Third level Diophantine formula

I saw a workout which was suggested to locate integers all integers $m,n$ pleasing $2m^2 + 5n^2 = 11(mn-11)$. I located them making use of the factorization $(m-5n)(2m-n)=-11\cdot 11$. Nonetheless, what sort of approaches there are to address the initial trouble, locate all $(m,n)\in\mathbb{Z}\times\mathbb{Z}$ pleasing $2m^2+5n^3=11(mn-11)$? I have not addressed several cubic Diophantine equations so I was simply asking yourself if there is some birational makeover to transform the formula to a Weierstrass kind of an elliptic contour.

2019-12-02 02:49:17
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The formula $2y^2+5x^3=11(xy−11)$ defines an elliptic contour. Offered any kind of elliptic contour, you can execute an adjustment of variables to place it right into Weierstrass kind, and also if the area of definition has particular various from 2 or 3, you can also place it right into the kind $y^2 = x^3 + ax + b$. In this certain instance, to locate the makeover is very easy : first range $x$ and also $y$ to remove make the coefficients of $x^3$ and also of $y^2$ equivalent to 1. After that write $y' = y + \alpha x$ to remove the $xy$ - term. This will certainly present an $x^2$ term. So currently, convert $x$ to remove the $x^2$ term (I have not in fact done the calculation).

A vital point to note is that while the idea of sensible factors does not rely on the version over $\mathbb{Q}$ that you are collaborating with (as long as you just transform variables over $\mathbb{Q}$), the idea of indispensable factors does rely on the indispensable version. A theory of Siegel claims that on any kind of Weierstrass version of an elliptic contour, there are just finitely several indispensable factors, and also there are bounds on their height in regards to the coefficients of the version. Yet regarding I recognize, to in fact locate the factors boils down to strength approaches : as an example if you take care of to calculate generators of the Mordell - Weil team, i.e. of the team of the sensible factors, and also the torsion subgroup, after that you simply examine all feasible straight mixes approximately the offered bound. There are far better approaches making use of elliptic logarithms, yet they are in theory extra entailed.

Modify : some even more details on your concrete contour : To get a Weierstrass version you can change $y$ by $y'/20$ and also $x$ by $-x'/10$ in the initial formula, causing $y'^2+11x'y' = x'^3 - 2^35^211^2$, validating your tip. The elliptic contour has ranking 1, as you claim, and also no torsion. The factor $P=(22,-88)$ creates all the sensible factors on the contour under the team regulation. Indispensable remedies of the initial formula represent factors on the Weierstrass version with indispensable $x$ - coordinate divisible by 10 and also indispensable $y$ - coordinate divisible by 20 (see our makeover). So the ignorant means of dismissing indispensable remedies to the initial formula is to examine all multiples of the generator $P=(22,-88)$ under the team regulation approximately the offered bound and also encourage on your own that no factor pleases the divisibility standards. Nonetheless, the bound of Baker, described in the wikipedia write-up, is massive and also the calculation could not in fact be viable. A perhaps extra encouraging strategy is to list the polynomial that calculates $Q\mapsto Q\oplus P$ and also see whether this procedure constantly presents greater and also greater that never ever terminate the numerators.

2019-12-03 04:09:27