<s>[INST] When will a parametric solution generate all possible solutions? I was looking for integer solutions to this equation: 
$$a^2+b^2+c^2=3d^2$$
And found a parametric solution. Given r, s, t:
$$a=r^2+s^2-t^2+2t(r+s)$$
$$b=r^2-s^2+t^2+2s(t-r)$$
$$c=-r^2+s^2+t^2+2r(t-s)$$
$$d=r^2+s^2+t^2$$
My question is will this generate every possible solution?
 [/INST] You almost got it. The complete rational parameterization to
$$a^2+b^2+c^2=3\tag1$$
is given by,
$$a=\frac{r^2+s^2-t^2-2t(r+s)}{r^2+s^2+t^2}\\b=\frac{r^2-s^2+t^2-2s(r+t)}{r^2+s^2+t^2}\\c=\frac{-r^2+s^2+t^2-2r(s+t)}{r^2+s^2+t^2}\tag2$$
Proof: For any solution $a,b,c$, one can find $r,s,t$ using the simple formulas,
$$r=\frac{1}{b-1}\\s=\frac{1}{c-1}\\t=\frac{a-1}{(b-1)(c-1)}\tag3$$
Example: Using Jagy's example, we have $a,b,c = \frac{1}{7}, \frac{5}{7}, 
\frac{11}{7}$. Plugging this into $(3)$, we get $r,s,t = \frac{-7}{2}, \frac{7}{4}, \frac{21}{4}$. Plugging this into $(2)$, we recover the $a,b,c$.
Note: Division by zero is easily avoided by using $(\pm1)^2+(\pm1)^2+(\pm1)^2=3$.
</s>