# proving two expressions can be squares

• Apr 13th 2012, 01:51 PM
gortwell
proving two expressions can be squares
Hi,

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

\$\displaystyle (A^2+B^2)(C^2+D^2)=25^2\$
\$\displaystyle (A^2+B^2)(C^2+D^2)+4(A^2-B^2)CD=19^2\$
\$\displaystyle \ (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

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

thank you

• Jun 5th 2012, 06:01 PM
Media_Man
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 \$\displaystyle (A^2+B^2)(C^2+D^2)\$ and the common difference is \$\displaystyle 4(B^2-A^2)CD\$ (reversing the order to indicate \$\displaystyle B > A\$). There is a parametric formula for progressions of three squares that says: \$\displaystyle X^2,Y^2,Z^2 = (a^2-2ab-b^2)^2,(a^2+b^2)^2,(-a^2-2ab+b^2)^2\$ for \$\displaystyle a<b\$, giving a common difference of \$\displaystyle a^2-2ab-b^2\$. (Click here for a wonderful proof that this captures all of them.) So, your question becomes:

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

You might start by isolating \$\displaystyle a\$ in the first equation, subbing it into the second, and attempting to isolate \$\displaystyle b\$. Not sure if this is possible directly because this gives you a quartic polynomial on \$\displaystyle b\$, but if you can, then the problem will reduce to proving when or whether your expression \$\displaystyle 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. (Nerd)
• Jun 12th 2012, 09:56 PM
richard1234
Re: proving two expressions can be squares
Assume that they are both perfect squares, i.e.

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

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

\$\displaystyle 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.