anonymous
  • anonymous
prove that (4n^2+1,144)=1
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
@ganeshie8
ganeshie8
  • ganeshie8
Notice 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
  • thomas5267
Is \(4n^2+1\not\equiv 0\pmod3\) equivalent to \(n^2+1\not\equiv 0\pmod3\)?

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More answers

ganeshie8
  • ganeshie8
definitely, they are equivalent
ganeshie8
  • ganeshie8
but im not sure if OP has been introduced to congruences..
thomas5267
  • thomas5267
All I can think of no matter \(n\equiv0,1,2\pmod3,\,n^2+1\not\equiv0\pmod3\).
thomas5267
  • thomas5267
What I mean is by brute force.
anonymous
  • anonymous
i 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.
ganeshie8
  • ganeshie8
Looks good!
thomas5267
  • thomas5267
Basically he is doing modular arithmetic lol.
ganeshie8
  • ganeshie8
Yes, but he thinks that he is using division algorithm... they are essentially same, just the notation is different
ganeshie8
  • ganeshie8
By 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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