A Product of a Sum of Two Squares is a Sum of Two Squares

Theorem 1: For all $a, b, c, d \in \mathbb{Z}$, $(a^2 + b^2)(c^2 + d^2) = (ad - bc)^2 + (ac + bd)^2$. In particular, the product of a sum of two squares is itself a sum of two squares.

The equality above actually holds for all $a, b, c, d \in \mathbb{R}$.

• Proof: We have that:

(1)

\begin{align} \quad (a^2 + b^2)(c^2 + d^2) = a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2 \end{align}

• And also:

(2)

\begin{align} \quad (ad - bc)^2 + (ac + bd)^2 &= a^2d^2 - 2abcd + b^2c^2 + a^2c^2 + 2abcd + b^2d^2 \\ &= a^2d^2 + b^2c^2 + a^2c^2 + b^2d^2 \\ \end{align}

• Comparing the two and we see that indeed:

(3)

\begin{align} \quad (a^2 + b^2)(c^2 + d^2) = (ad - bc)^2 + (ac + bd)^2 \end{align}

For example, consider $m = 13$ and $n = 20$. Observe that $m = 2^2 + 3^2$ and $n = 2^2 + 4^2$. By the theorem above, the product $mn = 260$ can also be written as a sum of two squares, and in particular:

(4)

\begin{align} \quad mn = 260 = (8 - 6)^2 + (4 + 12)^2 = 2^2 + 16^2 \end{align}