Sums of integers are squares

You can find 'triples' a, b, and c by taking any three square numbers, e.g. ${1, 4, 9}$ and then solving for a, b and c simultaneously, such that: $a+b = 1, a+c=4, b+c=9$

In this case: $a=-2, b=3, c=6$

This case actually doesn't work because $a = -2$ but nonetheless this gives you a general method. If your chosen three numbers are sufficiently far away from 0 (close to each other), negative numbers shouldn't be a problem. In fact, to avoid negatives, I think we need to choose three squares which satisfy the triangle inequality, that the sum of the two smaller squares is bigger than the largest square.

Now note that $2(a+b+c) = (1+4+9)$ So, then you need to find three sets of squares that sum to the same square number, to give your three sets. The big square also has to be even, or you'll get non-integer results. You also have to ensure that you get integer results :S

The 'minimal (a+b+c)' condition is covered by choosing the smallest square number that has this property, so we can stop when we find one solution.

So, we need the smallest even square that can be split into three distinct squares, in three distinct ways, where each way satisfies the triangle inequality!

I was hoping for help from The Encyclopedia of Integer Sequences but it only partially covers that situation, at https://oeis.org/A024803, and not well enough to find a suitable square by hand :(

The equation $x^2+y^2+z^2=2w^2$ has infinitely many solutions. For example, one family is

$x=3m^2+2mn-n^2$

$y=3m^2-2mn-n^2$

$z=4mn$

$w=3m^2+n^2$

A computer search gives $a=97$, $b=192$, $c=2112$; $d=720$, $e=801$, $f=880$; and $g=376$, $h=465$, $i=1560$.