Talk:Euler's sum of powers conjecture: Difference between revisions

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(Reading that code, I'm having trouble deciding whether it could work for arbitrarily large values of N.) --[[User:Rdm|Rdm]] ([[User talk:Rdm|talk]]) 17:21, 8 July 2015 (UTC)
 
ANSWER: Fermat's little theorem says x^p == x (mod p). Think of this as saying x^{p-1} == 1 OR x == 0 (mod p). This implies that x^{K(p-1)+1} == x (mod p) for any K. So we combine p=2, K=4 (trivial); p=5, K=1 (direct application of FlT; and p=3, K=2 (the case you are looking at). In every case we show x^5 == x. --[[User:TomHyer|Tom Hyer]]
 
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EchoLisp solution :
 
if N = 3p + 1 , N^5 = "243p^5 +405p^4 +270p^3 +90p^2 +15p +1 " , which is 1 (mod 3)
 
if N = 3p + 2 ; N^5 = "243p^5 +810p^4 +1080p^3 +720p^2 +240p +32 ", which is 2 (mod 3)
--[[User:G.Brougnard|G.Brougnard]] ([[User talk:G.Brougnard|talk]]) 22:10, 8 July 2015 (UTC)
 
: Ok, but why? Is that just a convenient way of hardcoding the desired result, or is there some generally valid rule here? You can fit a fifth degree polynomial such as this one to up to six points and - perhaps coincidentally - the N=1000 example has six answers. --[[User:Rdm|Rdm]] ([[User talk:Rdm|talk]]) 23:10, 8 July 2015 (UTC)
 
: Actually, I don't see any such expression in echolisp's solution. That solution does make sense to me - but I'm not seeing anything related to fermat's little theorem there, yet. Nor am I seeing anything like that polynomial. --[[User:Rdm|Rdm]] ([[User talk:Rdm|talk]]) 02:26, 9 July 2015 (UTC)
 
:: It was just a mean to demonstrate that '''N = 0,1,2 (mod 3) implies N^5 = 0,1,2 (mod 3)'''. I do'nt know any more. This is not used in EchoLisp solution to the task. In this '''talk''' 'Echolisp solution' only means I computed the powers of 3p+i polynomials with EchoLisp. I should have written 'EchoLisp solution to your question in this talk' .Sorry for the misunderstanding. --[[User:G.Brougnard|G.Brougnard]] ([[User talk:G.Brougnard|talk]]) 05:53, 9 July 2015 (UTC)
 
== What if we were to create a program where power was much larger than 5?(like 20+) ==
 
Clearly none of the optimisations provided in c++ would be able to tackle such a quantity without even considering that c++ isn't very good with large integers