Prob!

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Mathlover
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Prob!

Unread post by Mathlover » Thu Sep 15, 2011 9:24 pm

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

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Abdul Muntakim Rafi
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Re: Prob!

Unread post by Abdul Muntakim Rafi » Thu Sep 15, 2011 9:57 pm

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

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

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

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

:D
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Mathlover
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Re: Prob!

Unread post by Mathlover » Thu Sep 15, 2011 10:05 pm

thank u soooo much!!

sourav das
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Re: Prob!

Unread post by sourav das » Thu Sep 15, 2011 11:18 pm

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)
You spin my head right round right round,
When you go down, when you go down down......
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*Mahi*
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Re: Prob!

Unread post by *Mahi* » Thu Sep 15, 2011 11:32 pm

Factorisation!
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Use $L^AT_EX$, It makes our work a lot easier!

Nur Muhammad Shafiullah | Mahi

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Abdul Muntakim Rafi
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Re: Prob!

Unread post by Abdul Muntakim Rafi » Fri Sep 16, 2011 12:13 pm

I don't understand the signs... :( Explain them...
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sourav das
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Re: Prob!

Unread post by sourav das » Fri Sep 16, 2011 3:25 pm

$\prod_{i=1}^{n-1} (a_{i}^2 + a_{i+1}^2)=(a_{1}^2 + a_{2}^2)(a_{3}^2 + a_{4}^2).....(a_{n-1}^2 + a_{n}^2)$
You spin my head right round right round,
When you go down, when you go down down......
(-$from$ "$THE$ $UGLY$ $TRUTH$" )

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Tahmid Hasan
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Re: Prob!

Unread post by Tahmid Hasan » Fri Sep 16, 2011 4:25 pm

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

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*Mahi*
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Re: Prob!

Unread post by *Mahi* » Fri Sep 16, 2011 6:04 pm

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.
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Abdul Muntakim Rafi
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Re: Prob!

Unread post by Abdul Muntakim Rafi » Fri Sep 16, 2011 6:36 pm

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... :D

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
Man himself is the master of his fate...

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