Derivation of Diophantine Equation
Suppose integers $a,b,c,d \gt 0$ are the side lengths of a cyclic quadrilateral whose area and perimeter are equal. Let $s = (a+b+c+d)/2$ be the semiperimeter, so that by Brahmagupta's formula for area of a cyclic quadrilateral:
$$ 2s = \sqrt{(s-a)(s-b)(s-c)(s-d)} $$
A priori it is not evident that $s$ is an integer, although clearly $2s=a+b+c+d$ is. Squaring both sides and multiplying by $2^4$ gives:
$$ 64s^2 = (2s-2a)(2s-2b)(2s-2c)(2s-2d) $$
The product on the right side must have an even factor, since the left hand side is even. If any two factors are added, e.g.
$$ (2s-2a)+(2s-2b) = (-a+b+c+d)+(a-b+c+d) = 2(c+d) $$
the result is even, so that all four factors have the same parity. All four factors are (positive) even integers, and this implies $s$ is an integer.
Letting $w=s-a,x=s-b,y=s-c,z=s-d$, and noting $w+x+y+z=4s-2s=2s$, our first equation becomes $ w+x+y+z = \sqrt{wxyz} $. Thus:
$$ wxyz = (w+x+y+z)^2 $$
where each unknown is a positive integer.
Which solutions are realized by requisite cyclic quadrilaterals?
Conversely, suppose this equation has a positive integer solution $w,x,y,z$.
Will there be corresponding integer side lengths $a,b,c,d$ of a cyclic
quadrilateral with area equal to perimeter?
First we observe that $w+x+y+z$ must be even. If that sum were odd, at least
one of $w,x,y,z$ would be even, and their product would also be even.
Then (from the equation) the sum $w+x+y+z$ would be even as well.
Setting $w+x+y+z = 2s$ for some integer $s \gt 0$, we can define integers:
$$ a = s-w, b = s-x, c = s-y, d = s-z $$
and note that $a+b+c+d = 4s - (w+x+y+z) = 2s$. Moreover they satisfy the polygon inequality
requirements:
$$ a \lt b+c+d,\ b \lt a+c+d,\ c \lt a+b+d,\ d \lt a+b+c $$
because (for instance) $-a+b+c+d = 2s-2a = 2w \gt 0$ implies $a \lt b+c+d$.
If only we knew these were positive integers, then we could realize $a,b,c,d$ as
side lengths of a cyclic quadrilateral having (from the equation) area = perimeter.
In an appealling bit of symmetry:
Prop. $a,b,c,d$ are positive if and only if $w,x,y,z$ satisfy the
polygon inequalities $w \lt x+y+z,\ x \lt w+y+z,\ y \lt w+x+z,\ z \lt w+x+y$.
However this is not always so. Several of the "fundamental" solutions listed
by Will Jagy satisfy instead $w = x+y+z$, so that $a = s-w = (-w+x+y+z)/2 = 0$,
and these might be interpreted as "degenerate quadrilaterals" (triangles).
Consideration of the infinite families exhibited by Tito Piezas III shows
as Pell equation pairs $(p,q)$ tend to infinity, inevitably one of the
quadruple entries will dominate the sum of the others.
For example: taking $(p,q)=(9,4)$ satisfying $p^2 - 5q^2 = 1$ in Tito's second
listed family, we get this solution:
$$ w = 6*(p+q)^2, x = 6*(p-q)^2, y = z = 3 $$
where $w = 1014 \gt 156 = x+y+z$. As $(p,q)$ increase, so will the excess
of $w$ over $x+y+z$.
Thanks to Will Jagy's formulation of the condition for Hurwitz fundamental solutions:
$$ w \ge x \ge y \ge z,\; xyz \le 2(w+x+y+z) $$
there are only finitely many cyclic quadrilaterals with integer sides having area = perimeter. Up to permutation of the unknowns (equivalent to permuting the side lengths $a,b,c,d$ since $s$ is invariant), these fundamental solutions are exactly the ones that satisfy the polygon inequality.
The nine such solutions include five cases of "degenerate" quadrilaterals, where $w=x+y+z$ and thus $a = s-w = 0$, in the same sense that Heron's Rule for triangles is a limiting case of Brahmagupta's formula for area of cyclic quadrilateral:
(w,x,y,z)=(30,24,5,1) (a,b,c,d)=(0,6,25,29) triangle area = 60
(w,x,y,z)=(21,14,6,1) (a,b,c,d)=(0,7,15,20) triangle area = 42
(w,x,y,z)=(18,9,8,1) (a,b,c,d)=(0,9,10,17) triangle area = 36
(w,x,y,z)=(15,10,3,2) (a,b,c,d)=(0,5,12,13) triangle area = 30
(w,x,y,z)=(12,6,4,2) (a,b,c,d)=(0,6,8,10) triangle area = 24
Note that the last two solutions are Pythagorean (right triangles).
That leaves four solutions that give "nondegenerate" cyclic quadrilaterals:
(w,x,y,z)=(10,10,9,1) (a,b,c,d)=(5,5,6,14) cyc. quad. area = 30
(w,x,y,z)=(8,5,5,2) (a,b,c,d)=(2,5,5,8) cyc. quad. area = 20
(w,x,y,z)=(6,6,3,3) (a,b,c,d)=(3,3,6,6) cyc. quad. area = 18
(w,x,y,z)=(4,4,4,4) (a,b,c,d)=(4,4,4,4) cyc. quad. area = 16
Each of these has at least a pair of equal sides, so one way to permute the sides will give trapezoids or rectangles. However permuting the sides can also give cyclic quadrilaterals that are not trapezoidal or rectangular. With the exception of the final case (square), each solution gives us two noncongruent quadrilaterals that have the given side lengths.
Also note the "self-dual" property of the last three cases, in that (a,b,c,d) is a permutation of (w,x,y,z).
