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Topic: Homogeneous cubic equation with no rational (Read 939 times) |
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Michael Dagg
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Homogeneous cubic equation with no rational
« on: Oct 31st, 2006, 4:52pm » |
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Suppose f(x,y) is a homogeneous cubic equation over Z that has no nontrivial
roots modulo some prime p. (For example, x^3 - xy^2 + y^3 works for p=2,3).
Show that
f(x_0,x_1) + p f(x_2, x_3) = 0
has no rational solutions.
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Regards,
Michael Dagg
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Pietro K.C.
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Re: Homogeneous cubic equation with no ratio
« Reply #1 on: Dec 11th, 2006, 12:13am » |
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Suppose there was indeed a solution (q0,q1,q2,q3) in Q. Let M be the gcd of the denominators of q0,q1, and similarly for N and q2,q3. Write zi=Mqi for i=0,1 and zi=Nqi for i=2,3.
Then one may multiply both sides of
f(q0,q1) + p f(q2,q3) = 0
by M3N3 to get
N3 f(z0,z1) + pM3 f(z2,z3) = 0
since f is homogeneous. Consider the equation
N3 f(z0,z1) = -pM3 f(z2,z3).
Since f(x,y) is only ever divisible by p if x and y are, and it is homogeneous of degree 3, then the highest power of p dividing f(x,y) can only ever be a multiple of 3. Same goes for M3 and N3, more trivially (p is a prime). This means that p can only appear as a factor of the left side with exponent 3m; but it appears on the right with exponent 3n+1. Contradiction. Therefore the proposed equation has no rational roots.
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"I always wondered about the meaning of life. So I looked it up in the dictionary under 'L' and there it was --- the meaning of life. It was not what I expected." (Dogbert)
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Michael Dagg
Senior Riddler
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Posts: 500
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Re: Homogeneous cubic equation with no ratio
« Reply #2 on: Jan 5th, 2007, 5:55pm » |
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Dang, I didn't see this. I am not seeing NEW stuff on page
load or getting an email when there is a post. Nice approach!
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Regards,
Michael Dagg
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