Finding m and n if given a primitive Pythagorean triple

Have you ever been, while attending a rap concert, political gathering, or pro-wrestling event, asked to find numerical values for the parameters used to generate a particular primitive Pythagorean triple? And did you suffer embarrassment and scorn because you did not know how to perform such a simple task? If you answered yes to those two questions then this page is for you.

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This section in pdf form. triplesmn.pdf

Claim 1   If $a^2+b^2=c^2$ is a primitive Pythagorean triangle, where $a$ is odd, then each of the fractions

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

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

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

Proof.

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

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

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

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

$\qedsymbol$

Claim 2   If $a^2+b^2=c^2$ is a primitive Pythagorean triangle, where $a$ is even, then each of the fractions

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

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

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

Proof. Since $a,b,c$ is a PPT, there exists relatively prime, positive integers, $u$ and $v$ , $u > v$ , $u$ and $v$ of opposite parity such that

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

So,

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

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

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

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

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

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

Hence, we have,

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

$\qedsymbol$

Examples

$3^2+4^2=5^2$ is a primitive Pythagorean triple and 3 is odd. So, let $a=3$ . Then

$$\frac{5+3}{4}=\frac{5+4+3}{5+4-3}=\frac{4}{5-3}=\frac{3+5-4}{3+4-5}=\frac{2}{1} .$$

Hence

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

$20^2+21^2=29^2$ is a PPT and 21 is odd. So

$$\frac{29+21}{20}=\frac{29+20+21}{29+20-21}=\frac{20}{29-21}=\frac{21+29-20}{21+20-29}=\frac{5}{2} .$$

Hence,

$$20=2(5)(2),\quad 21=5^2-2^2 ,$$   and$$\quad 29=5^2+2^2 .$$

$4^2+3^2=5^2$ is a primitive Pythagorean triple and 4 is even. So, let $a=4$ . Then

$$\frac{5+4}{3}=\frac{5+3+4}{5+3-4}=\frac{3}{5-4}=\frac{4+5-3}{4+3-5}=\frac{3}{1} .$$

Thus

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

$20^2+21^2=29^2$ is a PPT and 20 is even. So let a=20. Then

$$\frac{29+20}{21}=\frac{29+21+20}{29+21-20}=\frac{21}{29-20}=\frac{20+29-21}{20+21-29}=\frac{7}{3} .$$

Therefore

$$20=\frac{7^2-3^2}{2}, \quad 21=(7)(3),$$   and$$\quad 29=\frac{7^2+3^2}{2} .$$

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