Classify all $n$ for which there exists a right triangle with coprime integer side lengths and such that its area equals $n$ times its perimeter.

Question

I know a characterization of primitive pythagorean triples: A right triangle has coprime integer side lengths if and only if those side lengths are: \begin{align*} m^2-k^2, \qquad 2mk, \qquad m^2+k^2 \end{align*} for some $m,k \in \mathbb{N}$ where $\gcd(m,k)=1$, $\, m>k$, and $\,m\not\equiv k$ (mod $2$). So essentially I'm looking for a classification of all $n\in \mathbb{N}$ for which there exist these $m,k$ where (plugging into the area/perimeter formulas for triangles): \begin{align*} (m^2-k^2)mk=n(2m^2 +2mk) \end{align*} And then solving for $n$: \begin{align*} n = \frac{(m+k)(m-k)mk}{2m(m+k)} = \frac{(m-k)k}{2} \end{align*} Since $m\not \equiv k$ (mod $2$), $\,2 \nmid (m-k)$, and thus $k$ must be even, and $m$ odd. Set $\frac{k}{2}=c$. Maybe an answer to this problem could be that a necessary and sufficient condition for $n$ is that: \begin{align*} n=(m-2c)c \end{align*} for some $m,c \in \mathbb{N}$ where $m$ is odd, $\,m>2c$, and $\gcd(m,c)=1$. Here's my issue with this answer though - it doesn't exactly help determine if some given $n$ has the desired property or not, at least not right off the bat... For instance, if I wanted to know if there's a coprime side-length right triangle with area equaling $147$ times its perimeter... well I don't know if $147=(m-2c)c\,$ for any $m,c\in \mathbb{N}$. What else can be done here?

Answer 1

Let $c$ be the highest power of $2$ that goes evenly into $n$, and let $m=\frac{n}{c}+2c$. Thus every $n$ has a corresponding $m$ and $c$.

Since this makes $k$ a power of $2$ and $m$ odd, they are coprime.

Answer 2

m is odd iff m-2c is odd

gcd(m,c)=1 iff gcd(m-2c,c)=1

Therefore we can formulate our final solution as n=pc where gcd(p,c)=1 and p is odd. Ofcourse for all n, this gives us p=1 and c=n. This gives us k=2n and m=2n+1. These are certainly coprime with m>k. And checking that they satisfy $$\begin{align*} (m^2-k^2)mk=n(2m^2 +2mk) \end{align*}$$ $$\impliedby(m-k)k=2n$$ Plugging in k=2n and m=2n+1, $$((2n+1)-(2n))(2n)=2n$$ So it seems n can be any natural number.

With side lengths, $4n+1$, $8n^2+4n+1$ and $8n^2+4n$

Answer 3

We seek a formula where area/perimeter ratio = R which can be any multiple of $1/2$. \begin{equation} R=\frac{mk-k^2}{2}\quad\implies k=\frac{m\pm\sqrt{m^2-8R}}{2}\quad\text{for}\quad \big\lceil\sqrt{8R}\big\rceil\le m \le (2R+1) \end{equation} The lower limit insures that $k\in \mathbb{N}$ and the upper limit ensures that $m> k$. $$R=1\implies \lceil\sqrt{8}\rceil=3\le m \le (2+1)=3 \qquad\land\qquad m\in\{ 3\}\implies k\in\{ 2,1\}$$ $$F(3,2)=(5,12,13)\quad\land\quad \frac{30}{30}=1\qquad\qquad\qquad F(3,1)=(8,6,10)\quad\land\quad \frac{24}{24}=1$$ $\\$ \begin{equation} R=3\implies \lceil\sqrt{8*3}\rceil=5\le m \le (2*3+1)=7\\ m\in\{ 5\}\implies k\in\{ 3,2\} \qquad\land\qquad m\in\{ 7\}\implies k\in\{ 6.1\} \end{equation} \begin{equation} f(5,3)=(16,30,34)\quad R=\frac{16*30/2}{16+30+34}=3\\ f(5,2)=(21,20,29)\quad R=\frac{21*20/2}{21+20+29}=3\\ f(7,6)=(13,84,85)\quad R=\frac{13*84/2}{13+84+85}=3\\ f(7,1)=(48,14,50)\quad R=\frac{48*14/2}{48+14+50}=3 \end{equation}

Note that when $m,k$ are of the same parity, the triples are not primitive, even if they are co-prime. Some claim that $k=2^n$ but a counterexample is $R=1.5\text{ where } m\in\{4\}\implies k\in\{3,1\}$. $$f(4,3)=(7,24,25)\quad\implies R=\frac{7*24/2}{7+24+25}=\frac{84}{56}=1.5$$ $$f(4,1)=(15,8,17) \quad\implies R=\frac{15*8/2}{15+8+17}=\frac{60}{40}=1.5$$

For your final question, using this technique, here are some triples where R=147.

$$f(35,21)=(784,1470,1666)\qquad f(35,14)=(1029,980,1421)\qquad f(49,42)=(637,4116,4165)\qquad f(49,7)=(2352,686,2450)\qquad f(55,49)=(624,5390,5426)\qquad f(55,6)=(2989,660,3061)\qquad f(101,98)=(597,19796,19805)\qquad f(101,3)=(10192,606,10210)\qquad f(149,147)=(592,43806,43810)\qquad f(149,2)=(22197,596,22205)\qquad f(295,294)=(589,173460,173461)\qquad f(295,1)=(87024,590,87026) $$