Pythagorean Triples

If x and y are the legs of a right triangle and z is the hypotenuse, then Pythagoras' theorem says $x^2 + y^2 = z^2$ . A triple of integers $\{x,y,z\}$ is a Pythagorean triple if it satisfies $x^2 + y^2 = z^2$ . (In what follows, I'll assume that x, y, and z are positive integers.)

For example $\{3,4,5\}$ is a Pythagorean triple, since $3^2 + 4^2 = 5^2$ . $\{6,8,10\}$ is also a Pythagorean triple, but there is a sense in which it's "redundant": $2\cdot \{3,4,5\} = \{6,8,10\}$ . If a Pythagorean triple is not a proper multiple of of another triple, it is said to be primitive. Thus, $\{x,y,z\}$ is a primitive Pythagorean triple if $(x,y,z) = 1$ .

The result I'll prove will show how you can generate all primitive Pythagorean triples.

Theorem.

(a) Suppose a and b are positive numbers, one is even and the other is odd, $a > b$ , and $(a,b) = 1$ . Then

$$x = 2ab, \quad y = a^2 - b^2, \quad z = a^2 + b^2$$

is a primitive Pythagorean triple.

(b) Suppose $\{x,y,z\}$ is a primitive Pythagorean triple. Then one of x, y is even and the other is odd. If x is even, then there are positive numbers a and b, such that one is even and the other is odd, $a > b$ , $(a,b) = 1$ , and

$$x = 2ab, \quad y = a^2 - b^2, \quad z = a^2 + b^2.$$

A similar statement holds if y is even.

Proof. (a)

$$x^2 + y^2 = (2ab)^2 + (a^2 - b^2)^2 = 4a^2b^2 + a^4 - 2a^2b^2 + b^4 = a^4 + 2a^2b^2 + b^4 = (a^2 + b^2)^2 = z^2.$$

Therefore, $\{x,y,z\}$ is a Pythagorean triple. I have to show it's primitive.

Suppose on the contrary that $p \mid x, y, z$ , where p is prime. One of a, b, is even and the other is odd, so y and z must be odd. On the other hand, x is even. Therefore, $p \ne 2$ .

Now $p \mid y$ and $p \mid z$ implies $p \mid y + z = 2a^2$ . Since $p \ne 2$ , $p \mid a^2$ . Since p is prime, $p \mid a$ .

Likewise, $p \mid y$ and $p \mid z$ implies $p \mid z - y = 2b^2$ . Since $p \ne 2$ , $p \mid b^2$ . Since p is prime, $p \mid b$ .

This is a contradiction, because $(a,b) = 1$ .

Therefore, $(x,y,z) = 1$ , and $\{x,y,z\}$ is a primitive Pythagorean triple.

(b) Suppose $\{x,y,z\}$ is a primitive Pythagorean triple, so $x^2 + y^2 = z^2$ and $(x,y,z) = 1$ . First, I'll show that one of x, y must be even and the other odd.

If both x and y are even, then $x^2 + y^2 = z^2$ is even, so z is even. This contradicts $(x,y,z) = 1$ .

Suppose both x and y are odd. Note that the square of an odd number is congruent to 1 mod 4:

$$(2m + 1)^2 = 4m^2 + 4m + 1 = 1 \mod{4}.$$

So $z^2 = x^2 + y^2 = 1 + 1 = 2 \mod{4}$ . This is impossible, because only 0, 1, and 4 are squares mod 4.

Therefore, one of x, y must be even and the other odd. Suppose x is even and y is odd. Note that $z^2 = x^2 + y^2$ must be odd, so z must be odd. This means that $z - y$ and $z + y$ are even. Then

$$x^2 = z^2 - y^2 \quad\hbox{implies}\quad \left(\dfrac{x}{2}\right)^2 = \left(\dfrac{z - y}{2}\right) \left(\dfrac{z + y}{2}\right),$$

and $\dfrac{x}{2}$ , $\dfrac{z - y}{2}$ , and $\dfrac{z + y}{2}$ are all integers.

Next, I'll show that $\left(\dfrac{z - y}{2},\dfrac{z + y}{2}\right) = 1$ . Suppose p is a prime and $p \bigm| \dfrac{z - y}{2},\dfrac{z + y}{2}$ . Then

$$p \bigm| \dfrac{z - y}{2} + \dfrac{z + y}{2} = z$$

$$p \bigm| \dfrac{z + y}{2} - \dfrac{z - y}{2} = y$$

$$p \bigm| \left(\dfrac{x}{2}\right)^2 \quad\hbox{so}\quad p \bigm| \dfrac{x}{2} \bigm| x$$

This contradicts $(x,y,z) = 1$ . Thus, $\left(\dfrac{z - y}{2},\dfrac{z + y}{2}\right) = 1$ .

