From: Edward Early Newsgroups: sci.math Subject: Help with a Diophantine equation (not homework) Date: 17 Mar 1996 21:01:10 GMT Organization: Gauss's Angels Can anybody solve this Diophantine equation: w^2 = 16f^2 + 24mf + 8m^2 + 8ef - 3e^2 + 4ce - 4c^2 where c <> e, c <> m, m <> 0, c <> 0, w <> 0? I can prove that there are infinitely many solutions, but I would really like to have a classification of all solutions, even one involving hellish functions such as p_1(n) which returns the n'th prime congruent to 1 mod 4 or upsilon(n) which returns the smallest positive number whose square is a multiple of n. Alternately, if anybody has the time and computing power to generate tables of solutions, I would appreciate that, too. As a warm-up you might want to try solving w^2 = 12ek - 3e^2 where w <> 0. This much simpler equation has infinitely many solutions which can be classified (hint: use the upsilon function mentioned above and look at e mod 4). Any useful information or questions about why on earth I even care about this equation (I'd be very impressed if anybody can guess where it comes from and what the significance of each variable is) should be e-mailed to early@tenet.edu Thanks, E pi=3.141592653589793238462643383279502884197169399375105820974944592307816 | Edward Early | Amateur mathematician, | If I have not seen as far as| | early@tenet.edu | music lover, off-duty | others it is because giants | | efedula@aol.com | superhero, and student | have stood on my shoulders. | 406286208998628;e=2.71828182845904523536;1+1=1.999999999999999999999999999 ============================================================================== From: rusin@vesuvius.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: Help with a Diophantine equation (not homework) Date: 25 Mar 1996 21:10:38 GMT Summary: Homogeneous quadratics are easy In article <4ihuim$q9b@geraldo.cc.utexas.edu>, Edward Early wrote: >Can anybody solve this Diophantine equation: >w^2 = 16f^2 + 24mf + 8m^2 + 8ef - 3e^2 + 4ce - 4c^2 >I can prove that there are infinitely many solutions, but I would really >like to have a classification of all solutions Well, OK, although maybe I understand "solution" a little differently from you. To me the problem is "easy" since this is a homogeneous rational quadratic. To see why, let's see what happens when I accept your proposal: >As a warm-up you might want to try solving >w^2 = 12ek - 3e^2 I would respond with two different suggestions. The first is to realize that this is a _homogeneous_ quadratic, and so the problem is more or less equivalent to solving in rational numbers the corresponding equation 1 = 12 xy - 3 x^2 (where of course I have set x=e/w, y=k/w). This equation is easy to solve because the original was a homogeneous _quadratic_ (note change of emphasis). Quadratic equations with a rational point have easily-parameterized rational solution sets. Geometrically, one draws all the lines through a fixed point (say, (1,1/3) ); each line meets a quadratic twice, so each of these lines leads to a unique additional point on the curve, and thus the points are parameterized by the lines. If the slope of the line is m, the line is the set of points (x,y) = (1,1/3) + t (1,m), which lies on the curve if 1 = 12 (1+t)(1/3 + m t) - 3 (1+t)^2, which require either t=0 or t= - (12m-2)/(12m-3). Thus the other point on this line is (x,y) = ( -1/(12m-3), (-12m^2+6m-1)/(12m-3) ) As m ranges over Q, this parameterizes the whole rational curve. If you wish, you may recover the integer solutions: write m=a/b in lowest terms and then deduce the possibilities for w, e, and k; for example, if b is not divisible by 2 or 3 we conclude w=3b(4a-b) r e=-b^2 r k=-(12a^2+6ab-b^2) r for any r. The cases with 2|b or 3 | b are similar but a bit harder. My other suggestion is to stick to a more polar form for the curve. Rather than w^2=12ek-3e^2, write this as w^2+3E^2-3K^2=0 where E=e-2k and K=2k. (In this form the question may be recast as one involving factorization in Z[sqrt(3)], but I won't pursue this.) Now we can turn to your main equation: >w^2 = 16f^2 + 24mf + 8m^2 + 8ef - 3e^2 + 4ce - 4c^2 I would try either or both of the suggestions made above. For example, the substitutions C=2c-e E=e-2f M=2m+3f W=w F=f turn this into the equation W^2 + C^2 + 2E^2 - 2M^2 = 6F^2. (I think there is a way to turn this now into a factorization question involving a quaternion-like algebra over Z, but I can't remember the details and don't need them.) Finding rational solutions is easy, using the previous techniques. To solve x^2 + y^2 + 2 z^2 - 2 v^2 = 6, look for the intersections of this hypersurface with lines through the rational point (2,0,1,0), for example. For each fixed m1, m2, m3, the line (2,0,1,0) + t(1, m1, m2, m3) meets the hypersurface when t = -2(1 + 2 m2)/(1 + m1^2 + 2 m2^2 - 2 m3^2). Substitute this in for t to get all possible rational values for (x,y,z,v) on the hypersurface, each just once, in terms of the (arbitrary) rational values m1, m2, and m3. (Well, we miss the other solutions with x=2, but they can be obtained similarly. Also be sure not to take m_i with 1+m1^2+2m2^2-2m3^2=0, of course.) If you just want lots of integers points on the hypersurface, you may easily compute them by writing m1=n1/n0, m2=n2/n0, and m3=n3/n0 and then F = (n0^2 + n1^2 + 2 n2^2 - 2 n3^2) W = 2F - 2 n0 (n0 + 2 n2) C = - 2 n1 (n0 + 2 n2) E = F - 2 n2 (n0 + 2 n2) M = - 2 n3 (n0 + 2 n2) Indeed, taking all integer values of the n_i in the above formulas, and then dividing by common factors of F, W, C, E, and M will give _all_ primitive integral solutions (with allowances as for the rational solutions). In order to get integral, rather than half-integral, values for f, w, c, e, and m, one must satisfy a couple of parity checks: M=0 and C=E mod 2; this will reduce the set of common divisors to be divided out. dave