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}}