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

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 $47$.
