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Topic: ask a Q for help!
Replies: 5   Last Post: May 24, 2010 2:53 AM

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 Torsten Hennig Posts: 2,381 Registered: 12/6/04
Re: ask a Q for help!
Posted: May 20, 2010 4:54 AM

> > hi,friend:
> > an equation (about r,k. m is a constant number):
> > r): 2m^2+3m+1=2rm+k,here m,r,k are all positive
> > integers.i test (m,r,k) many times,and find that

> r,k
> > have the only one solution: r=k=m+1. but i cannt
> > prove my guess.who can help me? thanks

>
> This is one equation in two unknowns - so it will
> in general have arbitrary many solutions for r and
> k.

Not arbitrary many because of the positivity
constraint - but at least more than one.
All solutions (r,k) are given by
(1,2m^2+m+1),(2,2m^2-m+1),(3,2m^2-3m+1),...
as long as 2m^2+(3-2*r)*m+1 remains positive.

> E.g. fix r>0 arbitrary such that 2rm < 2m^2+3m+1.
> Then (r,k=2m^2+3m+1-2rm) is a solution.
>
> Best wishes
> Torsten.

Date Subject Author
5/20/10 China -->hk
5/20/10 Torsten Hennig
5/20/10 Torsten Hennig
5/20/10 Mok-Kong Shen
5/24/10 China -->hk
5/24/10 China -->hk