Finding m and n if given a 120 degree primitive triple

If given a primitive triangle $a^2+b^2+ab=c^2$ , we know that one of $a$ and $b$ is equal to $m^2-n^2$ and the other equals $2mn+n^2$ where $m$ and $n$ are relatively prime, positive integers, $m > n$ , and $3\nmid m-n$ . But, how do we determine which of $a$ and $b$ equals $m^2-n^2$ ?

Lemma 2   If $a^2+b^2+ab=c^2$ is a primitive integer-sided 120 degree triangle, generated by the parameters $m$ and $n$ , then $a=m^2-n^2$ if and only if $3\nmid c+a-b$ .

Proof. $(\Rightarrow)\quad a=m^2-n^2$ . Then $c+a-b=\left(m^2+n^2+mn\right)+\left(m^2-n^2\right)-\left(2mn+n^2\right)=(2m+n)(m-n)$ . Since $3\nmid m-n$ then $m=3k+n+1$ or $3k+n+2$ for some integer $k$ . So, $2m+n=3(k+n)+2$ or $3(k+n+1)+1$ . Therefore $3\nmid c+a-b$

$(\Leftarrow)\quad 3\nmid c+a-b$ . Either $a=2mn+n^2$ or $a=m^2-n^2$ . If $a=2mn+n^2$ then $c+a-b=\left(m^2+n^2+mn\right)+\left(2mn+n^2\right)-\left(m^2-n^2\right)=3\left(n^2+mn\right)$ . Therefore $a=m^2-n^2$ . $\qedsymbol$

Theorem 6   Let $a^2+b^2+ab=c^2$ be a given primitive, positive integer-sided triangle. Assign labels $a$ and $b$ such that $3\nmid c+a-b$ ; from Lemma (2), this can be done in exactly one way. Then the fractions

$$\frac{c+a}{b},\quad \frac{c+b+2a}{c+b-a},\quad \frac{a+b}{c-a},$$ and $$\frac{2c+a-b}{2b+a-c}$$

(each fraction reduced to lowest terms) is equal to $\frac{m}{n}$ where

$$a=m^2-n^2 ,\quad b= 2mn+n^2,$$   and$$\quad c=m^2+n^2+mn.$$

Proof. Since $3\nmid c+a-b$ then $a=m^2-n^2$ . So,

$$\frac{c+a}{b} =\frac{\left(m^2+n^2+mn\right)+\left(m^2-n^2\right)}{2mn+n^2}= \frac{m(2m+n)}{n(2m+n)}=\frac{m}{n} ,$$

$$\frac{c+b+2a}{c+b-a} =\frac{\left(m^2+n^2+mn\right)+\left(2mn+n^2\right)+2\left(m^2-n^... ...ft(2mn+n^2\right)-\left(m^2-n^2\right)}=\frac{3m(m+n)}{3n(m+n)} =\frac{m}{n} ,$$

$$\frac{a+b}{c-a} =\frac{\left(m^2-n^2\right)+\left(2mn+n^2\right)} {\left(m^2+n^2+mn\right)-\left(m^2-n^2\right)}=\frac{m(m+2n)}{n(2n+m)}=\frac{m}{n} ,$$

$$\quad \frac{2c+a-b}{2b+a-c} =\frac{2\left(m^2+n^2+mn\right)+\left(m^2-n^2\right) -\left(2mn+n... ...+\left(m^2-n^2\right)-\left(m^2+n^2+mn\right)} =\frac{3m^2}{3mn}=\frac{m}{n} .$$

$\qedsymbol$

Theorem 7   Let $a^2+b^2+ab=c^2$ be a given primitive, positive integer-sided triangle. Assign labels $a$ and $b$ such that $3\mid c+a-b$ ; from Lemma (2), this can be done in exactly one way. Then the fractions

$$\frac{c+a}{b},\quad \frac{c+b+2a}{c+b-a},\quad \frac{a+b}{c-a},$$ and $$\frac{2c+a-b}{2b+a-c}$$

(each fraction reduced to lowest terms) is equal to $\frac{m}{n}$ where

$$a=\frac{m^2-n^2}{3} ,\quad b= \frac{2mn+n^2}{3},$$   and$$\quad c=\frac{m^2+n^2+mn}{3} .$$

