Pythagorean Triples

Introduction

A right‑angled triangle has one angle equal to $90^\circ$.
The side opposite the right angle is the hypotenuse.
A Pythagorean triple is a set of three positive integers $(a,b,c)$ satisfying $$a^2 + b^2 = c^2.$$
These triples describe all right‑angled triangles with integer side lengths.
This article explores:
- What Pythagorean triples are
- How to find them
- How to generate all of them using a simple formula
- Why the formula works

Right‑Angled Triangles and the Pythagorean Equation

For any right‑angled triangle: $$\text{(adjacent)}^2 + \text{(opposite)}^2 = \text{(hypotenuse)}^2.$$
If the sides are integers, we are solving the Diophantine equation: $$a^2 + b^2 = c^2.$$
We usually assume:
- $a$ and $b$ are the adjacent and opposite sides
- $c$ is the hypotenuse
- $a \le b < c$

What Is a Pythagorean Triple?

A triple $(a,b,c)$ of positive integers with $a^2 + b^2 = c^2$.
Examples:
- $(3,4,5)$
- $(5,12,13)$
- $(8,15,17)$
These are not coincidences—there is structure behind them.

Famous Examples and Patterns

Common small triples:
- $(3,4,5)$
- $(6,8,10)$
- $(9,12,15)$
- $(5,12,13)$
- $(7,24,25)$
Observations:
- Many triples seem to be “scaled‑up” versions of smaller ones.
- The hypotenuse is often just slightly larger than the longer leg.
- Odd numbers often appear in pairs like $(m^2 - n^2, 2mn)$.

Primitive vs. Non‑Primitive Triples

A triple is primitive if $\gcd(a,b,c) = 1$.
Otherwise it is non‑primitive.
Examples:
- $(3,4,5)$ is primitive.
- $(6,8,10)$ is non‑primitive because all numbers share a factor of $2$.
Every non‑primitive triple is a multiple of a primitive one.

Generating Triples by Scaling

If $(a,b,c)$ is a triple, then so is $(ka, kb, kc)$ for any integer $k \ge 1$.
Example:
- Start with $(3,4,5)$.
- Multiply by $k=3$ to get $(9,12,15)$.
This explains many of the “bigger” triples we see.

A First Method: Trial, Error, and Patterns

Try small values of $a$ and $b$ and check whether $a^2 + b^2$ is a perfect square.
Example:
- $3^2 + 4^2 = 9 + 16 = 25 = 5^2$.
Patterns that help:
- If $a$ is odd, $a^2$ is odd.
- If $a$ is even, $a^2$ is even.
- $c$ must be larger than both $a$ and $b$.
This method works but is slow.

A Better Method: Using Algebraic Rearrangements

Rewrite the equation: $$a^2 = c^2 - b^2 = (c-b)(c+b).$$ This suggests:

Choose two numbers $u = c-b$ and $v = c+b$.
Then $a^2 = uv$.
Since $u$ and $v$ have the same parity, we can set:
- $u = m-n$
- $v = m+n$
This leads naturally to Euclid’s formula.

Euclid’s Formula for All Primitive Triples

For integers $m > n > 0$:

Let $$a = m^2 - n^2$$ $$b = 2mn$$ $$c = m^2 + n^2.$$
Then $(a,b,c)$ is a Pythagorean triple.
If:
- $\gcd(m,n)=1$
- $m$ and $n$ have opposite parity (one even, one odd)
then the triple is primitive.

Examples:

$m=2$, $n=1$: $(3,4,5)$
$m=3$, $n=2$: $(5,12,13)$
$m=4$, $n=1$: $(15,8,17)$ (order of $a,b$ doesn’t matter)

Understanding Why Euclid’s Formula Works

Key ideas:

Start with the identity: $$(m^2 - n^2)^2 + (2mn)^2 = (m^2 + n^2)^2.$$
Expand both sides:
- Left side becomes $$m^4 - 2m^2n^2 + n^4 + 4m^2n^2 = m^4 + 2m^2n^2 + n^4.$$
- Right side is $$m^4 + 2m^2n^2 + n^4.$$
Both sides match exactly.
So the triple $$(m^2 - n^2,\; 2mn,\; m^2 + n^2)$$ always satisfies the Pythagorean equation.
This identity is the algebraic heart of the formula.

