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A good problemby Aditya89
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#1
Nov1905, 04:35 AM

P: 23

If m^2<a<b, & a*b=a perfect square, TPT: b>(m+1)^2



#2
Nov1905, 08:18 AM

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P: 1,994

a*b=a means a is zero or b=1?? This is surely not what you mean?
What does TPT mean? "To Prove True"? 


#3
Nov1905, 01:20 PM

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Huh, on furthur reflection I guess "& a*b=a perfect square" means "and a*b is a perfect square".
Is this a homework problem? In case it is, just a simple hint for now treat the cases where "a" is a perfect square and "a" is not a perfect square seperately. 


#4
Nov1905, 02:46 PM

P: 78

A good problem
I've understood the trouble... and I've got my picture but I think that I can't explain in a simple way... in sense that I'm not able to make it easy...



#5
Nov2105, 04:16 AM

P: 23

Sorry, a*b= k^2, k belongs to the naturals. Not that a*b=a. Really sorry!
Thanks for the hint, Shmoe! I may be clser to the solution by an another method. I'll let you know if I get it. BTW, TPT is "to prove that". 


#6
Nov2105, 04:38 AM

P: 23

Hey does anybody know how to create mathematical symbols while posting?



#7
Nov2105, 04:53 AM

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THnigs like [tex]ab=k^2, k \in \mathbb{N}[/tex]?
Try looking up the LaTeX sticky post in the physics section. Or trying seraching the site using the search function for such "frequently asked questions" 


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