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:

\(\ds \)

\(\)

\(\ds \prod_{j \mathop = 1}^{k + 1} \paren {a_j^2 + b_j^2 + c_j^2 + d_j^2}\)

\(\ds \)

\(=\)

\(\ds \paren {\prod_{j \mathop = 1}^k \paren {a_j^2 + b_j^2 + c_j^2 + d_j^2} } \paren {a_{k + 1}^2 + b_{k + 1}^2 + c_{k + 1}^2 + d_{k + 1}^2}\)

\(\ds \)

\(=\)

\(\ds \paren {r^2 + s^2 + t^2 + u^2} \paren {a_{k + 1}^2 + b_{k + 1}^2 + c_{k + 1}^2 + d_{k + 1}^2}\)

from the induction hypothesis: for some $r, s, t, u, \in \Z$

\(\ds \)

\(=\)

\(\ds w^2 + x^2 + y^2 + z^2\)

from the basis for the induction: for some $w, x, y, z, \in \Z$

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

Therefore:

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

$\blacksquare$