Area of Integer Heronian Triangle is Multiple of 6

== Theorem ==
Let $\triangle {ABC}$ be an integer Heronian triangle.
Then the area of $\triangle {ABC}$ is a multiple of $6$.
== Proof ==
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:
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.
==== 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.
==== 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.
In each valid case, we see that $2 \divides \AA^2$.
=== 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:
:$3 \divides \paren {a + b - c}$
==== Case $2$: One of $a, b, c$ is divisible by $3$ ====
 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:
:$3 \divides \paren {a + b + c}$
==== Case $3$: Two of $a, b, c$ is divisible by $3$ ====
, suppose $3 \divides a, b$.
Then:
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.
==== Case $4$: $a, b, c$ are all divisible by $3$ ====
We have:
:$3 \divides \paren {a + b + c}$
We see that in every valid case, $3 \divides 16 \AA^2$.
Hence $3 \divides \AA$, and thus $6 \divides \AA$.
== Sources ==
* : Bachet: $111$
Category:Heronian Triangles