Parity Conditions and Coprimality

To ensure primitiveness:

$m$ and $n$ must not share a common factor.
If both are odd:
- $a = m^2 - n^2$ is even
- $b = 2mn$ is even
- So the triple is not primitive.
If one is even and one is odd:
- Exactly one of $a$ or $b$ is even.
- This gives a primitive triple.

Classifying All Pythagorean Triples

Every Pythagorean triple is of the form: $$(ka, kb, kc)$$ where $(a,b,c)$ is primitive and generated by Euclid’s formula.

Thus:

Primitive triples ↔ pairs $(m,n)$ with:
- $m>n$
- $\gcd(m,n)=1$
- $m$ and $n$ of opposite parity
All triples ↔ multiply primitive ones by $k$.

This gives a complete classification.

Connections to Geometry and Number Theory

Geometry
- Integer right triangles appear in construction, surveying, and design.
- The unit circle equation $x^2 + y^2 = 1$ relates to normalized triples.
Number Theory
- Solving $a^2 + b^2 = c^2$ is a classic Diophantine problem.
- Links to:
  - Sums of two squares
  - Gaussian integers
  - Modular arithmetic
  - Primitive lattice points on circles
History
- Known to the Babylonians (tablet Plimpton 322).
- Studied by Euclid, Diophantus, and many later mathematicians.

Applications and Historical Notes

Applications
- Computer graphics (grid‑aligned distances)
- Cryptography (structure of integer solutions)
- Architecture and engineering
- Puzzle design and recreational mathematics
Historical Notes
- The triple $(3,4,5)$ appears in ancient Egyptian rope‑stretching.
- Euclid’s formula is over 2000 years old.
- Pythagorean triples are one of the earliest studied integer patterns.

Calculator

Coprime

Checks if a set of numbers are coprime with each other
- E.g. if $\gcd(a, b, c) = 1$

coprime(3, 4) coprime([3, 4, 5])

Generating triples

We can define a custom function for finding Pythagorean triples

pythagTriple(m, n) = [m^2 - n^2, 2*m*n, m^2 + n^2] pythagTriple(2, 1) pythagTriple(3, 2)

Exercises

Try these to reinforce the ideas:

