BdMO Online Forum

\[(a^{2}+b^{2})(c^{2}+d^{2})(e^{2}+f^{2})\]
express this equation as the sum of two squares.

\[(a^2c^2+a^2d^2+b^2c^2+b^2d^2) (e^2+f^2)\]
\[e^2(a^2c^2+a^2d^2+b^2c^2+b^2d^2)+f^2(a^2c^2+a^2d^2+b^2c^2+b^2d^2) \]
\[e^2((ac+bd)^2+(ad-bc)^2)+f^2((ac+bd)^2+(ad-bc)^2) \]

Now let $ac+bd=x,ad-bc=y$

\[e^2(x^2+y^2)+f^2(x^2+y^2) \]
\[e^2x^2+e^2y^2+f^2x^2+f^2y^2 \]
\[(ex+fy)^2+(ey-fx)^2\]

\[(e(ac+bd)+f(ad-bc))^2+(e(ad-bc)-f(ac+bd))^2\]

thank u soooo much!!

Proof the general version now,$\prod (a^{2}_{i}+a^{2}_{j})$ can be shown as sum of two squares. (Very easy with a simple trick)

Factorisation!

I don't understand the signs... Explain them...

$\prod_{i=1}^{n-1} (a_{i}^2 + a_{i+1}^2)=(a_{1}^2 + a_{2}^2)(a_{3}^2 + a_{4}^2).....(a_{n-1}^2 + a_{n}^2)$

here's the crux move-if two integers are sum of two squares,then their product is also sum of two squares.

Or use complex numbers to write $a^2+b^2=(a+bi)(a-bi)$ and use the fact that multiplications of any two numbers in the form $a+bi$ or $a-bi$ also has that form.So it follows directly.

Thanks for telling the meaning... Sourav

and mine is like tahmid's...

\[(a^2+b^2)(c^2+d^2)\]
can be written as
\[(ad+bc)^2+(ac-bd)^2\]

We can continue in this way...

And isn't the two lines I wrote enough to prove

if two integers are sum of two squares,then their product is also sum of two squares