Can a "sum of two squares" solution be the average of two other solutions? If $n = a^2 + b^2$ and $n$ has more than two unique solutions for $a$ and $b$ (where unique means that $a$ and $b$ are unsigned and unordered), then is it possible, impossible, or unknown whether one solution can be the average of two other solutions? For example, given three solutions:
$n = j^2 + k^2$
$n = s^2 + t^2$
$n = x^2 + y^2$
Can $j^2$ and $k^2$ can be the average of the other two solutions? Is this possible:
$j^2 = \frac{s^2 + x^2}{2}$
$k^2 = \frac{t^2 + y^2}{2}$
Where might I learn more? I ask because I've been playing with the "3x3 magic square of squares" problem and this seems to be a core question. Thanks!
 A: Observations towards a solution

*

*$ s^2 + x^2 = 2j^2$ has parametrized solutions for integers $a, b, k$ of
$$ s = k(a^2 + 2ab -b^2), x =k( -a^2 + 2ab + b^2), j = k(a^2 + b^2).$$

*

*This can be algebraically verified.

*In the event that the values are negative, just take the absolute value.

*Note: This is a complete classification, but we don't require this fact. See here for more details.



*Likewise, $ t = c^2 + 2cd - d^2,  y = -c^2 + 2cd + d^2$.

*Now, since $s^2 + t^2 = x^2 + y^2$, expanding the expression and simplifying yields
$$ ab(a^2 - b^2) = cd(d^2 - c^2).$$

*This has solutions like $ a = 6, b = 1,  c = 2, d = 5$.

*

*There are also other solution sets, like $ a = 7, b = 3, c = 7, d = 8$. (Obtained by setting $ a = c$ as a simplification.)

*I don't know how to get a complete classification of the solutions. These were mainly found by trial and error.



*This yields $n = 47^2 + 1^2 = 37^2 + 29^2 = 23^2 + 41^2, 47^2 + 23^2 = 2\times 37^2, 1^2 + 41^2 = 2 \times 29^2 $.

