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What is known about solutions in integers of the following equation ?

$$( m^2 + n^2 + q^2 )^2 = 36 ( u^2 + s^2 + t^2 )$$

I am asking this because I just recently have got these:

$$(a-b)^2 + (2a+b)^2 + (a+2b)^2 = 6 ( a^2 + b^2 + ab)$$


$$( a^2 + b^2 + ab)^2 = a^4 + b^4 + (ab)^2 + 2(a^3) b + 2a(b^3) + 2(ab)^2 $$

$$= (ab)^2 + a^4 + 2(a^3)b + (ab)^2 + (ab)^2 + 2a(b^3) + b^4$$ $$ = (ab)^2 + (a(a+b))^2 + (b(a+b))^2$$

so that I can get as a solution:

$$m = a-b\, ,\, n = 2a+b \,,\, q = a+2b \,, \,u = ab \,,\, s = a(a+b)\,,\,t= b(a+b)$$

I would like to know : what other solutions exist ?

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An integer $m$ is a sum of 3 squares if and only if it is NOT of the form $4^k (8r+7)$. Let $a$ be even and not of the form $4^k (8r+7)$. Then $a^2/36$ is also not of the form $4^k (8r+7)$. So, for every even $a$ not of the forbidden form, there are solutions (probably many of them) to $m^2+n^2+q^2=a$, $u^2+s^2+t^2 = a^2/36$. – David Speyer Oct 9 '12 at 13:29
@ David Speyer: That's true by Legendre's theorem , but what about writing the solution in a compact form (as I came up with one) ? Is this equation mentioned in any area of Mathematics ? – Souvik Dey Oct 10 '12 at 4:15
If you think about David's answer, it says there are lots of solutions, probably many, many more than you get from your 2-parameter family. I bet a little computation would turn up many solutions that don't fit your formula. – Gerry Myerson Apr 30 '13 at 10:48

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