Talk:Special pythagorean triplet: Difference between revisions
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(Final solution? Last guess? See the F# which seems to work even if this description is still confused!) |
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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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If you are going to pilfer problems from another site you should at least copy the actual problem description over, maybe with a slight rewording, and attempt to add some value to it, in this case perhaps some discussion (as above) of brute force vs. smarter. --[[User:Petelomax|Pete Lomax]] ([[User talk:Petelomax|talk]]) 11:42, 2 September 2021 (UTC)
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