Prove the pythagorean theorem

• March 9th 2010, 12:30 PM
Mukilab
Prove the pythagorean theorem
Prove that $a^2+b^2=c^2$ using:
$a=m^2-n^2$
$b=2mn$
$c=m^2+n^2$

So far I have got:
(m*m)(m*m)-(n*n)(n*n)+4(m*m)(n*n)=(m*m)(m*m)-(n*n)(n*n)

What can I do with 4(m*m)(n*n)?
Can I do (m*m)+(m*m)+(m*m)+(m*m)*(n*n)+(n*n)+(n*n)+(n*n)

???

~Cyf
• March 9th 2010, 12:45 PM
pickslides
Quote:

Originally Posted by Mukilab
Prove that $a^2+b^2=c^2$ using:

$a=m^2-n^2$
$b=2mn$
$c=m^2+n^2$

For this proof to work you must use the LHS to show the RHS

$a^2+b^2$

$(m^2-n^2)^2+(2mn)^2$

$(m^2-n^2)(m^2-n^2)+4m^2n^2$

$m^4-2n^2m^2+n^4+4m^2n^2$

$m^4+2n^2m^2+n^4$

Now factor this term and you will get $c^2$