Proof. Since $3\mid c+a-b$ there exists relatively prime, positive integers $u$ and $v$ , $u > v$ , $u \equiv v \pmod{3}$ such that

$$a=2uv+v^2,\quad b=u^2-v^2,$$   and$$\quad c=u^2+v^2+uv.$$

Then

$$\frac{c+a}{b}= \frac{u^2+v^2+uv+2uv+v^2}{u^2-v^2}=\frac{(u+v)(u+2v)}{(u+v)(u-v)}=\frac{u+2v}{u-v} .$$

$$\frac{c+b+2a}{c+b-a}= \frac{u^2+v^2+uv+u^2-v^2+2\left(2uv+v^2\right)}{u^2+v^2+uv+u^2-v^... ...u^2+5uv+2v^2}{2u^2-uv-v^2}=\frac{(2u+v)(u+2v)}{(2u+v)(u-v)}=\frac{u+2v}{u-v} .$$

$$\frac{a+b}{c-a}= \frac{2uv+v^2+u^2-v^2}{u^2+v^2+uv-2uv-v^2}=\frac{u(u+2v)}{u(u-v)}=\frac{u+2v}{u-v} .$$

$$\frac{2c+a-b}{2b+a-c}= \frac{2\left(u^2+v^2+uv\right)+2uv+v^2-u^2+v^2} {2\left(u^2-v^2\right)+2uv+v^2-u^2-v^2-uv}=\frac{(u+2v)^2}{(u+2v)(u-v)}=\frac{u+2v}{u-v} .$$

Let $m=u+2v$ and $n=u-v$ . Then

$$u=\frac{m+2n}{3}$$   and$$\quad v=\frac{m-n}{3} .$$

Hence, we have,

$$a = 2uv+v^2=\frac{m^2-n^2}{3} ,$$

$$b = u^2-v^2=\frac{2mn+n^2}{3} ,$$

$$\quad c =u^2+v^2+uv=\frac{m^2+n^2+mn}{3} .$$

$\qedsymbol$

Examples

$3^2+5^2+(3)(5)=7^2$ is a primitive 120 degree triangle. $3\nmid 7+3-5=5$ . Therefore $a=3$ . So,

$$\frac{7+3}{5}=\frac{7+5+6}{7+5-3}=\frac{3+5}{7-3}=\frac{14+3-5}{10+3-7}=\frac{2}{1} .$$

Hence $3=2^2-1^2$ , $5=2(2)(1)+1^2$ , and $7=2^2+1^2+(2)(1)$ .

$32^2+45^2+(32)(45)=67^2$ is primitive. $3\nmid 67+45-32=70$ . Therefore $a=45$ . Thus

$$\frac{67+45}{32}=\frac{67+32+90}{67+32-45}=\frac{45+32}{67-45}=\frac{134+45-32}{64+45-67} =\frac{7}{2} .$$

Hence $45=7^2-2^2$ , $32=2(7)(2)+2^2$ , and $67=7^2+2^2+(7)(2)$ .

$3^2+5^2+(3)(5)=7^2$ is a primitive 120 degree triangle. $3\mid 7+5-3=9$ . Therefore $a=5$ . So,

$$\frac{7+5}{3}=\frac{7+3+10}{7+3-5}=\frac{5+3}{7-5}=\frac{14+5-3}{6+5-7}=\frac{4}{1} .$$

Hence

$$5=\frac{4^2-1^2}{3},\quad 3=\frac{2(4)(1)+1^2}{3},$$   and$$\quad 7=\frac{4^2+1^2+(4)(1)}{3} .$$

$32^2+45^2+(32)(45)=67^2$ is primitive. $3\mid 67+32-45=54$ . Therefore $a=32$ . Thus

$$\frac{67+32}{45}=\frac{67+45+64}{67+45-32}=\frac{32+45}{67-32}=\frac{134+32-45}{90+32-67}=\frac{11}{5} .$$

Hence

$$32=\frac{11^2-5^2}{3},\quad 45=\frac{2(11)(5)+5^2}{3},$$   and$$\quad 67=\frac{11^2+5^2+(11)(5)}{3} .$$

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