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:

$\displaystyle \exists w, x, y, z \in \Z: \prod_{j=1}^n \left({a_j^2 + b_j^2 + c_j^2 + d_j^2}\right) = w^2 + x^2 + y^2 + z^2$

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 $n \in \N^*$, let $P \left({n}\right)$ be the proposition:

$\displaystyle \exists w, x, y, z \in \Z: \prod_{j=1}^n \left({a_j^2 + b_j^2 + c_j^2 + d_j^2}\right) = w^2 + x^2 + y^2 + z^2$

$P \left({1}\right)$ 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

$P \left({2}\right)$ is the case:

$\exists w, x, y, z \in \Z: \left({a_1^2 + b_1^2 + c_1^2 + d_1^2}\right) \left({a_2^2 + b_2^2 + c_2^2 + d_2^2}\right) = w^2 + x^2 + y^2 + z^2$

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 $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:

$\displaystyle \exists w, x, y, z \in \Z: \prod_{j=1}^k \left({a_j^2 + b_j^2 + c_j^2 + d_j^2}\right) = w^2 + x^2 + y^2 + z^2$

Then we need to show that it directly implies:

$\displaystyle \exists w, x, y, z \in \Z: \prod_{j=1}^{k+1} \left({a_j^2 + b_j^2 + c_j^2 + d_j^2}\right) = w^2 + x^2 + y^2 + z^2$

Induction Step

This is our induction step:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\)	\(\displaystyle \prod_{j=1}^{k+1} \left({a_j^2 + b_j^2 + c_j^2 + d_j^2}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({\prod_{j=1}^k \left({a_j^2 + b_j^2 + c_j^2 + d_j^2}\right)}\right) \left({a_{k+1}^2 + b_{k+1}^2 + c_{k+1}^2 + d_{k+1}^2}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({r^2 + s^2 + t^2 + u^2}\right) \left({a_{k+1}^2 + b_{k+1}^2 + c_{k+1}^2 + d_{k+1}^2}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	from the induction hypothesis: for some $r, s, t, u, \in \Z$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle w^2 + x^2 + y^2 + z^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	from the basis for the induction: for some $w, x, y, z, \in \Z$

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

Therefore:

$\displaystyle \forall n \in \Z_{>0}: \exists w, x, y, z \in \Z: \prod_{j=1}^n \left({a_j^2 + b_j^2 + c_j^2 + d_j^2}\right) = w^2 + x^2 + y^2 + z^2$

$\blacksquare$