proving two expressions can be squares

**gortwell** · April 13th 2012, 01:51 PM

Hi,

If it is known that $(A^2+B^2)(C^2+D^2)$ 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 $(A^2+B^2)(C^2+D^2)+4(A^2-B^2)CD$ can be a perfect square..........(1)
and $\ (A^2+B^2)(C^2+D^2)-4(A^2-B^2)CD$ 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

$(A^2+B^2)(C^2+D^2)=25^2$
$(A^2+B^2)(C^2+D^2)+4(A^2-B^2)CD=19^2$
$\ (A^2+B^2)(C^2+D^2)-4(A^2-B^2)CD$ is not a perfect square

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

$(A^2+B^2)(C^2+D^2)=485^2$
$(A^2+B^2)(C^2+D^2)+4(A^2-B^2)CD$ is not a perfect square
$\ (A^2+B^2)(C^2+D^2)-4(A^2-B^2)CD=617^2$

thank you

**Media_Man** · June 5th 2012, 06:01 PM

Well, recognize that what you want to construct is an arithmetic progression of three squares, where the middle square is $(A^2+B^2)(C^2+D^2)$ and the common difference is $4(B^2-A^2)CD$ (reversing the order to indicate $B > A$ ). There is a parametric formula for progressions of three squares that says: $X^2,Y^2,Z^2 = (a^2-2ab-b^2)^2,(a^2+b^2)^2,(-a^2-2ab+b^2)^2$ for $a<b$ , giving a common difference of $a^2-2ab-b^2$ . (Click here for a wonderful proof that this captures all of them.) So, your question becomes:

Find positive integers $A,B,C,D,a,b$ with $(A,B)=(C,D)=1, A<B\le C<D$ satisfying
$(A^2+B^2)(C^2+D^2)=(a^2+b^2)^2$ and $4(B^2-A^2)CD=a^2-2ab-b^2$

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

**richard1234** · June 12th 2012, 09:56 PM

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

$(A^2 + B^2)(C^2 + D^2) + 4(A^2 - B^2)CD = k^2$
$(A^2 + B^2)(C^2 + D^2) - 4(A^2 - B^2)CD = m^2$ .

Note the constraint, 0 < A < B, so $A^2 - B^2 < 0$ . Taking their difference (and switching order a bit),

$m^2 - k^2 = 8(B^2 - A^2)CD$

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

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