, 2026

Suppose that $n$ faces on
$D_2$ have a $1$ on them and $6-n$ faces have a $2$ on them.

We make a chart to enumerate the possible totals when the two dice are
rolled. The number rolled on $D_1$ is
shown in the left column and the number rolled on $D_2$ is shown across the top row. Inside
the chart, we track the sum and the number of times this sum could
occur.

$\boldsymbol{1}$
($\boldsymbol{n}$ times)
$\boldsymbol{2}$
($\boldsymbol{6-n}$ times)

$\boldsymbol{1}$
$2$
($n$ times)
$3$
($6-n$ times)

$\boldsymbol{2}$
$3$
($n$ times)
$4$
($6-n$ times)

$\boldsymbol{3}$
$4$
($n$ times)
$5$
($6-n$ times)

$\boldsymbol{4}$
$5$
($n$ times)
$6$
($6-n$ times)

$\boldsymbol{5}$
$6$
($n$ times)
$7$
($6-n$ times)

$\boldsymbol{6}$
$7$
($n$ times)
$8$
($6-n$ times)

Of these sums, $2$, $3$, $5$, and $7$ are prime.

These occur a total of $n + (6-n) + n + (6-n)
+ n + (6-n) + n = 18 + n times out of the possible 6 $\times$ 6 = 36$ outcomes from rolling the
two dice together.

Since the probability of having a prime sum is $\dfrac{23}{36}$, then $23$ of the $36$ outcomes give a prime sum, and so
$18 + n = 23$ or $n = 5$.
Since $ABCD$ is a square and
$AB$ is horizontal, then $CD$ is parallel to $AB$ and so is also horizontal.

Since $C$ and $D$ are on a parabola and $CD$ is horizontal, then $C$ and $D$ are equidistant from the axis of
symmetry.

Since the parabola has equation $y = x^2 -
4$, its$x$-intercepts are
$2$ and $-2$ and so its axis of symmetry has
equation $x = 0$.

Thus, we can say that $C$ and $D$ have $x$-coordinates $s$ and $-s$, respectively, for some $s > 0$.

This means that $A$ has coordinates
$(-s, 0)$ and $B$ has coordinates $(s, 0)$.

This means that the side length of square $ABCD$ is $s -
(-s) = 2s$.

Since the height and width of $ABCD$
are equal, then $C$ has coordinates
$(s, -2s)$ and $D$ has coordinates $(-s, -2s)$.

Since $C$ lies on the parabola with
equation $y = x^2 - 4$, then $-2s = s^2 - 4$ and so $s^2 + 2s - 4 = 0$.

By the quadratic formula, $$s = \dfrac{-2 \pm \sqrt{2^2 - 4(1)(-4)}}{2} = \dfrac{-2 \pm \sqrt{20}}{2} = \dfrac{-2 \pm 2\sqrt{5}}{2} = -1 \pm \sqrt{5}$$ Since $s > 0$, then $s = -1 + \sqrt{5}$.

This means that the area of square $ABCD$ is equal to $(2s)^2$ which equals $(-2+2\sqrt{5})^2$.

Expanding and simplifying, we obtain $4 + 20
- 8 $\sqrt{5}$ = 24 - $8\sqrt{5}$ = 24 - $\sqrt{320}\$.$