Assume we have a positive integer $n$ which is the sum of the squares of three integers — for example,
$$ n = 451 = 15^2 + 15^2 + 1^2. $$
As it turns out, this particular $n$ has are several other such representations, because
$$ n = 451 = 15^2 + 15^2 + 1^2 = 21^2 + 3^2 + 1^2 = 17^2 + 9^2 + 9^2 = 19^2 + 9^2 + 3^2. $$
I know there are ways of determining whether a given number has such a representation, calculating how many representations a given number has, and so on. What I want instead is transformations or algorithms that could be used to generate additional representations from a given one. For example, Euler gave $$a^2+b^2+c^2 = (2m-a)^2 + (2m-b)^2 + (2m-c)^2, \qquad a+b+c=3m$$ and related relations; several mathematicians (including Bradley) have given complete solutions (a.k.a. “parameterizations“).
Does anyone know of a connected graph of solutions/equations, analogous to the Tree of Pythagoras for Pythagorean triples, which would give an algorithm by which one could move from a given three-square representation to [all] others? In particular, a way to find the representation which satisfies a particular criteria (e.g., the one which contains the term with the smallest possible absolute value) would be most helpful.
I could try to build/compile it myself, but didn’t want to reinvent the wheel if it’s already been done!
Comment: Consider following relation:
$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)$
Rewrite this as:
$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac)$
This relation implies that for every $n=a^2+b^2+c^2$ there exists a number like $a+b+c=m$. m can be some certain numbers their squares are greater than n. For example for 451 we have numbers, $22^2=484$, $23^2=529$, $25^2=625$ etc. If n is odd then m must also be odd such that $2| m^2-n$. If n is even then m must also be even. For example for 451 only odd m are acceptable.
Suppose a is greatest number then the third equation can be :
$a\leq\lfloor \sqrt n\rfloor$
So we can arrange following system of Diophantine equations:
$\begin{cases} a^2+b^2+c^2=n\\a+b+c=m\\a\leq\lfloor \sqrt n\rfloor \end {cases}$
For example with $n=451$, $m^2=23^2=529$, $25^2=625$, $27^2=729$, $29^2=841$, $31^2=961$, $33^2=1089$, $35^2=1225$ and $37^2=1369$ can be checked.
$a\leq \lfloor \sqrt{451}\rfloor=21 $
$25=21+3+1$
$31=19+9+3=15+15+1$
$35=17+9+9$
By brute force we may find more solutions.