Semiperimeter of Integer Heronian Triangle is Composite

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The semiperimeter of an integer Heronian triangle is always a composite number.


Let $a, b, c$ be the side lengths of an integer Heronian triangle.

By Heron's Formula, its area is given by:

$\AA = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} } \in \N$

where the semiperimeter $s$ is given by:

$s = \dfrac {a + b + c} 2$

First we prove that $s$ is indeed an integer.

Aiming for a contradiction, suppose not.

Since $2 s = a + b + c \in \N$, $2 s$ must be odd.

Hence $2 s - 2 a, 2 s - 2 b, 2 s - 2 c$ are odd as well.


\(\ds 16 \AA^2\) \(=\) \(\ds 16 s \paren {s - a} \paren {s - b} \paren {s - c}\)
\(\ds \) \(=\) \(\ds 2 s \paren {2 s - 2 a} \paren {2 s - 2 b} \paren {2 s - 2 c}\)

Since $16 \AA^2$ is a product of odd numbers, it must be odd.

But then $\AA^2$ is not an integer, a contradiction.

Therefore $s \in \N$.


Now we show that $s$ is composite number.

Aiming for a contradiction, suppose not.

Then $s$ is either $1$ or prime.

Since $a, b, c \ge 1$, $s \ge \dfrac 3 2 > 1$.

Hence $s$ is prime.


$\AA^2 = s \paren {s - a} \paren {s - b} \paren {s - c}$

We have $s \divides \AA^2$.

By Prime Divides Power, $s^2 \divides \AA^2$.

Thus $s \divides \paren {s - a} \paren {s - b} \paren {s - c}$.

By Euclid's Lemma, $s$ divides some $s - x$.

However by Absolute Value of Integer is not less than Divisors:

$s \le s - x$

which is a contradiction.

Therefore $s$ is composite.