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.
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 $$ a^2+b^2+c^2=n=d^2+e^2+f^2. $$
When $ a = -d, b=-e, c=-f $ then it satisfies the equality.
$$ (-99)^2+(-100)^2+(-101)^2 = (99)^2+(100)^2+(101)^2 $$
$$ (-d)^2+(-e)^2+(-f)^2= d^2+e^2+f^2 $$
Where $ a≠b≠c≠d≠e≠f$
You can write a simple solution:
$p,t,k,q,s$ - any integer asked us.
There are a lot of methods, e.g. matrices are very useful.
But to avoid complicate constructions I will show you an easy way. After some calculations you can imagine how a possible general solution looks like.
We begin with $3$ variables $a,b,c$ . Here '$i$' is the imaginary unit.
We get $a^2+b^2=c^2$ with e.g. $a:=\alpha^2-\beta^2$ and $b=2\alpha\beta$ .
We continue with $4$ variables $a,b,c,d$ .
We get $a^2+b^2+c^2=d^2$ with e.g. $a:=\alpha^2+\beta^2-\gamma^2$ , $b=2\alpha\gamma$ and $c=2\beta\gamma$ .
If we like to have $a^2+b^2-c^2=d^2$ it’s enough to substitute $\beta$ by $i\beta$ .
So, your problem can be started with $f:=\alpha^2+\beta^2+\gamma^2+\delta^2+\epsilon^2=…$ and you will get
a formula for $a^2+b^2+c^2+d^2+e^2=f^2$ . To get $a^2+b^2+c^2-d^2-e^2=f^2$ you only
have to substitute e.g. $\delta$ by $i\delta$ and $\epsilon$ by $i\epsilon$ .