x^2 = 4*y^3 + 1
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From: James Waldby on 11 Aug 2010 19:08 On Wed, 11 Aug 2010 18:57:44 -0500, quasi wrote: > On Wed, 11 Aug 2010 22:43:19 +0000 (UTC), James Waldby wrote:> >>On Wed, 11 Aug 2010 16:58:54 -0500, quasi wrote:>> >>> On Wed, 11 Aug 2010 19:53:17 +0000 (UTC), James Waldby wrote:>>> >>>>On Wed, 11 Aug 2010 02:55:20 -0400, M.A.Fajjal wrote: >>>> >>>>> Is there any rational solution for >>>>> x^2 = 4*y^3 + 1 [snip] >>> What you showed above is that there are no non-trivial integer >>> solutions. >>> >>> But the problem was to find all _rational_ solutions. >>> >>> However, assuming you already knew that any rational solution must, in >>> fact, be an integer solution (but I don't think that's so elementary), >>> then your argument finishes the problem nicely. >> >>I might be mistaken, but isn't it elementary that if integer a is a cube >>of a rational, then it is a cube of an integer? > > Yes, but how do you use that to show that a supposed rational solution > to the original equation must be an integer solution? How, indeed. Moments after sending my reply above, I realized I had missed the point you made, that my algebra was based on an unwarranted assumption that x or y are integers. Of course I now agree with you, but unfortunately can't unpost my earlier reply... -- jiw
From: alainverghote on 12 Aug 2010 06:17 On 11 août, 21:53, James Waldby <n...(a)no.no> wrote: > On Wed, 11 Aug 2010 02:55:20 -0400, M.A.Fajjal wrote: > > Is there any rational solution for > > x^2 = 4*y^3 + 1 > > W Elliot and quasi both mentioned trivial solutions > x = +/- 1, y = 0, while T Murphy ruled out non-trivial > solutions via an elliptic curves argument. I think the > following simple algebra also rules out other solutions: > > x^2 = 4*y^3 + 1 ==> x^2 - 1 = 4*y^3 = (x-1)(x+1) = > (2a+1-1)(2a+1+1) (as 4y^3 is even, both x-1 and x+1 > must be even) ==> 4a(a+1) = 4y^3 ==> a*(a+1) = y^3. > Now a and a+1 have no common factors; hence both a and > a+1 must be integer cubes (of integers) which cannot be > if a != 0. > > -- > jiw Bonjour, That seems very nice! It is faster than an infinite descent approach, Alain
From: Timothy Murphy on 12 Aug 2010 07:50
alainverghote(a)gmail.com wrote: >> On Wed, 11 Aug 2010 02:55:20 -0400, M.A.Fajjal wrote: >> > Is there any rational solution for >> > x^2 = 4*y^3 + 1 >> >> W Elliot and quasi both mentioned trivial solutions >> x = +/- 1, y = 0, while T Murphy ruled out non-trivial >> solutions via an elliptic curves argument. I think the >> following simple algebra also rules out other solutions: >> >> x^2 = 4*y^3 + 1 ==> x^2 - 1 = 4*y^3 = (x-1)(x+1) = >> (2a+1-1)(2a+1+1) (as 4y^3 is even, both x-1 and x+1 >> must be even) ==> 4a(a+1) = 4y^3 ==> a*(a+1) = y^3. >> Now a and a+1 have no common factors; hence both a and >> a+1 must be integer cubes (of integers) which cannot be >> if a != 0. > That seems very nice! > It is faster than an infinite descent approach, I don't think there is really a choice. The OP asked if there were any _rational_ solutions. The argument above shows that there are no non-trivial _integer_ solutions. I think you would have to use infinite descent, or something equivalent, to show that the rank is 0. If you know the rank is 0 it follows that any solutions are integral, and you don't need to use infinite descent. |