Product of Sums of Four Squares

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Theorem

Let $a, b, c, d, w, x, y, z$ be numbers.

Then:

\(\ds \) \(\) \(\ds \left({a^2 + b^2 + c^2 + d^2}\right) \left({w^2 + x^2 + y^2 + z^2}\right)\)
\(\ds =\) \(\) \(\ds \left({a w + b x + c y + d z}\right)^2\)
\(\ds \) \(+\) \(\ds \left({a x - b w + c z - d y}\right)^2\)
\(\ds \) \(+\) \(\ds \left({a y - b z - c w + d x}\right)^2\)
\(\ds \) \(+\) \(\ds \left({a z + b y - c x - d w}\right)^2\)


Corollary

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 1

Taking each of the squares on the right hand side and multiplying them out in turn:

\(\ds \paren {a w + b x + c y + d z}^2\) \(=\) \(\ds a^2 w^2 + b^2 x^2 + c^2 y^2 + d^2 z^2\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds 2 \paren {a b w x + a c w y + a d w z + b c x y + b d x z + c d y z}\)


\(\ds \paren {a x - b w + c z - d y}^2\) \(=\) \(\ds a^2 x^2 + b^2 w^2 + c^2 z^2 + d^2 y^2\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds 2 \paren {-a b w x + a c x z - a d x y - b c w z + b d w y - c d y z}\)


\(\ds \paren {a y - b z - c w + d x}^2\) \(=\) \(\ds a^2 y^2 + b^2 z^2 + c^2 w^2 + d^2 x^2\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds 2 \paren {-a b y z - a c w y + a d x y + b c w z - b d x z - c d w x}\)


\(\ds \paren {a z + b y - c x - d w}^2\) \(=\) \(\ds a^2 z^2 + b^2 y^2 + c^2 x^2 + d^2 w^2\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds 2 \paren {a b y z - a c x z - a d w z - b c x y - b d w y + c d w x}\)


All the non-square terms cancel out with each other, leaving:

\(\ds \) \(\) \(\ds \paren {a w + b x + c y + d z}^2\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {a x - b w + c z - d y}^2\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {a y - b z - c w + d x}^2\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {a z + b y - c x - d w}^2\)
\(\ds \) \(=\) \(\ds a^2 w^2 + b^2 x^2 + c^2 y^2 + d^2 z^2\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds a^2 x^2 + b^2 w^2 + c^2 z^2 + d^2 y^2\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds a^2 y^2 + b^2 z^2 + c^2 w^2 + d^2 x^2\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds a^2 z^2 + b^2 y^2 + c^2 x^2 + d^2 w^2\)
\(\ds \) \(=\) \(\ds \paren {a^2 + b^2 + c^2 + d^2} \paren {w^2 + x^2 + y^2 + z^2}\)

$\blacksquare$


Proof 2

Let:

$\mathbf m = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$
$\mathbf n = -w \mathbf 1 + x \mathbf i + y \mathbf j + z \mathbf k$

be two quaternions.

Then:

\(\ds \size {\mathbf m} \size {\mathbf n}\) \(=\) \(\ds \size {\mathbf m \mathbf n}\) Quaternion Modulus of Product of Quaternions
\(\ds \leadsto \ \ \) \(\ds \paren {a^2 + b^2 + c^2 + d^2} \paren {\paren {-w}^2 + x^2 + y^2 + z^2}\) \(=\) \(\ds \size {\mathbf m \mathbf n}\) Definition of Quaternion Modulus
\(\ds \) \(=\) \(\ds \paren {-a w - b x - c y - d z}^2\) Definition of Quaternion Multiplication
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {a x - b w + c z - d y}^2\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {a y - b z - c w + d x}^2\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {a z + b y - c x - d w}^2\)
\(\ds \leadsto \ \ \) \(\ds \paren {a^2 + b^2 + c^2 + d^2} \paren {w^2 + x^2 + y^2 + z^2}\) \(=\) \(\ds \paren {a w + b x + c y + d z}^2\) simplifying
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {a x - b w + c z - d y}^2\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {a y - b z - c w + d x}^2\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {a z + b y - c x - d w}^2\)

$\blacksquare$

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