proving two expressions can be squares

Hi,

If it is known that is a perfect square

where A, B, C and D are integers, A and B are coprime, C and D are coprime and D > C >= B > A > 0

then *can *be a perfect square..........(1)

and *can *be a perfect square..........(2)

but can it be proven that (1) and (2) **can't** be perfect squares simultaneously (or examples to the contrary) ?

__Examples__

A = 1, B = 2, C = 2, D = 11

is not a perfect square

A = 1, B = 2, C = 58, D = 209

is not a perfect square

thank you

Re: proving two expressions can be squares

Well, recognize that what you want to construct is an arithmetic progression of three squares, where the middle square is and the common difference is (reversing the order to indicate ). There is a parametric formula for progressions of three squares that says: for , giving a common difference of . (Click here for a wonderful proof that this captures all of them.) So, your question becomes:

Find positive integers with satisfying

and

You might start by isolating in the first equation, subbing it into the second, and attempting to isolate . Not sure if this is possible directly because this gives you a quartic polynomial on , but if you can, then the problem will reduce to proving when or whether your expression evaluates to an integer. I know I'm being lazy, but that should give you something to chew on. (Nerd)

Re: proving two expressions can be squares

Assume that they are both perfect squares, i.e.

.

Note the constraint, 0 < A < B, so . Taking their difference (and switching order a bit),

What happens if 8CD is a perfect square? For example, try C = 32, D = 49. I'll let you play around with that.