I'm looking for positive integers $a, b, c,$ and $d$ such that $$ (ad - bc)(ac + bd) \: | \: abcd$$
One partial solution that I found is $$(a, \, b, \, c, \, d) = (2x + 1,\, 2x, \, 2x + 2, \, 2x + 1)$$
Are there other partial solutions?
Another way of looking at this problem is that I am trying to find positive integers $A$, $B$, $C$, and $D$ such that $(A-B)(C+D) | N$ where $AB=CD=N$. This is, in fact, equivalent to $A=ad, B=dc, C=ac,$ and $D=bd$.
One direction I went is this.
There must be some positive integer P such that \begin{align*} A - B &= \frac{N}{P}\\ A - B &= \frac{AB}{P}\\ \frac{1}{B} - \frac{1}{A} &= \frac{1}{P}\\ \frac{1}{A} + \frac{1}{P} &= \frac{1}{B} \end{align*}
It follows that, for some integers $r, s,$ and $t$ $$ A = r(r + s)t $$ $$ P = s(r + s)t $$ $$ B = rst $$ $$ A-B = r^2t$$ $$ N = r^2s(r+s)t^2$$
Similarly, there must be some positive integer Q such that \begin{align*} C + D &= \frac{N}{Q}\\ D + C &= \frac{CD}{Q}\\ \frac{1}{C} + \frac{1}{D} &= \frac{1}{Q} \end{align*}
It follows that, for some integers $u, v,$ and $w$ $$ C = u(u + v)w $$ $$ D = v(u + v)w $$ $$ Q = uvw $$ $$ C+D = (u+v)^2w$$ $$ N = uv(u+v)^2w^2$$
The requirement that $(A-B)(C+D) | N$ can be expressed as $r^2t(u+v)^2w | uv(u+v)^2w^2$ which simplifies to $$ r^2t | uvw $$
It looks so suggestive. But I haven't found anything useful yet.
Another thing I noticed is that $$(A - B)^2 + (C + D)^2 = (A + B)^2 + (C - D)^2 = A^2 + B^2 + C^2 + D^2$$
which is very pretty but doesn't seem to help any.
