Generating all Pythagorean Triples

This section in pdf form. triples2.pdf

If $(a,b,c)$ is a positive integer solution to the equation

$$a^2+b^2=c^2$$ (1)

then $(a,b,c)$ is a Pythagorean triple. If $a,b$ , and $c$ have no common divisors greater than $1$ , then $(a,b,c)$ is a primitive Pythagorean triple (PPT). Similarly, if $a^2+b^2=c^2$ where $(a,b,c)$ is a PPT then $a^2+b^2=c^2$ is primitive. In the following list only the second triangle is not primitive.

1. $3^2+4^2=5^2$
2. $6^2+8^2=10^2$
3. $5^2+12^2=13^2$
4. $7^2+24^2=\left(5^2\right)^2$
5. $\left(15^2\right)^2+272^2=353^2$

Clearly, if $k$ divides any two of $a, b$ , and $c$ it divides all three. And if $a^2+b^2=c^2$ then $k^2a^2+k^2b^2=k^2c^2$ . That is, for a positive integer $k$ , if $(a,b,c)$ is a Pythagorean triple then so is $(ka,kb,kc)$ . Hence, to find all Pythagorean triples, it's sufficient to find all primitive Pythagorean triples.

Let $a,b$ , and $c$ be relatively prime positive integers such that $a^2+b^2=c^2$ . Set

$$\frac{m}{n}=\frac{c+a}{b}$$

reduced to lowest terms, That is, $\gcd(m,n)=1$ . From the triangle inequality $m > n$ . Then

$$\frac{m}{n} b-a=c.$$ (2)

Squaring both sides of (2) and multiplying through by $n^2$ we get

$$m^2b^2-2mnab+n^2a^2=n^2a^2+n^2b^2;$$

which, after canceling and rearranging terms, becomes

$$b\left(m^2-n^2\right)=a(2mn).$$ (3)

There are two cases, either $m$ and $n$ are of opposite parity, or they or both odd. Since $\gcd(m,n)=1$ , they can not both be even.

Case 1. $m$ and $n$ of opposite parity, i.e., $m \not\equiv \pm n ( mod 2 )$ . So 2 divides b since $m^2-n^2$ is odd. From equation (2), $n$ divides $b$ . Since $\gcd(m,n)=1$ then $\gcd(m,m^2-n^2)=1$ , therefore $m$ also divides $b$ . And since $\gcd(a,b)=1$ , $b$ divides $2mn$ . Therefore $b=2mn$ . Then

$$a=m^2-n^2, \quad b=2mn, , c=\frac{m}{n} 2mn-(m^2-n^2)=m^2+n^2.$$ (4)

Case 2. $m$ and $n$ both odd, i.e., $m \equiv \pm n ( mod 2 )$ . So 2 divides $m^2-n^2$ . Then by the same process as in the first case we have

$$a=\frac{m^2-n^2}{2}, \quad b=mn,\quad and \quad c=\frac{m^2+n^2}{2}.$$ (5)

The parametric equations in (4) and (5) appear to be different but they generate the same solutions. To show this, let

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

Then $m=u+v$ , and $n=u-v$ . Substituting those values for $m$ and $n$ into (5) we get

$$a=2uv, b=u^2-v^2, \quad c=u^2+v^2$$ (6)

where $u > v$ , $gcd(u,v)=1$ , and $u$ and $v$ are of opposite parity. Therefore (6), with the labels for a and b interchanged, is identical to (4). Thus since $\left\{m^2-n^2,2mn,m^2+n^2\right\}$ , as in (4), is a primitive Pythagorean triple, we can say that $(a,b,c)$ is a primitive pythagorean triple if and only if there exists relatively prime, positive integers $m$ and $n$ , $m > n$ , such that $a=m^2-n^2, b=2mn,$ and $c=m^2+n^2$ . And $(a,b,c)$ is a Pythagorean triple if and only if

$$a=k\left(m^2-n^2\right),\quad b=k(2mn),$$   and$$\quad c=k\left(m^2+n^2\right)$$

where $k$ is a positive integer.

Alternatively, $(a,b,c)$ is a Pythagorean triple if and only if there exists relatively prime, positive integers $u$ and $v$ , $u > v$ , $u \equiv v \pmod{2}$ such that

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

Another Method (using Gaussian Integers).

