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Prove
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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For a,b integers greater than 1, if \(\sqrt{2ab} \in \mathbb{Z}\) then \(a^2n^4+b^2\) is not prime for n>1.
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I guess this is either going to be hard or easy I'm not sure which it is cause I know the answer lol
amilapsn
  • amilapsn
use \[a^2+b^2=(a+b)^2-2ab\]

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Yep that'll do it. :D
amilapsn
  • amilapsn
:D
ganeshie8
  • ganeshie8
Clever!
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I came up with this after reading this http://www.cut-the-knot.org/blue/McWarter.shtml to try to generalize this question at the end. :P
ganeshie8
  • ganeshie8
spoiler : \(\color{white}{\sqrt{2ab} \in \mathbb{Z}\implies ab=2c^2 \\~\\a^2n^4+b^2 = (an^2)^2+b^2 = (an^2+b)^2-2an^2b = (an^2+b)^2-(2cn)^2}\)
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Yeah that's exactly how I did, it, with c and everything haha.
ganeshie8
  • ganeshie8
little challenge : prove that \(5\) is a divisor whenever \(5\nmid n\)
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Oooh interesting I'm thinking.
ganeshie8
  • ganeshie8
sry that is a blunder, doesn't work...
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I am still interested in how you thought it would have worked though :D
ganeshie8
  • ganeshie8
let \(a=2mx^2\) and \(b=my^2\) and let \(5\nmid n\), then : \[a^2n^4+b^2 = 4m^2x^4n^4+m^2y^4\equiv 4m^2+m^2\equiv 0\pmod{5}\]
ganeshie8
  • ganeshie8
similarly a fake proof can be made for the other case : \(a=mx^2\) and \(b=2my^2\)
ganeshie8
  • ganeshie8
btw \(m\) is the square free factor, that must match in both \(a\) and \(b\) for \(2ab\) to be a perfect square
ganeshie8
  • ganeshie8
Here is the revised challenge (not a challenge anymore) : prove that \(5\) is a divisor whenever \(5\nmid n,a,b\)
ganeshie8
  • ganeshie8
i lost interest immediately because of the constraint, \(5\nmid n,a,b\)

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