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Mr.Math

  • 4 years ago

Find all integers a,b for which \(a^4+4b^4\) is a prime.

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  1. FoolForMath
    • 4 years ago
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    a=b=1 is the only possible solution.

  2. Mr.Math
    • 4 years ago
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    Could you prove your claim? :-P

  3. Hero
    • 4 years ago
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    Yeah FFM

  4. FoolForMath
    • 4 years ago
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    Yes I definitely can ... why you don't believe me MR.Math? :(

  5. Mr.Math
    • 4 years ago
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    lol, I do believe you! Where did I say that I don't? :-(

  6. Hero
    • 4 years ago
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    FFM, stop stalling and post the proof :P

  7. FoolForMath
    • 4 years ago
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    hehe, okay here it goes, \[ a^4+4b^4= ((a+b)^2+b^2) \times ((a-b)^2+b^2)) \] Now, \( ((a+b)^2+b^2) \gt 1\) (always) so for \( a^4+4b^4 \) to be prime \((a-b)^2+b^2) =1\) and this can only happen when \(a=b=1\) (QED)

  8. Mr.Math
    • 4 years ago
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    Awesome! I've always believed in you son :-)

  9. FoolForMath
    • 4 years ago
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    Haha, how old are you man? :D

  10. Mr.Math
    • 4 years ago
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    Very old.

  11. Hero
    • 4 years ago
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    FFM, how do you come up with such approaches to proofs?

  12. FoolForMath
    • 4 years ago
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    Hero, I really wonder that myself ...

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