Talk:Special pythagorean triplet: Difference between revisions
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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 |
the smallest value of i<sup>2</sup>-g<sup>2</sup> is when g=(999-n)/2 and i=1000-g |
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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 |
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 |
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n must be even |
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2(g+i) must be a factor of n<sup>2</sup> |
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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. |
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--[[User:Nigel Galloway|Nigel Galloway]] ([[User talk:Nigel Galloway|talk]]) 14:11, 31 August 2021 (UTC) |
--[[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) |
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) |
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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 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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Revision as of 14:07, 1 September 2021
For those paying "attention to timings" and more importantly for those paying attention to other comments I have made on these Euler tasks about filling RC with solutions worse than I would expect from a schoolboy with a pencil, considering n2+g2=i2 and n+g+i=1000 n<g<i note the following:
the largest value n can take is 332 with g=333 and i=335 the smallest value of i2-g2 is when g=(999-n)/2 and i=1000-g if i2-g2 is greater than n2 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 n2 given these conditions the solution is n, ((g+i)/2)-n2/2(g+i), ((g+i)/2)+n2/2(g+i) asserting that n<g.
--Nigel Galloway (talk) 14:11, 31 August 2021 (UTC)
Isn't the task to print out abc? Also, isn't using Euclid's formula (as in the XPL0 solution) going to be faster - less values to try ? --Tigerofdarkness (talk) 17:34, 31 August 2021 (UTC)
The task description says there is only one triple with a + b + c = 1000, so it must be a primitive one. --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 n2=(g+x)2-(g-x)2. (g+x)2-(g-x)2 is 4xg. Therefore 4g must be a factor of n2, which implies that n is even. To determine x I divide n2 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.--Nigel Galloway (talk) 14:06, 1 September 2021 (UTC)