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anonymous

  • one year ago

prove that (4n^2+1,144)=1

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  1. anonymous
    • one year ago
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    @ganeshie8

  2. ganeshie8
    • one year ago
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    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 \(4n^2+1\)

  3. thomas5267
    • one year ago
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    Is \(4n^2+1\not\equiv 0\pmod3\) equivalent to \(n^2+1\not\equiv 0\pmod3\)?

  4. ganeshie8
    • one year ago
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    definitely, they are equivalent

  5. ganeshie8
    • one year ago
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    but im not sure if OP has been introduced to congruences..

  6. thomas5267
    • one year ago
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    All I can think of no matter \(n\equiv0,1,2\pmod3,\,n^2+1\not\equiv0\pmod3\).

  7. thomas5267
    • one year ago
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    What I mean is by brute force.

  8. anonymous
    • one year ago
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    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.

  9. ganeshie8
    • one year ago
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    Looks good!

  10. thomas5267
    • one year ago
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    Basically he is doing modular arithmetic lol.

  11. ganeshie8
    • one year ago
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    Yes, but he thinks that he is using division algorithm... they are essentially same, just the notation is different

  12. ganeshie8
    • one year ago
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    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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