From: rusin@vesuvius.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: Number Theory Date: 8 Sep 1998 13:37:42 GMT In article <6t2d58$m6$1@news.nevada.edu>, Troy Kessler wrote: >I found that the solution of two pell equations is useful in the >triangle=square problem. (When is a triangle number square.) My question >is: >Is there a general solution to > n n > x - d* y =+-1 >x^3-7y^3=1 has the solution x=2 y=1 >x^3-43y^3=-1 has the solution x=7 y=2 >x^4-5y^4=1 has the solution x=3 y=2 >Do these equations have more solutions? If they do have more solutions is >there a way to generate new solutions? The short answer to your general question is "no". When n=1 your problem is trivial and as you observed, when n=2 there is a fairly efficient method for solving these equations. If d is a perfect n-th power, then your equation amounts to x^n - z^n = 1 which has no solutions in integers; otherwise x^2-d y^2 is irreducible in Q[x,y]. If F(x,y) is an irreducible binary form of degree greater than 2 (yours is of degree n) then the equation F(x,y) = t is called a Thue equation. It has at most finitely many solutions in integers. The solutions can be found by a method similar in spirit to the continued fraction algorithm for n=2: one attempts to approximate a root of F. Some more details are at 96/thue The situation is a little different if you want to solve these equations with _rational_ numbers rather than simply integers. Note that in this setting even the case d=1 is difficult, this being Fermat's Last Theorem! Still, some things are known. For n=1 the problem is still trivial. For n=2, the problem x^2-dy^2=1 is trivially solvable for all d. The problem x^2 - d y^2 = -1 is solvable rationally iff -1 is a square mod d; assuming it _is_ solvable, you can construct a solution very quickly; see 97/2squares.comp (Solving the more general problem x^2 - d y^2 = n quickly is a bit more challenging but as it turns out can be solved very rapidly too. Details available upon request.) From any one rational solution of a quadratic equation, all the others may be obtained with a rational parameterization; see 96/quadratics The case n=3 is quite subtle. In some cases (d=m^3, d=2m^3) there is an easy solution. In other cases, it appears to be true that if there is any solution, there will exist infinitely many, and you can easily construct infinitely many more from the first. But determining those values of d for which there is any solution at all, and determining the fundamental solutions in those cases, is challenging. I don't think a good characterization of the solvable values of d exists. See 97/2cubes Finally, by Falting's theorem, there are at most a finite number of rational solutions for any particular d<>0 and n>3. The proof is not constructive, and as far as I know there is no really food way to decide whether any solutions exist, nor to find them when they do. The specific question of which triangular numbers are also square is a classic Pell equation; see e.g. 95/pell.djr dave [URL updated 1999/01 -- djr]