Convert the following into sum of four squares:
(a) $6a^2b^2$
(b) $7a^2b^2$
(c) $na^2b^2$
where $n$ is a positive integer.
We'll be using the Sum of Four Squares Identity (8.10) is
$$ \begin{align} (a^2 + b^2 + c^2 + d^2) \times (x^2 + y^2 + z^2 + w^2) & = (ax + by + cz + dw)^2 + (az− bw− cx + dy)^2 \\ \\ & + (ay− bx + cw− dz)^2 + (aw + bz− cy− dx)^2 \end{align} $$
(a)
$$ 6a^2b^2 = (2^2 + 1^2 +1^2) \times (ab)^2 = (2ab)^2 + (ab)^2 + (ab)^2$$
(b)
$$ 7a^2b^2 = (2^2 + 1^2 +1^2 + 1^2) \times (ab)^2 = (2ab)^2 + (ab)^2 + (ab)^2 + (ab)^2 $$
(c) Since $n$ is a positive integer, Lagrange’s Sum of Four Squares Theorem (8.14), it can be written as the sum of four squares,
$$ n = n_1^2 + n_2^2 + n_3^2 + n_4^2 $$
And so
$$ na^2b^2 = (n_1^2 + n_2^2 + n_3^2 + n_4^2) \times (ab)^2 = (n_1ab)^2 + (n_2ab)^2 + (n_3ab)^2 + (n_4ab)^2 $$