Let

$$a^2+b^2=c^2$$ (7)

where $a, b$ , and $c$ are pairwise, relatively prime, positive integers. Let $z=a+bi$ and $\bar{z}=a-bi$ , where $i=\sqrt{-1}$ . Then $z$ and $\bar{z}$ are Gaussian Integers, and $\bar{z}$ is the conjugate of $z$ . Note that

$$a=\frac{z+\bar{z}}{2},\quad b=\frac{z-\bar{z}}{2i} , \quad c=\sqrt{z\bar{z}}.$$ (8)

Since $\gcd(a,b)=1$ then $\gcd(z,\bar{z})=1$ . Hence each of $z$ and $\bar{z}$ is a square. That is, there exists integers $m$ and $n$ such that $(m+ni)^2=z$ and $(m-ni)^2=\bar{z}$ . So, from equation (8),

$$a =\frac{(m+ni)^2+(m-ni)^2}{2}=m^2-n^2,$$

$$b =\frac{(m+ni)^2-(m-ni)^2}{2i}=2mn,$$

$$\quad c =\sqrt{(m+ni)^2(m-ni)^2}=m^2+n^2.$$

Since $a$ is a positive integer, $m$ and $n$ must be positive integers, $m > n$ . And since $\gcd(a,b)=1$ , $m$ and $n$ must be relatively prime and of opposite parity.

Equation (8) illustrates a method for finding primitive Pythagorean triangles where the hypotenuse is to a power. We have the identity,

$$\left\vert \frac{z^{2k}+\bar{z}^{2k}}{2}\right\vert^2+\left\vert ... ...c{z^{2k}-\bar{z}^{2k}}{2i}\right\vert^2 =\Bigl(\left(z\bar{z}\right)^k\Bigr)^2.$$ (9)

The absolute values are necessary since the terms on the left, depending on $k$ , may, or may not, be positive.

Example

Let $a=3, b=4$ , and $k=3$ . Then, from equation (9), we have,

$$\left\vert\frac{(3+4i)^6+(3-4i)^6}{2}\right\vert^2+ \left\vert\frac{(3+4i)^6-(3-4i)^6}{2i}\right\vert^2 =\Bigl(\bigl((3+4i)(3-4i)\bigr)^3\Bigr)^2$$

$$= 11753^2+\bigl\vert-10296\bigr\vert^2=\left(25^3\right)^2.$$

Still another method -- the difference method.

Let $(a,b,c)$ be a primitive Pythagorean triple. Without loss of generality, let $a$ be odd. Equation (7) can be written as,

$$a^2=c^2-b^2.$$

Set $w=c+b$ and $\bar{w}=c-b$ . Thus,

$$a=\sqrt{w\bar{w}} ,\quad c=\frac{w+\bar{w}}{2}, \quad b=\frac{w-\bar{w}}{2}.$$ (10)

Since $(a,b,c)$ is primitive, $\gcd(c,b)=\gcd(w,\bar{w})=1$ . This implies that each of $w$ and $\bar{w}$ is a square. So, since any odd positive integer greater than 1 can be written as the difference of two positive integer squares, there exists positive integers $m$ and $n$ , $m > n$ , such that $(m+n)^2=w$ and $(m-n)^2=\bar{w}$ . Hence, from equation (10),

$$a = \sqrt{(m+n)^2(m-n)^2}=m^2-n^2,$$

$$c = \frac{(m+n)^2+(m-n)^2}{2}=m^2+n^2,$$

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

And, since $\gcd(a,b)=1$ , $m$ and $n$ must be relatively prime and of opposite parity.

Equation (10) illustrates an efficient method for finding primitive Pythagorean triples where the odd leg is to a power. We have the identity,

$$\Bigl(\left(w\bar{w}\right)^k\Bigr)^2=\left(\frac{w^{2k}+\bar{w}^{2k}}{2}\right)^2 -\left(\frac{w^{2k}-\bar{w}^{2k}}{2}\right)^2 .$$ (11)

Example

Let $c=5, b=4$ , and $k=3$ . Then, from equation (11), we have,

$$\Bigl(\bigl((5+4)(5-4)\bigr)^3\Bigr)^2 =\left(\frac{(5+4)^6+(5-4)^6}{2}\right)^2 -\left(\frac{(5+4)^6-(5-4)^6}{2}\right)^2.$$

$$\quad \left(9^3\right)^2 =265721^2-265720^2.$$

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