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

Question

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

Solutions in the answers.

$\ \\ \ \\ \ \\ \ \\$

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.

Answer 1

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}

Answer 2

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 $B=0$

Answer 3

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} $

Answer 4

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:

$$ (\sin\alpha\sin\gamma \pm \cos\alpha\cos\gamma)^2 + (\sin\alpha\cos\gamma \mp \cos\alpha\sin\gamma)^2 = 1 \\ \iff\quad \cos^2 (\alpha \mp \gamma) + \sin^2 (\alpha \mp \gamma) = 1 \quad\quad $$

Answer 5

\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}

Answer 6

\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}