

Title: Integral Solutions Post by l4z3r on Aug 29^{th}, 2008, 4:46am a function is defined as: f: Z(+) > Z f(m,n) = (n^{3} + 1)/ (mn  1) where Z(+) denotes the set of positive integers and Z the set of integers. Find all the solutions for (m,n) EDIT: f(m,n) not f(x) 

Title: Re: Integral Solutions Post by towr on Aug 29^{th}, 2008, 5:23am Shouldn't the x in f(x) come into it somewhere? 

Title: Re: Integral Solutions Post by l4z3r on Aug 29^{th}, 2008, 5:42am ah. meant f(m,n). sorry. 

Title: Re: Integral Solutions Post by SMQ on Aug 29^{th}, 2008, 5:58am So, in other words, "find all http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/m.gif, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/n.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbz.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/supplus.gif such that (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/n.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sup3.gif + 1) / (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/m.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/n.gif  1) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbz.gif", right? SMQ 

Title: Re: Integral Solutions Post by l4z3r on Aug 29^{th}, 2008, 7:00am yes. hint:[hide]use n^{3}+1 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/equiv.gif 1(mod3) and mn1 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/equiv.gif 1 (mod n) (number theory)[/hide] 

Title: Re: Integral Solutions Post by Eigenray on Aug 29^{th}, 2008, 11:53am If (m,n) is a solution, then [hide](m, (m^{2}+n)/(mn1))[/hide] is also; then use the ideas that appear [link=http://www.ocf.berkeley.edu/~wwu/cgibin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1214032409]here[/link] (and which should appear [link=http://www.ocf.berkeley.edu/~wwu/cgibin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1214617577]here[/link]). Or, you can proceed more directly by writing n^{3}+1 = [hide](mn1)((anm)n1)[/hide] and bounding. 

Title: Re: Integral Solutions Post by l4z3r on Aug 30^{th}, 2008, 5:34am hmm. good one. I agree with the first part. [hide]If (m,n) is a solution, then (m, (m2+n)/(mn1)) is also[/hide] but, instead of [hide] (mn1)((anm)n1)[/hide] i feel a better alternative would be [hide](kn1)(mn1)[/hide]. Gives the answer in lesser steps, i think. 

Powered by YaBB 1 Gold  SP 1.4! Forum software copyright © 20002004 Yet another Bulletin Board 