Talk:Special pythagorean triplet: Difference between revisions
Note that the required triplet isn't primitive contra to my earlier suggestion...
(Note that the required triplet isn't primitive contra to my earlier suggestion...) |
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The task description says there is only one triple with a + b + c = 1000, so it must be a primitive one. --[[User:Tigerofdarkness|Tigerofdarkness]] ([[User talk:Tigerofdarkness|talk]]) 18:14, 31 August 2021 (UTC)
::...or not - of course that doesn't follow and the triplet isn't a primitive one. Doh! --[[User:Tigerofdarkness|Tigerofdarkness]] ([[User talk:Tigerofdarkness|talk]]) 17:23, 6 September 2021 (UTC)
I have added three observations to the list above which remove the requirement for searching which I think may need explanation. For a given n when z-n is even identify a value g such that n+(g-1)+(g+1)=z (eg 200,399,401). The problem may be rewritten as n<sup>2</sup>=(g+x)<sup>2</sup>-(g-x)<sup>2</sup>. (g+x)<sup>2</sup>-(g-x)<sup>2</sup> is 4xg. Therefore 4g must be a factor of n<sup>2</sup>. To determine x I divide n<sup>2</sup> by 4g. Let me work this for n=200. g=400 {(1000-n)/2}. 4g=1600. 40,000/1600=25 so the solution is 200,400-25,400+25. When z-n is odd similar logic applies identifying g such that n+g+(g+1)=z and solving for x: n<sup>2</sup>=(g+1+x)<sup>2</sup>-(g-x)<sup>2</sup>=4gx+2x+2g+1.--[[User:Nigel Galloway|Nigel Galloway]] ([[User talk:Nigel Galloway|talk]]) 14:06, 1 September 2021 (UTC)
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