Theorem: Proper Integer Heronian Triangle whose Area is 24

There exists exactly one proper integer Heronian triangle whose area equals $24$.
That is, the obtuse triangle whose sides are of length $4$, $13$ and $15$.

Proof:
First we show that the $\tuple {4, 13, 15}$ triangle is actually Heronian.
Heron's Formula gives us that:
:$\AA = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$
where:
:$\AA$ denotes the area of the triangle
:$a$, $b$ and $c$ denote the lengths of the sides of the triangle
:$s = \dfrac {a + b + c} 2$ denotes the semiperimeter of the triangle.
Hence:
{{begin-eqn}}
{{eqn | l = s
      | r = \frac {4 + 13 + 15} 2
      | c = 
}}
{{eqn | r = 16
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }
      | r = \sqrt {16 \paren {16 - 4} \paren {16 - 13} \paren {16 - 15} }
      | c = 
}}
{{eqn | r = \sqrt {16 \times 12 \times 3 \times 1}
      | c = 
}}
{{eqn | r = \sqrt {576}
      | c = 
}}
{{eqn | r = 24
      | c = 
}}
{{end-eqn}}
It can be constructed by taking a $9-12-15$ Pythagorean triangle and removing a $5-12-13$ Pythagorean triangle from it:
:500px
Let $\tuple {a, b, c}$ be the sides of such a triangle.
Using Heron's Formula, we have:
{{begin-eqn}}
{{eqn | l = 24
      | r = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }
}}
{{eqn | ll= \leadsto 
      | l = 576
      | r = s \paren {s - a} \paren {s - b} \paren {s - c}
}}
{{end-eqn}}
Note that:
:$\paren {s - a} + \paren {s - b} + \paren {s - c} = 3 s - a - b - c = s$
Hence by substituting $x = s - a$, $y = s - b$, $z = s - c$:
:$x y z \paren {x + y + z} = 576$
By Semiperimeter of Integer Heronian Triangle is Composite, $s$ is an integer.
Hence $s, x, y, z \in \N_{>0}$.
{{WLOG}} suppose $x \le y \le z$.
Then:
{{begin-eqn}}
{{eqn | l = 576
      | r = x y z \paren {x + y + z}
}}
{{eqn | o = \ge
      | r = x^3 \paren {3 x}
}}
{{eqn | r = 3 x^4
}}
{{eqn | ll= \leadsto
      | l = x
      | o = \le
      | r = \sqrt [4] {\frac {576} 3}
}}
{{eqn | o = \approx
      | r = 3.72
}}
{{end-eqn}}
so we need to check $1 \le x \le 3$.
Note that for fixed $x$:
{{begin-eqn}}
{{eqn | l = 576
      | r = x y z \paren {x + y + z}
}}
{{eqn | o = >
      | r = x y^2 \paren {2 y}
}}
{{eqn | r = 2 x y^3
}}
{{eqn | ll= \leadsto
      | l = y
      | o = \le
      | r = \sqrt [3] {\frac {576} {2 x} }
}}
{{end-eqn}}
For $x = 1$:
:$1 \le y \le \sqrt [3] {\frac {576} 2} \approx 6.60$
For $x = 2$:
:$2 \le y \le \sqrt [3] {\frac {576} 4} \approx 5.24$
For $x = 3$:
:$3 \le y \le \sqrt [3] {\frac {576} 6} \approx 4.58$
Finally, for fixed $x$, $y$:
{{begin-eqn}}
{{eqn | l = x y z \paren {x + y + z}
      | r = 576
}}
{{eqn | ll= \leadsto
      | l = z^2 + \paren {x + y} z - \frac {576} {x y}
      | r = 0
}}
{{eqn | ll= \leadsto
      | l = z
      | r = \frac {- x - y + \sqrt {\paren {x + y}^2 + \frac {2304} {x y} } } 2
      | c = Quadratic Formula
}}
{{end-eqn}}
Since $z \in \Z$:
:$\sqrt {\paren {x + y}^2 + \dfrac {2304} {x y} } = 2 z + x + y \in \Z$
We check the value of $\sqrt {\paren {x + y}^2 + \dfrac {2304} {x y} }$ for the valid values of $x, y$.
We do not need to check $y = 5$ since $5 \nmid 576$.
We have:
:$\tuple {1, 1}: \sqrt {\paren {1 + 1}^2 + \dfrac {2304} {1 \times 1} } = \sqrt {2308} \notin \Z$
:$\tuple {1, 2}: \sqrt {\paren {1 + 2}^2 + \dfrac {2304} {1 \times 2} } = \sqrt {1161} \notin \Z$
:$\tuple {1, 3}: \sqrt {\paren {1 + 3}^2 + \dfrac {2304} {1 \times 3} } = \sqrt {784} = 28$
:$\tuple {1, 4}: \sqrt {\paren {1 + 4}^2 + \dfrac {2304} {1 \times 4} } = \sqrt {601} \notin \Z$
:$\tuple {1, 6}: \sqrt {\paren {1 + 6}^2 + \dfrac {2304} {1 \times 6} } = \sqrt {433} \notin \Z$
:$\tuple {2, 2}: \sqrt {\paren {2 + 2}^2 + \dfrac {2304} {2 \times 2} } = \sqrt {592} \notin \Z$
:$\tuple {2, 3}: \sqrt {\paren {2 + 3}^2 + \dfrac {2304} {2 \times 3} } = \sqrt {409} \notin \Z$
:$\tuple {2, 4}: \sqrt {\paren {2 + 4}^2 + \dfrac {2304} {2 \times 4} } = \sqrt {324} = 18$
:$\tuple {3, 3}: \sqrt {\paren {3 + 3}^2 + \dfrac {2304} {3 \times 3} } = \sqrt {292} \notin \Z$
:$\tuple {3, 4}: \sqrt {\paren {3 + 4}^2 + \dfrac {2304} {3 \times 4} } = \sqrt {241} \notin \Z$
Now we calculate $z$ for $\tuple {x, y} = \tuple {1, 3}, \tuple {2, 4}$.
:$\tuple {1, 3}: z = \dfrac {- 1 - 3 + 28} 2 = 12$
:$\tuple {2, 4}: z = \dfrac {- 2 - 4 + 18} 2 = 6$
Using:
:$a = s - x = x + y + z - x = y + z$
:$b = s - y = x + z$
:$c = s - z = x + y$
the possible sets of side lengths are:
:$\tuple {15, 13, 4}, \tuple {10, 8, 6}$
But note that the $\tuple {6, 8, 10}$ triangle is right-angled, and hence not a proper Heronian Triangle.
Hence the result.
{{qed}}