pythagorean triples exercise

Jeffrey Gaynor jgaynor at ncsa.uiuc.edu
Fri Oct 22 21:40:08 CEST 2010


As I indicated, generating such triples is easy. What you found is the edge case that

2*i*j = 200  => 100 = i*j

so (i,j) = (100,1) or (50,2) (25,4), (20,5) or (10,10). The maximal value are i = 100, j = 1. The other sides are

i^2 - j^2 = 10,000 - 1 = 9999

i^2 + j^2 = 10,000 + 1 = 10,001

...and there you have your figures. A real proof consists of a bit more, but nobody wants to read it and there is no easy way to notate it in plain text.


----- Original Message -----
From: "Mel" <mwilson at the-wire.com>
To: python-list at python.org
Sent: Friday, October 22, 2010 2:20:47 PM
Subject: Re: pythagorean triples exercise

MRAB wrote:
> On 22/10/2010 13:33, Baba wrote:

>> only a has an upper limit of 200
>>
> Really? The quote you gave included "whose small sides are no larger
> than n". Note: "sides", plural.

Strangely, there does seem to be a limit.  Fixing one side at 200, the 
largest pythagorean triple I have found is (200, 9999, 10001.0).  So far my 
math has not been up to explaining why.

	Mel

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