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Math Help - Factorize:

  1. #1
    Member great_math's Avatar
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    Factorize:

    Can a^4+b^4 be factorized unlike a^2+b^2 ?

    I think it can not be but there is a battle going on with me and my friend about this one.. So i just want to check it.
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  2. #2
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by great_math View Post
    Can a^4+b^4 be factorized unlike a^2+b^2 ?

    I think it can not be but there is a battle going on with me and my friend about this one.. So i just want to check it.
    How would you factor that? I think your friend is mixing up

    a^2-b^2 = (a+b)(a-b)
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by great_math View Post
    Can a^4+b^4 be factorized unlike a^2+b^2 ?

    I think it can not be but there is a battle going on with me and my friend about this one.. So i just want to check it.
    If we were dealing with a^3+b^3, a^5+b^5 and so on (the odd exponents), it can be factored.

    Since we are dealing with a^2+b^2 or a^4+b^4, it won't factor.

    On the other hand, it is interesting to note that a^6+b^6, a^{10}+b^{10} and others can be factored, whereas a^8+b^8, a^{16}+b^{16}, a^{32}+b^{32}, etc. cannot be factored....


    In general, a^k+b^k;~~k\geq 2 can only be factored when k\neq 2,~4,~8,~16,~32,...,~2^n;~~\forall~n\in\mathbb{N}

    --Chris
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  4. #4
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    Quote Originally Posted by great_math View Post
    Can a^4+b^4 be factorized unlike a^2+b^2 ?

    I think it can not be but there is a battle going on with me and my friend about this one.. So i just want to check it.
    They can't be factorised using REAL factors. But they can be factorised using complex factors eg. a^2 + b^2 = (a + ib)(a - ib).
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  5. #5
    Member great_math's Avatar
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    Can a^4+b^4 be factorized using complex numbers ?
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    Quote Originally Posted by great_math View Post
    Can a^4+b^4 be factorized using complex numbers ?
    Yes:

     \left(a - \left[ \frac{\sqrt{2} + i \sqrt{2}}{2}\right] b\right) \, \left(a - \left[ \frac{\sqrt{2} - i \sqrt{2}}{2}\right] b\right) \left(a <br />
+ \left[ \frac{\sqrt{2} - i \sqrt{2}}{2}\right] b\right) \, \left(a + \left[ \frac{\sqrt{2} + i \sqrt{2}}{2}\right] b\right).
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  7. #7
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    Quote Originally Posted by great_math View Post
    Can a^4+b^4 be factorized using complex numbers ?
    a^4+b^4=\left(a+\dfrac12 \sqrt{b} + \dfrac12 \cdot i \cdot \sqrt{b}  \right)\left(a+\dfrac12 \sqrt{b} - \dfrac12 \cdot i \cdot \sqrt{b}  \right)  \left(a-\dfrac12 \sqrt{b} + \dfrac12 \cdot i \cdot \sqrt{b}  \right) \left(a-\dfrac12 \sqrt{b} - \dfrac12 \cdot i \cdot \sqrt{b}  \right)


    EDIT: Too late ... again!
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  8. #8
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    or on real numbers: a^4 + b^4 = (a^2 + b^2 - \sqrt{2}ab)(a^2 +b^2 + \sqrt{2}ab). now try to factorize a^8 + b^8 over real numbers! how many factors do you think a^{2^n} + b^{2^n} will have over real numbers?
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  9. #9
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    Quote Originally Posted by NonCommAlg View Post
    or on real numbers: a^4 + b^4 = (a^2 + b^2 - \sqrt{2}ab)(a^2 +b^2 + \sqrt{2}ab). now try to factorize a^8 + b^8 over real numbers!
    Yes, there's a nice re-arrangement that makes it fall out nicely.

    For the OP: a^4 + b^4 = (a^4 + b^4 + 2a^2 b^2) - 2a^2 b^2 = (a^2 + b^2)^2 - (\sqrt{2} \, a b)^2 = \, ....
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