+a_3b_4-a_4b_3)^2 +(a_1b_3 - a_2b_4 + a_3b_1 + a_4b_2)^2 + (a_1b_4 + a_2b_3 - a_3b_2 + a_4b_1)^2$$

Factor $1638$ as products of any small factors you know how to represent as sum of 4 squares. Repeat apply the formula will allow you to represent $1638$ itself as sum of 4 squares.

For example, let's say we have factored $1638$ as $2\cdot 3^2 \cdot 7 \cdot 13$, we have:

$$\begin{align} & 2\cdot 3^2 \cdot 7 \cdot 13\\ = & (1^2+1^2+0^2+0^2)(1^2+1^2+1^2+0^2)^2(2^2+1^2+1^2+1^2)(3^2+2^2+0^2+0^2)\\ = & (0^2 + 2^2 + 1^2 + 1^2)(1^2+1^2+1^2+0^2)(2^2+1^2+1^2+1^2)(3^2+2^2+0^2+0^2)\\ = & ((-3)^2 + 1^2 + 2^2 + 2^2)(2^2+1^2+1^2+1^2)(3^2+2^2+0^2+0^2)\\ = & ((-11)^2+(-1)^2+2^2 + 0^2)(3^2+2^2+0^2+0^2)\\ = & ((-31)^2 + (-25)^2 + 6^2 + 4^2)\\ \end{align}$$

This give you a non-trivial representation of $1638$ as $31^2 + 25^2 + 6^2 + 4^2$.

In general, there are many representations of a number as a sum of 4 squares. There is a theorem:

The total number of representations of a positive integer $n$ as the sum of four squares, representations that differ only in order and sign being counted as distinct, is eight times the sum of the divisors of $n$ that are not multiple of $4$.

The above representation is only $1$ out of $8 \sum_{d\mid 1638, 4 \nmid d} d = 34944$ ways of representing $1638$ as sum of 4 squares.

$\endgroup$
2
$\begingroup$

$1638=2\cdot3^2\cdot7\cdot13=(1^2+1^2)3^2(2^2+1^2+1^2+1^2)(3^2+2^2)$ Now use your technique for taking the product of two sums of two squares to a sum of two squares.

$\endgroup$
2
$\begingroup$

Just as the Gaussian integers are a modern method to prove every prime not congruent to $3$ (mod $4$) is a sum of two squares ( since $\mathbb{Z}[i]$ is a Euclidean domain), a similar method works for the so-called Hurwitz quaternions $\mathbb{H} = \{ \frac{a+bi + cj +dk}{2}: a,b,c,d \in \mathbb{Z}, a \equiv b \equiv c \equiv d ({\rm mod} 2) \}$.$ \mathbb{H}$ is not a commutative ring, but behaves sufficiently like a Euclidean ring that a similar proof shows that every prime $p$ is a sum of $4$ integer squares. A full proof can be found in I.N. Herstein's "Topics in Algebra", but here is an outline: given any odd prime $p \in \mathbb{N}$, we can express $-1$ as a sum of two squares in $\mathbb{Z}/p\mathbb{Z}$. This means that there are integers $a,b$ such that $p |(a^{2}+b^{2}+1).$ This means that $p$ is not an irreducible element in $\mathbb{H}$, and this leads to the fact that $p = x^{2}+y^{2}+z^{2}+w^{2}$ for integers $x,y,z,w$. Once we know that every prime has such an expression, it follows (as noted in other answers and comments) that every positive integer is a sum of $4$ integer squares. However, it should be noted that, in practice, this may not be the most efficient way to express a given positive integer as a sum of $4$ integer squares.

$\endgroup$
1
$\begingroup$

Hint: Using the same condition as your previous question. If you can express the two numbers which add up to 1638 as a sum of two squares. You are through.:)

$\endgroup$
-2
$\begingroup$

This formula seems wrong. Try letting $a_1=b_1$, $a_2=b_2$, $a_3=b_3$, $a_4=a_4$. The result should be $(a_1^2+a_2^2+a_3^2+a_4^2)$.

Euler's four-square identity says that the product of two numbers, each of which is a sum of four squares, is itself a sum of four squares. Specifically:

$$(a_1^2+a_2^2+a_3^2+a_4^2)(b_1^2+b_2^2+b_3^2+b_4^2)= (a_1 b_1 + a_2 b_2 + a_3 b_3 + a_4 b_4)^2 + (a_1 b_2 - a_2 b_1 + a_3 b_4 - a_4 b_3)^2 + (a_1 b_3 - a_2 b_4 - a_3 b_1 + a_4 b_2)^2 + (a_1 b_4 + a_2 b_3 - a_3 b_2 - a_4 b_1)^2$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.