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From: "China -->hk" on 20 May 2010 00:04
hi,friend:
an equation (about r,k. m is a constant number): 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
From: Torsten Hennig on 20 May 2010 00:24
> 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.
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.
From: Mok-Kong Shen on 20 May 2010 04:43
China -->hk wrote:
> an equation (about r,k. m is a constant number): 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

(1,1,4)
From: Torsten Hennig on 20 May 2010 00:54
> > 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.
From: "China -->hk" on 23 May 2010 22:52
thank you very much
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