From: rusin@vesuvius.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: A Diophantine problem involving tangents. Date: 5 Oct 1998 15:57:28 GMT Ray Steiner wrote: >The following problem was posed to me by M. Gianotto. It appears to be >quite difficult. >Does anyone have an idea how to attack it? >Let D be an angle between 0 and pi/2 with the property that tan D is a >rational number >but not an integer. Let p be an odd prime number and suppose that >y^p= tan pD/tan D. Can y ever be rational? >More generally, what can one say about y if y^p= tan kD/tan D, where k is >a natural number >larger than 1? Probably this problem can be resolved for any particular p and k but I don't think we have the tools to do all cases at once. Let x= tan(D); then for any k, f_k=tan(kD)/tan(D) is a rational function of x. Indeed, f_1 = 1 and f_{k+1} = (1 + f_k)/(1 - x^2 f_k). So the equation y^p = f_k(x) determines an algebraic variety of dimension 1 in affine 2-space, that is, a planar algebraic curve. To each curve we may associate a number called the _genus_, which is significant for number theory: 1. When the genus is zero, if there are any rational points at all, there are infinitely many, and they may be explicitly presented as (x,y) = rational functions of a rational parameter m 2. When the genus is 1, the set of rational points may be given the structure of a finitely-generated abelian group; no parameterization of the solutions is possible, but a pattern for generating all solutions may be given (modulo some potential computational pitfalls which we don't know we can always surmount). 3. When the genus is 2 or more, there are at most finitely many solutions. The bound is not effective in general, that is, we can't tell in advance how far to look for solutions; in some cases knowledge of the Jacobian of the curve (another finitely-generated abelian group) can limit the solution set more effectively. For each (k,p) it is possible to compute the genus, and I suspect one could computed it generally as a function of k and p but I was too lazy to do so. Relying instead on software I found that the genus increased fairly rapidly with k and p and in particular was almost always of genus greater than 1 ; as noted above this leads to the conclusion that for almost every pair (k,p) there are at most a finite number of rational solutions to y^p = f_k(x). Some cases are of small genus: 1. Naturally if p=1 or k=1 the curve has genus zero and the solution set is given explicitly by the defining equation. 2. If k=p=2, we have the equation y^2=2/(1-x^2). Setting y=1/z gives x^2+2z^2=1, whose rational solutions are all given by (x,z) = ((2m^2-1)/(2m^2+1), 2m/(2m^2+1)) for m rational. 3. More generally, when k=2 the substitution y=1/z transforms the curve into the curve x^2 = 1 - 2 z^p. For p>4 this is hyperelliptic (in particular, has genus > 1) but for p=3 and p=4 it is of genus 1. For p=3, this is an elliptic curve with rank 0, that is, the only points are the three torsion points -- one at infinity and the other two with z=0 (x=+-1). For p=4, this is becomes an elliptic curve once we pick, say, (x,z)=(0,1) to be the identity element of the group. The elliptic curve has one other torsion point (namely (x,z)=(0,-1) of order 2) and a free rank of 1, with generator (x,z)=(7/9,2/3). Negation in the group, and addition of a torsion element, correspond to changing the signs of x and z respectively. The first few multiples of the generator are (-7967/12769, 84/113) (-3262580153/3263037129, 6214/57123) (-60912456065182847/68970122119586689, -151245528/262621633) "and so on". 4. When p=2 we similarly have (elliptic or) hyperelliptic curves which are equivalent to Y^2=(numerator of f_k)*(denominator of f_k). Again these are all finite solution sets except when the genus is 1, which occurs only for k=3: from the equation 2 2 -3 + x y = --------- 2 -1 + 3 x we deduce 3x^4 - 10x^2 + 3 must be a square. But this is not square for any rational x; indeed, modulo 16 it only takes on the values 3, 11, and 12 while all squares are congruent to 0, 1, 4, or 9. (More precisely: this homogeneous space has no rational points because it doesn't have any points in the 2-adic completion of Q.) 5. The data suggest there are no other combinations (k,p) which describe a curve of genus less than 2, so that all other combinations lead at most to finitely many rational points. I have not attempted to compute the ranks of their Jacobians so as to peform a complete search for rational points. A few promising combinations would be those which are of genus 2: (k,p) = (2,5) (2,6) (3,3) (4,2) of genus 3: (k,p) = (2,7) (2,8) (3,4) (5,2) (I believe those lists are complete.) 6. The original question concerned the cases k=p; here the genera begin p=3,g=2; p=4,g=7; p=5,g=12; p=6,g=22; p=7,g=30; ... I don't have the expertise to work with curves of genus 30; I suspect no one does without taking advantage of some special structure in the curve which I have not noticed. (The fact that p is supposed to be prime and that x is supposed to be positive and non-integral have not entered this discussion at all.) On must ask why this question is considered "natural" enough to pursue. It seems to me that perhaps asking when f_k(x) = f_k(y) would be a better candidate, unless the original question did indeed arise in some other investigation. For what it's worth, I believe I saw this problem before, possibly in this newsgroup, but I haven't any idea what responses were generated. dave ============================================================================== Remark: Other computed genera include (k,p,g)=(2,p,floor((p-1)/2)), (3,p,p-1), (k,2,k-2), (4,3,5), (5, 3, 6). Computations done with CASA package for maple. ============================================================================== From: rusin@vesuvius.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: A Diophantine problem involving tangents. Date: 6 Oct 1998 05:45:17 GMT Earlier today I discussed this Diophantine equation: >>More generally, what can one say about y if y^p= tan kD/tan D, where k is >>a natural number >>larger than 1? I noted that for each (k,p) this describes a curve C(k,p) in the plane, typically of high genus. I should have pointed out that there are obvious epimorphisms C(k,p) -> C(k,d) whenever d|p, and moreover there are coverings C(k,p) -> C~(k,p) where C~ is the curve described by replacing x^2 by x in the equation defining C. (It is easily checked by induction on k that C(k,p) is defined by a polynomial which is _even_ in x=tan(D).) This, perhaps, is why the problem was originally posed with the stipulation that p is prime; my previous response >(The fact that p is supposed to be prime [...] >[has] have not entered this discussion at all.) was a bit wide of the mark. (I don't see how to reduce the curve C(n,k) to any other C(h,k'), however; we can't assume that k is prime.) The reason to invoke these morphisms is that they give an upper bound on the size of the rational locus of C itself. For example, when we showed >3. More generally, when k=2 [...] >[...] For p=3, this is an elliptic curve with rank 0, that is, the only >points are the three torsion points -- one at infinity and the other two >with z=0 (x=+-1). it then follows that the curve C(2,3n) also has no rational points except the trivial ones. Likewise we conclude C(3,2n) has no rational points for any n, since we showed C(3,2) does not. A similar argument uses the observation that C(2,5) covers the curve Y^2=5X^4 - 6X^2 + 5 (which is an elliptic curve having only the four torsion points with X=+-1) to conclude that C(2,5) -- and thus C(2,5n) for all n --- has only the obvious rational points. Another similar argument, this time noting that C(4,2) covers Y^2=2X^4-2X^2+1, falters since this latter elliptic curve has rank 1; there's no obvious way to know how many of the infinitely many points on this elliptic curve are in the image of C(4,2). The remaining curve (3,3) of genus 2 does not seem to admit any obvious coverings of a curve of genus 1. dave