

Title: Cubic Diophantine Equation Post by Barukh on Oct 31^{st}, 2008, 10:15am Let p = 2^{n} + 1 be a prime number. How many integer solutions mod p has the following equation: Note: The equation was changed. 

Title: Re: Cubic Diophantine Equation Post by Eigenray on Oct 31^{st}, 2008, 3:36pm Well, to start with, it is the integer closest to p which is congruent to [hide]the coefficient of x^{p1} in (x^{3} 

Title: Re: Cubic Diophantine Equation Post by Barukh on Oct 31^{st}, 2008, 9:45pm Sorry, I misstated the problem (which made it much harder IMHO). Let's try to go with the easier one first... Sorry for inconvenience. 

Title: Re: Cubic Diophantine Equation Post by Eigenray on Oct 31^{st}, 2008, 10:25pm Yes that is much easier. But why is p a Fermat prime? It's enough that p is not 1 mod 3. Or did you have a different proof in mind? Theorem: For a cubic polynomial f(x), the number of solutions to y^{2} http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/equiv.gif f(x) mod p is congruent, mod p, to the coefficient of x^{p1} in f(x)^{(p1)/2}. But I've seen this result before so it would feel like cheating to give the proof right away. Does someone else want to try? 

Title: Re: Cubic Diophantine Equation Post by Barukh on Nov 1^{st}, 2008, 12:26am on 10/31/08 at 22:25:17, Eigenray wrote:
No, your condition is sufficient, and the proof I had in mind uses this condition. My formulation is a special case of that. Quote:
I haven't heard about this theorem before, but after seeing it, it does make sense, and probably is based on [hide]Euler criterion for quadratic residues[/hide]. 

Title: Re: Cubic Diophantine Equation Post by Eigenray on Nov 1^{st}, 2008, 6:31pm It suddenly hit me that there's a simpler solution that I didn't notice because I had been thinking about the harder problem: [hide]everything is a cube[/hide]. 

Title: Re: Cubic Diophantine Equation Post by Barukh on Nov 2^{nd}, 2008, 8:31am on 11/01/08 at 18:31:18, Eigenray wrote:
If I get you right, yes, that's the solution I had in mind. Very nice! 

Title: Re: Cubic Diophantine Equation Post by Eigenray on Nov 2^{nd}, 2008, 11:50am Here are two related problems: how many solutions are there to: (1) y^{2} = x^{3} + ax mod p, p a Mersenne prime ;) (2) y^{2} = x^{3} + ax^{2} mod p. Both can be answered using the theorem I quoted, but there are also more direct(?) proofs. 

Title: Re: Cubic Diophantine Equation Post by Barukh on Nov 3^{rd}, 2008, 11:38pm on 11/02/08 at 11:50:56, Eigenray wrote:
Assuming a http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ne.gif 0 mod p, I get the following: 1) p 2) p  (a/p), where the last is Legendre symbol (http://en.wikipedia.org/wiki/Legendre_symbol#Definition). 

Title: Re: Cubic Diophantine Equation Post by Eigenray on Nov 4^{th}, 2008, 11:11pm Yep. I thought it was interesting how the three problems can be solved individually using quite distinct arguments, or all using that one theorem I quoted. 

Powered by YaBB 1 Gold  SP 1.4! Forum software copyright © 20002004 Yet another Bulletin Board 