Re: Solving Diophantine Equations

*To*: mathgroup at smc.vnet.net*Subject*: [mg61468] Re: Solving Diophantine Equations*From*: "Diana" <diana53xiii at earthlink.remove13.net>*Date*: Wed, 19 Oct 2005 02:17:41 -0400 (EDT)*References*: <32592565.1129236555309.JavaMail.root@elwamui-darkeyed.atl.sa.earthlink.net> <9F2DCCEC-34BF-4F53-9475-F686A911F260@akikoz.net> <E39C9EDB-A88A-42F3-9AD9-7DA09C7D790A@yhc.att.ne.jp> <dio1s4$suo$1@smc.vnet.net> <diprnq$hn2$1@smc.vnet.net> <divhhk$gb0$1@smc.vnet.net> <dj2746$bh9$1@smc.vnet.net>*Reply-to*: "Diana" <diana53xiii at earthlink.remove13.net>*Sender*: owner-wri-mathgroup at wolfram.com

Ray, Thanks so much. I have printed it out and will use it. Diana M. "Ray Koopman" <koopman at sfu.ca> wrote in message news:dj2746$bh9$1 at smc.vnet.net... > Diana wrote: >> Ray, >> >> I am a beginning - intermediate Mathematica user. >> >> The algorithm you and Andrzej have coded works great. Could I trouble you >> to >> explain briefly what the code is doing? To the un-initiated, the >> sophisticated code is hard to interpret. >> >> As was mentioned in this thread, I am trying to corroborate the findings >> of >> Pingzhi Yuan in 2004. >> >> I hope to use your improvements towards the equations of twenty other >> Diophantine equation articles, as well, for a thesis. >> >> Thanks, >> >> Diana > > Here's an improved gg, with some comments. > > gg2[a_, b_, c_:5] := Block[{y, r1}, Reap[ > Do[y = 2; > While[y^(n-1) < (y - 1)a + y, > r1 = (y^n - 1)/(y - 1) - 1; > Do[If[(x+1)x == r1, Sow[{x, y, n}]], > {x, y^((n-1)/2) + Boole[n==3], (y^(n-1) - y)/(y - 1)}]; > y++], > {n, c, b, 2}]][[2,1]]] > > gg2[10^5,30]//Timing > {6.91 Second,{{5,2,5},{90,2,13}}} > > The arguments are > a = upper bound for x > b = upper bound for n > c = lower bound for n; defaults to 5 > > Only y & r1 are declared explicitly as local variables, because n & x > are created implicitly as local variables by the Do statements. > > There are three nested loops. The outermost is Do[...,{n,c,b,2}]. > Note that there is no check that c is odd. > > The middle loop is y=2;While[y^(n-1)<(y-1)a+y,...;y++]. The final y > is the largest value for which the corresponding upper bound for x, > (y^(n-1) - y)/(y - 1), is less than a. > > The innermost loop is Do[...,{x,xmin,xmax}], > with xmin = y^((n-1)/2) + Boole[n==3] > and xmax = (y^(n-1) - y)/(y - 1). > Note that xmin is adjusted so that the test x!=y can be omitted. > > Inside the x-loop, Sow[{x,y,n}] saves {x,y,n} whenever > (x^3-1)/(x-1) == (y^n-1)/(y-1). Note that the much of the arithmetic > involved in the comparison is done outside the x-loop. > Note also that (x+1)x is faster than x*x+x. > > Finally, Reap[...][[2,1]] returns what was Sown. > If no solutions were found then there will be an error message > "Part::partw: Part 1 of {} does not exist." > and the result will be {Null,{}}[[2,1]]. > If that bothers you, change [[2,1]] to simply [[2]], > which will give the solutions with an extra level of nesting > and thus avoid the error message when there are no solutions. >