Problem: Let $a > b > c > d > 0$ be integers satisfying
\[ ac + bd = (b+d+a-c)(b+d-a+c).  \]
Prove that $ab + cd$ is not prime.

\end{enumerate}

Solution: The problem condition is equivalent to
\[ ac + bd = (b+d)^2 - (a-c)^2 \]
or
\[ a^2-ac+c^2 = b^2+bd+d^2. \]

Let us construct a quadrilateral $WXYZ$ such that
$WX = a$, $XY = c$, $YZ = b$, $ZW = d$,
and \[ WY = \sqrt{a^2-ac+c^2} = \sqrt{b^2+bd+d^2}.\]
Then by the law of cosines, we obtain $\angle WXY = 60\dg$
and $\angle WZY = 120\dg$.
Hence this quadrilateral is cyclic.

\begin{center}
  \begin{asy}
    pair X = dir(110);
    pair W = dir(210);
    pair Y = dir(330);
    pair Z = dir(280);
    draw(unitcircle, blue);
    draw(W--X--Y--Z--cycle);
    draw(W--Y, red);

    pair a = midpoint(W--X);
    pair c = midpoint(X--Y);
    pair b = midpoint(Y--Z);
    pair d = midpoint(Z--W);

    dot("$X$", X, dir(X));
    dot("$W$", W, dir(W));
    dot("$Y$", Y, dir(Y));
    dot("$Z$", Z, dir(Z));

    label("$a$", a, dir(a));
    label("$b$", b, -dir(b));
    label("$c$", c, dir(c));
    label("$d$", d, -dir(d));
    label(minipage("\begin{align*} & \sqrt{a^2-ac+c^2} \\ =& \sqrt{b^2+bd+d^2}\end{align*}"), (W+Y)/2, dir(90), red);
  \end{asy}
\end{center}

By the more precise version of Ptolemy's theorem,
we find that
\[ WY^2 = \frac{(ab+cd)(ad+bc)}{ac+bd}. \]

Now assume for contradiction that that $ab+cd$ is a prime $p$.
Recall that we assumed $a > b > c > d$.
It follows, for example by rearrangement inequality, that
\[ p = ab+cd > ac+bd > ad+bc. \]
Let $y = ac+bd$ and $x = ad+bc$ now.
The point is that \[ p \cdot \frac xy \]
can never be an integer if $p$ is prime and $x < y < p$.
But $WY^2 = a^2-ac+c^2$ is clearly an integer, and this is a contradiction.

Hence $ab+cd$ cannot be prime.

\begin{remark*}
  It may be tempting to try to apply the more typical form of
  Ptolemy to get $ab+cd = WY \cdot XZ$;
  the issue with this approach is that $WY$ and $XZ$ are usually not integers.
\end{remark*}