$(a^2+b^2)\cdot (c^2+d^2) = z^2+1$ then $z = ac+bd$ ?

Question

I have the following equation : $$(a^2+b^2)\cdot (c^2+d^2) = z^2+1$$ with $a, b, c, d, z \in \mathbb{N}$

Then how can I prove that : $z = ac+bd$ ?

I tried Brahmagupta identity but it doesn't seem to work...

Answer 1

The statement is false. Counter-example: $$(9^2+7^2)(4^2+1^2) = 2210 = 47^2 + 1 \quad\text{ but }\quad 47 \ne \begin{cases} 43 &= 9\cdot 4 + 7\cdot 1\\ 37 &= 9\cdot 1 + 7\cdot 4 \end{cases}$$

Please note that $47$ is the smallest value of $z$ where $z^2+1$ contains three distinct prime factors of the form $4k+1$. This allow one to express one of $a^2+b^2$ or $c^2+d^2$ as sum of two squares in more than one way. As an example, the first factor above has following two representations: $$9^2 + 7^2 = 11^2+3^2$$ and $11\cdot 4 + 3\cdot 1$ do equal to $47$.

Answer 2

$$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2=z^2+1$$

$$z=ac+bd$$

$$ad=bc\pm1$$