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        riddles >> putnam exam (pure math) >> Integer Polynomial
        
(Message started by: JP05 on Aug 10^th, 2006, 12:42pm)

Title: Integer Polynomial
Post by JP05 on Aug 10^th, 2006, 12:42pm

Let K,R,N be integers. Prove that 2^(1/2) K^2 - R pi^2 + N = 0 has no solutions.

Title: Re: Integer Polynomial
Post by JP05 on Aug 10^th, 2006, 12:47pm

Correction: Let K,R,N be integers > 0.

Title: Re: Integer Polynomial
Post by towr on Aug 10^th, 2006, 3:23pm

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.

Title: Re: Integer Polynomial
Post by JP05 on Aug 10^th, 2006, 4:57pm

What you are saying makes sense but that is how I got it.

Title: Re: Integer Polynomial
Post by Michael_Dagg on Aug 10^th, 2006, 5:23pm

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.