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 March 7th, 2013, 06:14 AM #1 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Diophantine equation It seems like $x^3 + y^3= d x y$ has all the integer solutions either $x= y = \frac{d}{2}$ and/or $x= \frac{y}{2}$ assumed x < y. Can one give a proof/counterexample?
 March 7th, 2013, 08:14 AM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Diophantine equation (x, y, d) = (6, 18, 56). There are infinitely many such counterexamples; probably the counterexamples are asymptotically dominant with any reasonable measure.
March 7th, 2013, 09:48 AM   #3
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Re: Diophantine equation

Okay, thanks. I was being fooled by the behavior of small d's.

Quote:
 Originally Posted by CRGreathouse probably the counterexamples are asymptotically dominant with any reasonable measure.
Can you elaborate a bit more?

 March 7th, 2013, 10:51 AM #4 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Diophantine equation If you find all (x, y, d) satisfying 0 < x <= y < M and the equation, I guess that the percentage which meet your conditions have asymptotic density 0.
March 7th, 2013, 11:47 AM   #5
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Re: Diophantine equation

Quote:
 Originally Posted by mathbalarka Can you elaborate a bit more?
dxy-x^3=y^3;
Let x=ky where k - rational number;
dky^2-(ky)^3=y^3 or dk=y(k^3+1);
k(d-yk^2)=y;
If d=2y then k=1;
d=(9/2)y then k=2;
d=(28/3)y then k=3; etc.

 March 7th, 2013, 12:18 PM #6 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Diophantine equation The only thing that matters in this problem is that y | x^3 and x | y^3, so just make sure that they have the same primes and they're within a multiple of 3 of each other. So if one is divisible by (exactly) 2^4, the other needs to have 2 raised to a power between 2 = ceil(4/3) and 12 = 4*3.
 March 7th, 2013, 04:54 PM #7 Senior Member   Joined: Nov 2010 From: Berkeley, CA Posts: 174 Thanks: 35 Math Focus: Elementary Number Theory, Algebraic NT, Analytic NT Re: Diophantine equation I'm not sure if this will be helpful, but your equation can be written in parametric form as $x= \frac{dt}{1 + t^3}$ $y= \frac{dt^2}{1 + t^3}$ For the derivation of these equations see, for example, the Wikipedia article on the Folium of Descartes. In particular, y = tx. In CRG's solution (6, 18, 56), we have that t = 3.

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