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FoolForMath
 3 years ago
Best ResponseYou've already chosen the best response.7a=b=1 is the only possible solution.

Mr.Math
 3 years ago
Best ResponseYou've already chosen the best response.1Could you prove your claim? :P

FoolForMath
 3 years ago
Best ResponseYou've already chosen the best response.7Yes I definitely can ... why you don't believe me MR.Math? :(

Mr.Math
 3 years ago
Best ResponseYou've already chosen the best response.1lol, I do believe you! Where did I say that I don't? :(

Hero
 3 years ago
Best ResponseYou've already chosen the best response.0FFM, stop stalling and post the proof :P

FoolForMath
 3 years ago
Best ResponseYou've already chosen the best response.7hehe, okay here it goes, \[ a^4+4b^4= ((a+b)^2+b^2) \times ((ab)^2+b^2)) \] Now, \( ((a+b)^2+b^2) \gt 1\) (always) so for \( a^4+4b^4 \) to be prime \((ab)^2+b^2) =1\) and this can only happen when \(a=b=1\) (QED)

Mr.Math
 3 years ago
Best ResponseYou've already chosen the best response.1Awesome! I've always believed in you son :)

FoolForMath
 3 years ago
Best ResponseYou've already chosen the best response.7Haha, how old are you man? :D

Hero
 3 years ago
Best ResponseYou've already chosen the best response.0FFM, how do you come up with such approaches to proofs?

FoolForMath
 3 years ago
Best ResponseYou've already chosen the best response.7Hero, I really wonder that myself ...
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