Formula for finding sum of three squares that equal a fourth square

Question

I am investigating the sum of three squares that equal a fourth square value.

I am only interested in positive integers greater than zero.

Specifically i am collecting formulae that identify values of these combinations.

Of course i have the following example:-

for any positive integer x the following is true and gives the 4 square values

$x, (x + 1), x(x + 1), (x(x + 1) + 1)$

$x^2 + (x + 1)^2 +(x(x + 1))^2 = (x(x + 1) + 1)^2$

what other formulae exist that can identify 3 unique integers that when squared give a fourth square value?

Answer 1

I can't remember exactly where I came across it, but it is a fact that any natural number $n$ can be written as the sum of three squares iff $n≠4^m(8k+7)$ for some positive integers $m$ and $k$. If $n$ is square and of this form, then $4^m=8k+7$. The left hand side is always even if $m$ is an integer greater than $0$. Meanwhile, the product of any integer with an even one is always even, and the sum of an even integer and an odd integer is itself is always odd. In other words, the left hand side is always even while the right hand side is always odd, so there can be no integer values of $m$ and $k$ such that $4^m(8k+7)$ is square; therefore, every square integer can be written as the sum of three squares.

I know this doesn't actually address your goal of collecting formulae that help you in determining the square partitions of a square number, but I noticed that you didn't provide any clue that you know (i.e. have proved) that this is always possible, so I thought I'd shore that up for you. In terms of the question itself, I think you could probably do something with the fact that every square number $n$ is the sum of the first $n$ odd integers...

Answer 2

Your formula can be generalised to:

$$a^2+(a+b)^2 + \left(a^2+ab+\frac{b^2-1}{2}\right)^2 = \left(a^2+ab+\frac{b^2+1}{2}\right)^2$$

where your case is $b=1$. Of course $b$ needs to be odd for this to be an integer solution.

This can be found using the fact that $\left(\frac{x+1}2\right)^2 - \left(\frac{x-1}2\right)^2 = x$ and then applying it to $x=a^2+(a+b)^2$.

The above can be simplified a little by setting $b=c-a$ to get $$a^2+c^2 + \left(\frac{a^2+c^2-1}2\right)^2 = \left(\frac{a^2+c^2+1}2\right)^2$$ The integers $a$ and $c$ need to have opposite parity for this to provide integer solutions.

Answer 3

One way of finding such formulae is to write $a^2+b^2=d^2-c^2=(d+c)(d-c)$. Now you can choose anything for $a$ and $b$ and factorise it.

For instance, $a=n$ and $b=n+3$; then $(d+c)(d-c)= 2n^2+6n+9$, so let $d+c=2n^2+6n+9$ and $d-c=1$, so $d=n^2+3n+5$ and $c=n^2+3n+4$.

You have to be careful: choosing $a=n$ and $b=n+2$ gives $(d+c)(d-c)= 2n^2+4n+4$, so picking $d+c=2n^2+4n+4$ and $d-c=1$ does not give integer values of $d$ and $c$, so instead choose $d+c=n^2+2n+2$ and $d-c=2$; then $d=\frac{1}{2}n^2+n+2$ and $c=\frac{1}{2}n^2+n$, so you need even $n=2r$, say, and then $a=2r$, $b=2(r+1)$, $c=2r(r+1)$ and $d=2(r^2+r+1)$, which is just doubling your example.

A nicer example is $a=6n$, $b=8n+5$, Then you can choose $d+c=20n^2+16n+5$ and $d-c=5$, so $c=10n^2+8n$, $d=10n^2+8n+5$. When $n=1$ this gives $6^2+13^2+18^2=23^2$,

Answer 4

The 2.1.3 Diophantine equation $$A^2 = B^2+C^2+D^2$$ has been studied extensively. The complete solution in integers (a.k.a. “parameterization”) was first published by Catalan in 1885; you can find that solution, along with some other [incomplete] solutions in this paper by Ayoub B. Ayoub.

C. J. Bradley gives a lovely — and quite general — approach to this class of equation in this paper.

Answer 5

Let $a,b,c$ be integers. Then,

$$(a^2-b^2-c^2)^2+(2ab)^2+(2ac)^2=(a^2+b^2+c^2)^2$$

Check

Edit: I learned today that this is Lebesgue's formula. Unfortunately it does not produce all solutions. Catalan's formula does.