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

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a=b=1 is the only possible solution.
Could you prove your claim? :-P
Yeah FFM

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Yes I definitely can ... why you don't believe me MR.Math? :(
lol, I do believe you! Where did I say that I don't? :-(
FFM, stop stalling and post the proof :P
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)
Awesome! I've always believed in you son :-)
Haha, how old are you man? :D
Very old.
FFM, how do you come up with such approaches to proofs?
Hero, I really wonder that myself ...

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