Proving b>(m+1)^2 for Perfect Square Conditions

[Aditya89]

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

 

[shmoe]

a*b=a means a is zero or b=1?? This is surely not what you mean?

What does TPT mean? "To Prove True"?

 

[shmoe]

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.

 

[armandowww]

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...

 

[Aditya89]

Sorry for the confusion!

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".

 

[Aditya89]

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

 

[matt grime]

THnigs like $$ab=k^2, k \in \mathbb{N}$$?

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"