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MathGroup Archive 2005

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Re: Re: Solving Diophantine Equations

  • To: mathgroup at smc.vnet.net
  • Subject: [mg61429] Re: [mg61413] Re: Solving Diophantine Equations
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Wed, 19 Oct 2005 02:16:07 -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> <200510180645.CAA11275@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

I should perhaps add one small comment. One reason why I used the  
While loop with the lower bound fro x, is that exactly that bound  
appears later in the Do loop as the upper value of the iterator. The  
problem is that the Do loop will not work at all if this upper value  
exceeds a certain very large integer given by Developer` 
$MaxMachineInteger . If you arrange the loops differently it would be  
quite easy  for this number to be exceeded. So one should always set  
the bound a to a number less than Developer`$MaxMachineInteger.


Andrzej Kozlowski
Tokyo, Japan


On 18 Oct 2005, at 15:45, Ray Koopman wrote:

> 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.
>
>






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