From: rusin@vesuvius.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: Equations like FLT Date: 2 Dec 1998 08:29:14 GMT Andrew John Walker wrote: >Does anyone have any pointers or information on equations like those >in Fermat's last theorem, except where the exponent doesn't have to be >equal for all terms? Eg 11^2+2^2=5^3 >I'm especially interested in which if any forms have been proved >not solvable. The basic rule of thumb is that x^m + y^n = z^r should only have lots of solutions if 1/m + 1/n + 1/r >= 1. (I guess very loosely you can think of this as a pigeonhole argument: the number of triples (x^m , y^n, z^r) with all coordinates less than N is roughly N^(1/m + 1/n + 1/r); for each triple (X, Y, Z) the sum X + Y - Z is roughly N^1, so you expect to have X + Y - Z = 0 (or any other number) lots of times only if the messy exponent is at least equal to 1.) For example, m=n=2 gives an equation which is easy to solve for any r: we know full well which integers can be written as a sum of two squares. The case m=r=2 is even easier... Without two 2's in the exponents you can only have 1/m + 1/n + 1/r > 1 in the cases {m,n,r} = {2, 3, 3} {2, 3, 4} {2, 3, 5} and only have 1/m + 1/n + 1/r = 1 in the cases {2, 3, 6} {2, 4, 4} {3, 3, 3}. [These triples of numbers should be familiar to people studying groups generated by reflections and so on...] The first three sets yield equations with infinitely many solutions. There are no solutions to x^m + y^n = z^r if {m,n,r}={3,3,3}: Euler (Fermat's Last Theorem for the exponent 3). There are no solutions with {2,4,4}: Fermat. (Proof by infinite descent, so as to prove FLT with exponent 4). There are almost no solutions with exponents {2,3,6}: indeed, dividing by the 6th power, we would have a rational solution to one of the equations Y^2 = X^3 + 1 or Y^2 = X^3 - 1 . These both describe elliptic curves, each of which has rank 0 over the rationals. The second has a torsion group of order 2: the only rational point is (1,0). The first has a torsion group of order 6, including points with one coordinate equal to 0 and the nontrivial solutions (X, Y) = (2, +-3). When 1/m + 1/n + 1/r < 1, it is known that there are at most a finite number of solutions to x^m + y^n = z^r with x,y,z coprime. In fact, only nine such solutions are known (to me anyway!): Sorted by exponents they are {2,3,7}: 2^7 + 17^3 = 71^2 {2,3,7}: 17^7 + 76271^3 = 21063928^2 {2,3,7}: 1414^3 + 2213459^2 = 65^7 {2,3,7}: 9262^3 + 15312283^2 = 113^7 {2,3,8}: 3^8 + 96222^3 = 30042907^2 {2,3,8}: 33^8 + 1549034^2 = 15613^3 {2,3,9}: 7^3 + 13^2 = 2^9 {2,4,5}: 3^5 + 11^4 = 122^2 {2,4,5}: 2^5 + 7^2 = 3^4 You can get some more pointers about Fermat's equation and generalizations at: index/11D41.html More about this particular generalization is at index/14JXX.html (For each {m,n,r} the equation describes an algebraic surface.) dave ============================================================================== From: gerry@mpce.mq.edu.au (Gerry Myerson) Newsgroups: sci.math Subject: Re: Equations like FLT Date: Thu, 03 Dec 1998 09:38:17 +1100 In article <742toq$e00$1@gannett.math.niu.edu>, rusin@vesuvius.math.niu.edu (Dave Rusin) wrote: => When 1/m + 1/n + 1/r < 1, it is known that there are at most a finite => number of solutions to x^m + y^n = z^r with x,y,z coprime. In fact, only => nine such solutions are known (to me anyway!): Sorted by exponents they are => => {2,3,7}: 2^7 + 17^3 = 71^2 => {2,3,7}: 17^7 + 76271^3 = 21063928^2 => {2,3,7}: 1414^3 + 2213459^2 = 65^7 => {2,3,7}: 9262^3 + 15312283^2 = 113^7 => {2,3,8}: 3^8 + 96222^3 = 30042907^2 => {2,3,8}: 33^8 + 1549034^2 = 15613^3 => {2,3,9}: 7^3 + 13^2 = 2^9 => {2,4,5}: 3^5 + 11^4 = 122^2 => {2,4,5}: 2^5 + 7^2 = 3^4 You've left out the trivialish one: {2,3,942}: 1^942 + 2^3 = 3^2. Gerry Myerson (gerry@mpce.mq.edu.au)