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Topic: Integer Polynomial (Read 505 times) |
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JP05
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Let K,R,N be integers. Prove that 2^(1/2) K^2 - R pi^2 + N = 0 has no solutions.
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JP05
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Correction: Let K,R,N be integers > 0.
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towr
wu::riddles Moderator
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Re: Integer Polynomial
« Reply #2 on: Aug 10th, 2006, 3:23pm » |
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I have the feeling there's something more missing from the question.
Because [sqrt]2/pi2 isn't rational, which is sufficient cause to say the left side of the equation can never be an integer, let alone 0.
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| « Last Edit: Aug 10th, 2006, 3:23pm by towr » |
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Wikipedia, Google, Mathworld, Integer sequence DB
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JP05
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What you are saying makes sense but that is how I got it.
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Michael Dagg
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Re: Integer Polynomial
« Reply #4 on: Aug 10th, 2006, 5:23pm » |
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pi * pi = K^2 sqrt(2)/R + N/R
says that pi would lie in the ring Q( sqrt(2) ) and hence would satisfy
a quadratic polynomial with integer coefficients. Well, it doesn't.
You could get away with merely knowing that pi is not constructible this
way (i.e. lying in a chain of quadratic extensions of Q), which is equivalent to
the question of whether one can "square the circle"; but as far as I know
there is no easy proof of that either.
You could try to prove only the very weak claim that pi is not in this
quadratic extension field Q( sqrt(2) ). Since this claim is weaker than
the other two, it might have an easier proof. But I don't know of any.
I recall another problem in this forum much like this one and is in fact
equivalent to it.
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| « Last Edit: Aug 10th, 2006, 5:38pm by Michael Dagg » |
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Regards,
Michael Dagg
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