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Triangles with integer sides (a,b,c), perimeter (P), area (A), inradius (r), and circumradius (R) necessarily have sides which are even, and a perimeter divisible by 4 and I'd like to prove this but cannot as yet. I've search the web and this site but have found nothing about this.

Here's a short list:

$$ \left( \begin{array}{ccccccc} \text{a} & \text{b} & \text{c} & \text{P} & \text{A} & \text{R} & \text{r} \\ 6 & 8 & 10 & 24 & 24 & 5 & 2 \\ 12 & 16 & 20 & 48 & 96 & 10 & 4 \\ 10 & 24 & 26 & 60 & 120 & 13 & 4 \\ 18 & 24 & 30 & 72 & 216 & 15 & 6 \\ 16 & 30 & 34 & 80 & 240 & 17 & 6 \\ 14 & 30 & 40 & 84 & 168 & 25 & 4 \\ 24 & 32 & 40 & 96 & 384 & 20 & 8 \\ 30 & 30 & 48 & 108 & 432 & 25 & 8 \\ \end{array} \right) $$

So far I understand these triangles must necessarily be Heronian Triangles (those with integer sides and area). And being Heronian, the perimeter must be even and the area divisible by 6. I've worked with the formulas below and are not successful showing the sides are positive and perimeter divisible by 4 and was wondering if someone could help me with this?

$$ \begin{align} s&=\frac{a+b+c}{2}\\ A&=\sqrt{s(s-a)(s-b)(s-c)}\\ R&=\frac{abc}{4A}\\ r&=\frac{A}{s} \end{align} $$

Dominic
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What an interesting observation. Here is a proof based upon a well known result for Heronian triangles.

The Brahmagupta formula for the sides of any Heronian triangle says that they are proportional to $n(m^2+k^2), m(n^2+k^2), (m+n)(mn-k^2)$ where $m,n,k$ are positive integers with no common factor. Then the perimeter is $P=2mn(m+n)$ and the circumradius is $R=\frac{(m^2+k^2)(n^2+k^2)}{4k}.$

Scaling to make $R$ obviously integeral we have $$a=4kn(m^2+k^2)$$ $$b=4km(n^2+k^2)$$ $$c=4k(m+n)(mn-k^2)$$ $$P=8kmn(m+n)$$$$R=(m^2+k^2)(n^2+k^2)$$

If $k$ is either the only odd one of $k,m,n$ or the only even one, then $R$ is odd and so any scaling which leaves $R$ integral keeps the sides divisible by $8$ and the perimeter divisible by $32$.

If $m,n,k$ are all odd then each side is divisible by $8$ and the perimeter is divisible $16$, whereas $R$ is divisible by $4$ but not by $8$. So any scaling which leaves $R$ integral keeps the sides even and the perimeter divisible by $4$.

If $k,m$ are odd and $n$ is even (similarly with $m$ and $n$ interchanged) then $R$ is singly even so any scaling which leaves $R$ integral keeps the sides even and the perimeter divisible by $8$.

The final case is where $m,k$ are even and $n$ is odd (similarly with $m$ and $n$ interchanged). Let $m=2^uM,k=2^uK$ where at least one of $M,K$ is odd. Then $a,b,c$ are all divisible by $2^{2u+2}$ and P is divisible by $2^{2u+3}$ whereas $R$ is not divisible by $2^{2u+2}$ . So again the sides remain even and $P$ stays divisible by $4$ when the factor of proportionality is considered.

  • Thanks a bunch for that S. Dolan. I'm working through it but lil tough for me. I'll get it in a few days I'm sure. :) Oh, and I'll check it off then too. – Dominic Dec 07 '19 at 13:29
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    That's fine. It's perhaps worth mentioning that the odd/even results depend upon the parity of numbers of the form $x^2+y^2$. These numbers are odd if $x,y$ are of opposite parity and singly even (divisible by 2 but not 4) if $x,y$ are both odd. –  Dec 07 '19 at 16:15