Product of Sums of Four Squares/Corollary

Theorem
Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n, c_1, c_2, \ldots, c_n, d_1, d_2, \ldots, d_n$ be integers.

Then:
 * $\ds \exists w, x, y, z \in \Z: \prod_{j \mathop = 1}^n \paren {a_j^2 + b_j^2 + c_j^2 + d_j^2} = w^2 + x^2 + y^2 + z^2$

That is, the product of any number of sums of four squares is also a sum of four squares.

Proof
Proof by induction:

For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
 * $\ds \exists w, x, y, z \in \Z: \prod_{j \mathop = 1}^n \paren {a_j^2 + b_j^2 + c_j^2 + d_j^2} = w^2 + x^2 + y^2 + z^2$

$\map P 1$ is true, as this just says:
 * $\exists w, x, y, z \in \Z: a^2 + b^2 + c^2 + d^2 = w^2 + x^2 + y^2 + z^2$

which is trivially true.

Basis for the Induction
$\map P 2$ is the case:
 * $\exists w, x, y, z \in \Z: \paren {a_1^2 + b_1^2 + c_1^2 + d_1^2} \paren {a_2^2 + b_2^2 + c_2^2 + d_2^2} = w^2 + x^2 + y^2 + z^2$

which follows from Product of Sums of Four Squares.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:
 * $\ds \exists w, x, y, z \in \Z: \prod_{j \mathop = 1}^k \paren {a_j^2 + b_j^2 + c_j^2 + d_j^2} = w^2 + x^2 + y^2 + z^2$

Then we need to show that it directly implies:
 * $\ds \exists w, x, y, z \in \Z: \prod_{j \mathop = 1}^{k + 1} \paren {a_j^2 + b_j^2 + c_j^2 + d_j^2} = w^2 + x^2 + y^2 + z^2$

Induction Step
This is our induction step:

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\ds \forall n \in \Z_{>0}: \exists w, x, y, z \in \Z: \prod_{j \mathop = 1}^n \paren {a_j^2 + b_j^2 + c_j^2 + d_j^2} = w^2 + x^2 + y^2 + z^2$