# A good problem

1. Nov 19, 2005

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

2. Nov 19, 2005

### 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"?

3. Nov 19, 2005

### 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.

4. Nov 19, 2005

### 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...

5. Nov 21, 2005

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".

6. Nov 21, 2005

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