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. We set out to eliminate $s$ and simplify as best possible: {{begin-eqn}} {{eqn | l = \AA       | r = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }       | c =  }} {{eqn | l = \AA^2       | r = \dfrac {a + b + c} 2 \paren {\dfrac {a + b + c} 2 - a} \paren {\dfrac {a + b + c} 2 - b} \paren {\dfrac {a + b + c} 2 - c}       | c = substituting for $s$ and squaring }} {{eqn | ll= \leadsto       | l = 16 \AA^2       | r = \paren {a + b + c} \paren {-a + b + c} \paren {a - b + c} \paren {a + b - c}       | c = multiplying through by $16$ and simplifying }} {{eqn | r = 2 a^2 b^2 + 2 b^2 c^2 + 2 c^2 a^2 - a^4 - b^4 - c^4       | c = multiplying out and simplifying }} {{eqn | ll= \leadsto       | l = \paren {4 \AA}^2 + \paren {b^2 + c^2 - a^2}       | r = \paren {2 b c}^2       | c = factorising }} {{end-eqn}} This is now in the form $p^2 + q^2 = r^2$. From Solutions of Pythagorean Equation, $\tuple {p, q, r}$ has the parametric solution: :$\tuple {m^2 - n^2, 2 m n, m^2 + n^2}$ There are two steps to showing $6 \divides \AA$: === Step $1$: $2 \divides \AA$ === By Euclid's Lemma for Prime Divisors, we just need to show: :$2 \divides \AA^2 = s \paren {s - a} \paren {s - b} \paren {s - b}$ By Semiperimeter of Integer Heronian Triangle is Composite, $s \in \N$. There are $3$ cases: ==== Case $1$: There are sides with odd and even lengths ==== Under this condition, one of $s - x$ will be even. Then $\AA^2$ is also even. {{qed|lemma}} ==== Case $2$: All sides are of odd length ==== Suppose all sides are of odd length. Then the perimeter is also odd. But then the semiperimeter cannot be an integer. This is a contradiction. {{qed|lemma}} ==== Case $3$: All sides are of even length ==== Suppose all sides are of even length. If the semiperimeter is even, the result follows. Therefore we consider the case where $s$ is odd. Then $x := s - a$, $y := s - b$ and $z := s - c$ are also odd. Note that $x + y + z = 3 s - a - b - c = s$. Hence $\AA^2 = x y z \paren {x + y + z}$. Note that each of $x, y, z$ must be equivalent to $\pm 1 \pmod 4$. If all of $x, y, z \equiv 1 \pmod 4$: :$x + y + z \equiv -1 \pmod 4$ If two of $x, y, z \equiv 1 \pmod 4$: :$x + y + z \equiv 1 \pmod 4$ If one of $x, y, z \equiv 1 \pmod 4$: :$x + y + z \equiv -1 \pmod 4$ If none of $x, y, z \equiv 1 \pmod 4$: :$x + y + z \equiv 1 \pmod 4$ In any of the above cases, we have: :$\AA^2 \equiv -1 \pmod 4$ which is impossible by Square Modulo 4. This is a contradiction. {{qed|lemma}} In each valid case, we see that $2 \divides \AA^2$. {{qed|lemma}} === Step $2$: $3 \divides \AA$ === By Euclid's Lemma for Prime Divisors, we just need to show: :$3 \divides 16 \AA^2 = \paren {a + b + c} \paren {-a + b + c} \paren {a - b + c} \paren {a + b - c}$ We split the problem into $4$ cases: ==== Case $1$: None of $a, b, c$ are divisible by $3$ ==== If $a \equiv b \equiv c \pmod 3$: :$a + b + c \equiv 3 a \equiv 0 \pmod 3$ Hence: :$3 \divides \paren {a + b + c}$ If two of the lengths have the same remainder when divided by $3$, say $a \equiv b \not \equiv c \pmod 3$: :$a \equiv b \equiv -c \pmod 3$ Hence: :$a + b - c \equiv 3 a \equiv 0 \pmod 3$ Thus: :$3 \divides \paren {a + b - c}$ {{qed|lemma}} ==== Case $2$: One of $a, b, c$ is divisible by $3$ ==== {{WLOG}} suppose that number is $a$. If $b \equiv c \pmod 3$: :$a + b - c \equiv a \equiv 0 \pmod 3$ Hence :$3 \divides \paren {a + b - c}$ If $b \not \equiv c \pmod 3$: :$b \equiv -c \pmod 3$ Hence: :$a + b + c \equiv -c + c \equiv 0 \pmod 3$ Thus: :$3 \divides \paren {a + b + c}$ {{qed|lemma}} ==== Case $3$: Two of $a, b, c$ is divisible by $3$ ==== {{WLOG}}, suppose $3 \divides a, b$. Then: {{begin-eqn}} {{eqn | l = 16 \AA^2       | r = \paren {a + b + c} \paren {-a + b + c} \paren {a - b + c} \paren {a + b - c} }} {{eqn | o = \equiv       | r = c \cdot c \cdot c \cdot \paren {-c}       | rr = \pmod 3 }} {{eqn | o = \equiv       | r = -c^4       | rr = \pmod 3 }} {{eqn | o = \equiv       | r = -1       | rr = \pmod 3       | c = Square Modulo 3 }} {{end-eqn}} By Square Modulo 3, $16 \AA^2 = \paren {4 \AA}^2$ is not a square. This is a contradiction. Therefore this case is not valid. {{qed|lemma}} ==== Case $3$:  $a, b, c$ are all divisible by $3$ ==== We have: :$3 \divides \paren {a + b + c}$ {{qed|lemma}} We see that in every valid case, $3 \divides 16 \AA^2$. Hence $3 \divides \AA$, and thus $6 \divides \AA$. {{qed}}