Integer solutions to $a^{2}+b^{2}+c^{2} = d^{2}+e^{2}+f^{2}$, where $a, b, c, d, e, f \neq 0$

Question

How does one approach something like this? Is there an equivalent Legendre's three-square theorem for the sum of three squares in two different ways?

It seems like the only way to approach it would be via some computer power.

Answer 1

There is a method to find the different primitive representations of an integer $n$ as the sum of three squares, see here. This gives solutions to the above equation. For example, we have $150$ different representations of $n=225=a^2+b^2+c^2$, and $96$ primitive ones among them. So we can choose each pair of these solutions to obtain $$ a^2+b^2+c^2=n=d^2+e^2+f^2. $$

Answer 2

When $ a = -d, b=-e, c=-f $ then it satisfies the equality.

Consider

$$ (-99)^2+(-100)^2+(-101)^2 = (99)^2+(100)^2+(101)^2 $$

Similarly

$$ (-d)^2+(-e)^2+(-f)^2= d^2+e^2+f^2 $$

Where $ a≠b≠c≠d≠e≠f$

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So infinite integral solutions are possible. (Not positive integral).

Answer 3

Equation:

$$a^2+b^2+c^2=x^2+y^2+z^2$$

You can write a simple solution:

$$a=q^2+s^2+k^2-p^2-t^2$$

$$b=q^2+s^2+k^2-p^2+t^2+2pt-2kt-2st-2qt$$

$$c=q^2+s^2+k^2+p^2-t^2+2tp-2kp-2sp-2qp$$

$$x=p^2+t^2+k^2-q^2-s^2-2kt-2pk+2sk+2qk$$

$$y=p^2+t^2-k^2-q^2+s^2-2ts-2ps+2ks+2qs$$

$$z=p^2+t^2-k^2+q^2-s^2-2tq-2pq+2kq+2sq$$

$p,t,k,q,s$ - any integer asked us.