$a$ sum of $2$ squares and $b$ sum of $3$ squares imply $a^2b$ is sum of $3$ squares. I have been given the following problem to solve and am having a hard time finding a solution. I feel that one way, perhaps the only way, is to use a polynomial identity but I can't determine it.
Let $a,b$ two positive integers. If $a$ is sum of $2$ squares and $b$ is sum of $3$ squares then $a^2b$ is sum of three squares.
NOTE.-The sums are mean "primitive" (i.e. with coprime solutions).

Any positive integer $n$ that is not divisible by 4, also $n \neq 7 \pmod 8,$ can be written primitively as the sum of three squares.  The number of such primitive representations is a certain multiple of $h(-4n).$  This number is the count of primitive binary quadratic forms of discriminant $-4n.$  We always have $h(-4n) \geq 1$   because there is always the form $x^2 + n y^2.$
Oh, permutations of a triple are considered separate, also negating one or more variables...a triple (x,y,z)  with $ x > y > z > 0$  therefore counts as $48$ representations. Anyway,
For $n > 1$ and $ n \equiv 1 \pmod 8,$ $\; \; R_{0}(n) = 12 h(-4n).$
For $ n \equiv 3 \pmod 8,$ $ \; \; R_{0}(n) = 8 h(-4n).$
For $ n \equiv 5 \pmod 8,$ $ \; \; R_{0}(n) = 12 h(-4n).$
For $ n \equiv 2 \pmod 8,$ $ \; \; R_{0}(n) = 12 h(-4n).$
For $ n \equiv 6 \pmod 8,$ $ \; \;R_{0}(n) = 12 h(-4n).$
