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Thread: Prove the pythagorean theorem

  1. #1
    Senior Member Mukilab's Avatar
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    Prove the pythagorean theorem

    Prove that $\displaystyle a^2+b^2=c^2$ using:
    $\displaystyle a=m^2-n^2$
    $\displaystyle b=2mn$
    $\displaystyle c=m^2+n^2$

    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
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  2. #2
    Master Of Puppets
    pickslides's Avatar
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    Quote Originally Posted by Mukilab View Post
    Prove that $\displaystyle a^2+b^2=c^2$ using:


    $\displaystyle a=m^2-n^2$
    $\displaystyle b=2mn$
    $\displaystyle c=m^2+n^2$
    For this proof to work you must use the LHS to show the RHS

    $\displaystyle a^2+b^2$

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

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

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

    $\displaystyle m^4+2n^2m^2+n^4$

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