Brahmagupta-Fibonacci two-square identity

Statement

In any commutative unital ring, if $x$ and $y$ can each be written as a sum of two squares, so can $xy$.

More concretely, if $x=a^{2}+b^{2}$ and $y=c^{2}+d^{2}$, then $xy=(ac+bd)^{2}+(ad-bc)^{2}$. In other words, for all $a,b,c,d$ in a commutative ring:

$\!(a^{2}+b^{2})(c^{2}+d^{2})=(ac-bd)^{2}+(ad+bc)^{2}$

Interpretations

In terms of complex numbers

This identity is motivated by complex numbers. In fact, it is precisely the statement that the modulus-squared operation for complex numbers is multiplicative. To see this note that:

• $a^{2}+b^{2}$ is the modulus-squared of the complex number $a+ib$
• $c^{2}+d^{2}$ is the modulus-squared of the complex number $c+id<math>.*<math>(ac-bd)^{2}+(ad+bc)^{2}$ is the modulus-squared of the complex number $(a+ib)(c+id)$.

In terms of trigonometry

This is based on the interpretation in terms of complex numbers, and relates the formula to the angle sum formulas for sine and cosine. Fill this in later

Related facts

Applications

• Positive integer is a sum of two squares iff it has no prime divisor that is 3 mod 4 with odd multiplicity