x^2 = 4*y^3 + 1
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From: James Waldby on 11 Aug 2010 15:53 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
From: Timothy Murphy on 11 Aug 2010 16:30 achille wrote: > On Aug 11, 5:29 pm, Timothy Murphy <gayle...(a)eircom.net> wrote: >> 3. The Nagell-Lutz theorem says that a point of finite order >> has integer coordinates. >> >> Also y^2 | 3D , where D is the discriminant, ie >> > I'm confused, shouldn't it be y^2 | D instead of 3D ??? I'm sure you're right, I'm away from home (in Italy) and don't have any books handy. I calculated that if the elliptic curve has equation y^2 = x^3 + bx + c (ie the coeff of x^2 vanishes) then y^2 | D, and then it follows that in the general case y^2 | 3D. But the stronger condition probably holds even if a is not 0.
From: quasi on 11 Aug 2010 17:58 On Wed, 11 Aug 2010 19:53:17 +0000 (UTC), James Waldby <no(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. 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. quasi
From: James Waldby on 11 Aug 2010 18:43 On Wed, 11 Aug 2010 16:58:54 -0500, quasi wrote: > On Wed, 11 Aug 2010 19:53:17 +0000 (UTC), James Waldby <no(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. > > 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? Eg: Suppose a = (c/d)^3, with a,c,d integers; (c,d)=1; d > 1. From a*d^3 = c^3, c^3 | a because (c,d)=1. Then a = r*c^3 for some integer r, so c^3 = a*d^3 = r * c^3 * d^3, whence r * d^3 = 1 so that d = 1, contradicting assumption that d > 1. -- jiw
From: quasi on 11 Aug 2010 19:57
On Wed, 11 Aug 2010 22:43:19 +0000 (UTC), James Waldby <no(a)no.no> wrote: >On Wed, 11 Aug 2010 16:58:54 -0500, quasi wrote: > >> On Wed, 11 Aug 2010 19:53:17 +0000 (UTC), James Waldby <no(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. >> >> 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? quasi |