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Title: Homogeneous cubic equation with no rational Post by Michael_Dagg on Oct 31st, 2006, 4:52pm 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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Title: Re: Homogeneous cubic equation with no ratio Post by Pietro K.C. on Dec 11th, 2006, 12:13am 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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Title: Re: Homogeneous cubic equation with no ratio Post by Michael_Dagg on Jan 5th, 2007, 5:55pm 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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