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anonymous
 one year ago
prove that (4n^2+1,144)=1
anonymous
 one year ago
prove that (4n^2+1,144)=1

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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Notice that \(144 = 12^2 = 2^43^2\) \(4n^2+1\) is always odd, so you just need to show that \(3\) does not divide \(4n^2+1\)

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1Is \(4n^2+1\not\equiv 0\pmod3\) equivalent to \(n^2+1\not\equiv 0\pmod3\)?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2definitely, they are equivalent

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2but im not sure if OP has been introduced to congruences..

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1All I can think of no matter \(n\equiv0,1,2\pmod3,\,n^2+1\not\equiv0\pmod3\).

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1What I mean is by brute force.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i think we should solve this question using this solution . if n=3k then 4n^2+1 is not divisible by 3. if n=3k+1 then 4n^2+1=4(3k+1)^2+1=4*9k^2+4*6k+4+1=3k'+5 that is not divisible by 3. if n=3k+2 then 4n^2+1=4(3k+2)^2+1=4*9k^2+4*12k+16+1=3k'+17=3k''+2 that is not divisible by 3.

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1Basically he is doing modular arithmetic lol.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Yes, but he thinks that he is using division algorithm... they are essentially same, just the notation is different

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2By division algorithm, any integer can be expressed uniquely as \(3k+r\) where \(0\le r\lt 3\) so it is sufficient to check the \(3\) cases : \(n=3k,3k+1, 3k+2\) they cover all the integers, \(n\).
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