Proving $(ac+bd)^2+(ad-bc)^2=(a^2+b^2)(c^2+d^2)$ with various solutions. 
$(ac+bd)^2+(ad-bc)^2=(a^2+b^2)(c^2+d^2)$

Solutions in the answers.

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Edit) Since this question is closed, I'll add more contexts for this question.
This identity is called "Brahmagupta-Fibonacci identity", which the comment says.
This identity has a special feature, that the form of the expression maintains from LHS to RHS.
Also, for addition, we can expanse this identity to:
$$ (a^2+nb^2)(c^2+nd^2)=(ac\pm nbd)^2+n(ad\mp bc)^2. $$
or:
$$ X=xz-Cyw, Y=axw+a'yz+BYw. \\ (ax^2+Bxy+a'Cy^2)(a'z^2+Bzw+aCw^2)=aa'X^2+BXY+CY^2 $$
, from the answer of @Will Jagy.
This can be proved by various solutions, for example, just multiplying out this identity, or with trigonometric functions, or with the imaginary number "$i$".
I want you to prove this identity with more solutions.
 A: Simpler Solution.
\begin{align} & (ac+bd)^2+(ad-bc)^2 \\ = \; & (ac)^2+2abcd+(bd)^2+(ad)^2-2abcd+(bc)^2 \\ = \; & a^2c^2+b^2d^2+a^2d^2+b^2c^2 \\ = \; & (a^2+b^2)(c^2+d^2) \end{align}
A: Here is Gauss composition, using Dirichlet's method. This version is in David A. Cox, Primes of the Form $x^2 + n y^2,$   from the proof of Proposition 3.8. In the first edition, this is page 49(but has a typing error, corrected in the second edition).
Everything is integers. We are given $a, a', B,C$ and integer variables $x,y,z,w.$
By defining $$ X = xz - Cyw $$
$$Y = axw + a' yz  + B yw,$$
we find
$$ \left(ax^2 + Bxy + a'Cy^2  \right)  \left(a'z^2 + Bzw+aCw^2  \right) =  aa'X^2 + BXY + C Y^2 $$
Your version has $a=a'=C=1$   and $B=0$
A: There are two possible factorizations.
Here are both of them.
$\begin{array}\\
(a^2+b^2)(c^2+d^2)
&=a^2c^2+a^2d^2+b^2c^2+b^2d^2\\
&=a^2c^2+b^2d^2+a^2d^2+b^2c^2\\
&=a^2c^2+b^2d^2\pm2a^2b^2c^2d^2+a^2d^2+b^2c^2\mp2a^2b^2c^2d^2\\
&=(ac\pm bd)^2+(ad\mp bc)^2
\qquad\text{The two signs are different}\\
\end{array}
$
A: The identity is trivially satisfied if $\,a=b=0\,$ or $\,c=d=0\,$. Otherwise $\,a^2+b^2 \ne 0\,$ and $\,\left(\frac{a}{\sqrt{a^2+b^2}}\right)^2 + \left(\frac{b}{\sqrt{a^2+b^2}}\right)^2 = 1\,$, so there exists an angle $\,\alpha\,$ such that $\,\frac{a}{\sqrt{a^2+b^2}} = \sin \alpha\,$, $\,\frac{b}{\sqrt{a^2+b^2}} = \cos \alpha\,$. Similarly, define $\,\gamma\,$ such that $\,\frac{c}{\sqrt{c^2+d^2}} = \sin \gamma\,$ and $\,\frac{d}{\sqrt{c^2+d^2}} = \cos \gamma\,$.
After dividing by $\,(a^2+b^2)(c^2+d^2) \ne 0\,$, the identity can then be written as:
$$
(\sin\alpha\sin\gamma \pm \cos\alpha\cos\gamma)^2 + (\sin\alpha\cos\gamma \mp \cos\alpha\sin\gamma)^2 = 1
\\ \iff\quad \cos^2 (\alpha \mp \gamma) + \sin^2 (\alpha \mp \gamma) = 1 \quad\quad
$$
A: \begin{align} & (a^2+b^2)(c^2+d^2) \\ = \; & \Big(a^2-(-b^2)\Big)\Big(c^2+(-d^2)\Big) \\ = \; & \Big(a^2-(ib)^2\Big)\Big(c^2-(id)^2\Big) \\ = \; & (a+ib)(a-ib)(c+id)(c-id) \\ = \; & (a+ib)(c-id)(a-ib)(c+id) \\ = \; & \Big(ac+i(bc-ad)+bd\Big)\Big(ac-i(bc-ad)+bd\Big) \\ = \; & \Big((ac+bd)+i(bc-ad)\Big)\Big((ac+bd)-i(bc-ad)\Big) \\ = \; & (ac+bd)^2-\Big(i(bc-ad)\Big)^2 \\ = \; & (ac+bd)^2+(bc-ad)^2 \end{align}
A: \begin{align}& (a^2+b^2)(c^2+d^2) \\= \; & (ac)^2+(bc)^2+(ad)^2+(bd)^2 \\= \; & \Big((ac)^2+(bd)^2\Big)+\Big((ad)^2+(bc)^2\Big) \\= \; & \Big((ac)^2+2(ac)(bd)+(bd)^2\Big)+\Big((ad)^2-2(ad)(bc)+(bc)^2\Big) \\= \; & (ac+bd)^2+(ad-bc)^2\end{align}
