wu :: forums - Print Page


    
      
        wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> putnam exam (pure math) >> Integral Solutions
        
(Message started by: l4z3r on Aug 29^th, 2008, 4:46am)

Title: Integral Solutions
Post by l4z3r on Aug 29^th, 2008, 4:46am

a function is defined as:

f: Z(+) --> Z

f(m,n) = (n³ + 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 n³+1 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/equiv.gif 1(mod3) and mn-1 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, (m²+n)/(mn-1))[/hide] is also; then use the ideas that appear [link=http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1214032409]here[/link] (and which should appear [link=http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1214617577]here[/link]).

Or, you can proceed more directly by writing n³+1 = [hide](mn-1)((an-m)n-1)[/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)/(mn-1)) is also[/hide]

but, instead of [hide] (mn-1)((an-m)n-1)[/hide] i feel a better alternative would be [hide](kn-1)(mn-1)[/hide]. Gives the answer in lesser steps, i think.