Now $\left(\dfrac{x}{2}\right)^2 = \left(\dfrac{z - y}{2}\right) \left(\dfrac{z + y}{2}\right)$ expresses a product of two relatively prime integers as a perfect square. By the Fundamental Theorem of Arithmetic, each of the numbers on the right must be a perfect square:

$$\dfrac{z - y}{2} = a^2, \quad \dfrac{z + y}{2} = b^2.$$

Note that $(a,b) = 1$ , for if p is prime and $p \mid a, b$ , then $p \bigm| \dfrac{z - y}{2}, \dfrac{z + y}{2}$ .

If a and b are both odd or both even, then $z = a^2 + b^2$ and $y = b^2 - a^2$ are both even, contrary to assumption. Hence, one of a, b, is odd and the other is even.

Finally,

$$\left(\dfrac{x}{2}\right)^2 = a^2b^2, \quad\hbox{so}\quad x^2 = 4a^2b^2, \quad\hbox{and}\quad x = 2ab.\quad\halmos$$

Example. Take $a = 21$ . Choose a number b that is less than 21, such that b has different parity than 21 (so b is even), and such that $(a,b) = 1$ . For example, let $b = 16$ . Then

$$a^2 - b^2 = 185, \quad 2ab = 672, \quad a^2 + b^2 = 697.$$

Since

$$185^2 + 672^2 = 485809 = 697^2,$$

$\{185, 672, 697\}$ is a primitive Pythagorean triple.

Example. You can use the theorem to generate all primitive Pythagorean triples. To do this, fix the bigger number a. Then consider b's less than a such that b is of different parity than a and such that $(a,b) = 1$ . These requirements on b eliminate many possibilities. For each pair of numbers a and b, the formulas in the theorem give the elements x, y, and z of the triple.

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & a & & b & & $x = 2ab$ & & $y = a^2 - b^2$ & & $z = a^2 + b^2$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 2 & & 1 & & 4 & & 3 & & 5 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 3 & & 2 & & 12 & & 5 & & 13 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 4 & & 1 & & 8 & & 15 & & 17 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 4 & & 3 & & 24 & & 7 & & 25 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 5 & & 2 & & 20 & & 21 & & 29 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 5 & & 4 & & 40 & & 9 & & 41 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 6 & & 1 & & 12 & & 35 & & 37 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 6 & & 5 & & 60 & & 11 & & 61 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

For example, consider $a = 6$ . Then b must be less than 6, relatively prime to 6, and odd. Thus, the only possibilities are $b = 1$ and $b = 5$ , and these give the last two cases above.

Example. Let $\{x,y,z\}$ be a Pythagorean triple. Show that one of x, y, z is divisible by 5.

Mod 5 the only squares are 0, 1, and 4.

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & x & & $x^2 \mod{5}$ & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & 0 & & 0 & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & 1 & & 1 & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & 2 & & 4 & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & 3 & & 4 & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & 4 & & 1 & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} }} $$

Suppose neither x nor y is divisible by 5. Then $x^2$ and $y^2$ can be either 1 or 4 mod 5. Consider the possibilities for $z^2 = x^2 + y^2 \mod{5}$ :

$$z^2 = 1 + 1 = 2 \quad \hbox{is not a square } \mod{5}$$

$$z^2 = 1 + 4 = 0 \quad \hbox{implies } 5 \mid z$$

$$z^2 = 4 + 1 = 0 \quad \hbox{implies } 5 \mid z$$

$$z^2 = 4 + 4 = 3 \quad \hbox{is not a square } \mod{5}$$

In the only cases which are possible, z is divisible by 5.

Thus, one of x, y, z must be divisible by 5.

Primitive Pythagorean Triples

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & a & & b & & $a^2 - b^2$ & & $2ab$ & & $a^2 + b^2$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 2 & & 1 & & 3 & & 4 & & 5 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 3 & & 2 & & 5 & & 12 & & 13 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 4 & & 1 & & 15 & & 8 & & 17 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 4 & & 3 & & 7 & & 24 & & 25 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 5 & & 2 & & 21 & & 20 & & 29 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 5 & & 4 & & 9 & & 40 & & 41 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 6 & & 1 & & 35 & & 12 & & 37 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 6 & & 5 & & 11 & & 60 & & 61 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 7 & & 2 & & 45 & & 28 & & 53 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 7 & & 4 & & 33 & & 56 & & 65 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 7 & & 6 & & 13 & & 84 & & 85 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 8 & & 1 & & 63 & & 16 & & 65 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 8 & & 3 & & 55 & & 48 & & 73 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 8 & & 5 & & 39 & & 80 & & 89 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 8 & & 7 & & 15 & & 112 & & 113 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 9 & & 2 & & 77 & & 36 & & 85 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 9 & & 4 & & 65 & & 72 & & 97 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 9 & & 8 & & 17 & & 144 & & 145 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 10 & & 1 & & 99 & & 20 & & 101 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 10 & & 3 & & 91 & & 60 & & 109 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 10 & & 7 & & 51 & & 140 & & 149 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 10 & & 9 & & 19 & & 180 & & 181 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

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