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

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

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

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

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

FoolForMathBest ResponseYou've already chosen the best response.7
hehe, 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)
 2 years ago

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

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

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

FoolForMathBest ResponseYou've already chosen the best response.7
Hero, I really wonder that myself ...
 2 years ago
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