Talk:Special pythagorean triplet: Difference between revisions
Some further observations
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the smallest value of i<sup>2</sup>-g<sup>2</sup> is when g=(999-n)/2 and i=1000-g
if i<sup>2</sup>-g<sup>2</sup> is greater than n<sup>2</sup> then there can be no solution for this n with a smaller g
n must be even
2(g+i) must be a factor of n<sup>2</sup>
given these conditions the solution is n, ((g+i)/2)-n<sup>2</sup>/2(g+i), ((g+i)/2)+n<sup>2</sup>/2(g+i) asserting that n<g.
--[[User:Nigel Galloway|Nigel Galloway]] ([[User talk:Nigel Galloway|talk]]) 14:11, 31 August 2021 (UTC)
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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)
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 identify a value g such that n+(g-1)+(g+1)=1000 (eg 3,498,499). 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>, which implies that n is even. 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.--[[User:Nigel Galloway|Nigel Galloway]] ([[User talk:Nigel Galloway|talk]]) 14:06, 1 September 2021 (UTC)
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