Verify that $(8,15,17)$ is a Pythagorean triple.
Solution
Verify that $(8,15,17)$ is a Pythagorean triple.
- Compute:
  - $8^2 = 64$
  - $15^2 = 225$
  - $17^2 = 289$
- Check: $$8^2 + 15^2 = 64 + 225 = 289 = 17^2.$$
- So $8^2 + 15^2 = 17^2$, and $(8,15,17)$ is a Pythagorean triple.
Determine whether $(9,40,41)$ is primitive.
Solution
Determine whether $(9,40,41)$ is primitive.
- First check it is a triple:
  - $9^2 = 81$
  - $40^2 = 1600$
  - $41^2 = 1681$ $$9^2 + 40^2 = 81 + 1600 = 1681 = 41^2.$$
- Now check $\gcd(9,40,41)$:
  - $\gcd(9,40) = 1$ (since $9$ is $3^2$ and $40$ has prime factors $2$ and $5$).
  - $\gcd(9,41) = 1$ (41 is prime and not $3$).
  - $\gcd(40,41) = 1$ (consecutive integers).
- So $\gcd(9,40,41) = 1$, and the triple is primitive.
Generate a triple using Euclid’s formula with $m=5$, $n=2$.
Solution
Generate a triple using Euclid’s formula with $m=5$, $n=2$.
Using: $$a = m^2 - n^2,\quad b = 2mn,\quad c = m^2 + n^2,$$ we get:
- $a = 5^2 - 2^2 = 25 - 4 = 21$
- $b = 2 \cdot 5 \cdot 2 = 20$
- $c = 5^2 + 2^2 = 25 + 4 = 29$
So $(21,20,29)$ is a Pythagorean triple (often written as $(20,21,29)$).
Quick check: $$20^2 + 21^2 = 400 + 441 = 841 = 29^2.$$
Find all triples obtained by scaling $(3,4,5)$ with $k=2,3,4$.
Solution
Find all triples obtained by scaling $(3,4,5)$ with $k=2,3,4$.
- For $k=2$: $$(2\cdot 3, 2\cdot 4, 2\cdot 5) = (6,8,10)$$
- For $k=3$: $$(3\cdot 3, 3\cdot 4, 3\cdot 5) = (9,12,15)$$
- For $k=4$: $$(4\cdot 3, 4\cdot 4, 4\cdot 5) = (12,16,20)$$
Each is a Pythagorean triple because $$(k\cdot 3)^2 + (k\cdot 4)^2 = k^2(3^2 + 4^2) = k^2 \cdot 5^2 = (k\cdot 5)^2.$$
Show that if $(a,b,c)$ is primitive, then exactly one of $a$ or $b$ is even.
Solution
Show that if $(a,b,c)$ is primitive, then exactly one of $a$ or $b$ is even.
- Suppose both $a$ and $b$ are odd.
  - Then $a^2 \equiv 1 \pmod{4}$ and $b^2 \equiv 1 \pmod{4}$.
  - So $a^2 + b^2 \equiv 1 + 1 = 2 \pmod{4}$.
  - But any square is $0$ or $1 \pmod{4}$, so $c^2$ cannot be $2 \pmod{4}$—contradiction.
- Suppose both $a$ and $b$ are even.
  - Then $2$ divides $a$ and $b$, and from $a^2 + b^2 = c^2$, $2$ also divides $c$.
  - So $\gcd(a,b,c) \ge 2$, contradicting primitiveness.
- Therefore, $a$ and $b$ cannot both be odd or both be even.
- So exactly one of $a$ or $b$ is even.
Find a pair $(m,n)$ that generates the triple $(7,24,25)$.
Solution
Find a pair $(m,n)$ that generates the triple $(7,24,25)$.
We want: $$a = m^2 - n^2,\quad b = 2mn,\quad c = m^2 + n^2 = 25.$$
- Solve $m^2 + n^2 = 25$ with integers:
  - Possible pairs: $(m,n) = (4,3)$ or $(3,4)$ (up to sign).
- Try $m=4$, $n=3$:
  - $m^2 - n^2 = 16 - 9 = 7$
  - $2mn = 2 \cdot 4 \cdot 3 = 24$
- This matches $(7,24,25)$.
So one suitable pair is $(m,n) = (4,3)$.
Explain why $m$ and $n$ cannot both be even in Euclid’s formula.
Solution
Explain why $m$ and $n$ cannot both be even in Euclid’s formula.
- If $m$ and $n$ are both even, write $m = 2m'$, $n = 2n'$.
- Then:
  - $a = m^2 - n^2 = 4m'^2 - 4n'^2 = 4(m'^2 - n'^2)$
  - $b = 2mn = 2 \cdot 2m' \cdot 2n' = 8m'n'$
  - $c = m^2 + n^2 = 4m'^2 + 4n'^2 = 4(m'^2 + n'^2)$
- So $a$, $b$, and $c$ are all multiples of $4$.
- Hence $\gcd(a,b,c) \ge 4$, so the triple is not primitive.
- For generating primitive triples, $m$ and $n$ cannot both be even.
List all primitive triples with $c < 30$.
Solution
List all primitive triples with $c < 30$.
We use Euclid’s formula with $m>n$, $\gcd(m,n)=1$, and opposite parity.
Check small $m$:
- $m=2$, $n=1$:
  - $(3,4,5)$, $c=5 < 30$, primitive.
- $m=3$, $n=2$:
  - $(5,12,13)$, $c=13 < 30$, primitive.
- $m=4$, $n=1$:
  - $(15,8,17)$ → $(8,15,17)$, $c=17 < 30$, primitive.
- $m=4$, $n=3$:
  - $\gcd(4,3)=1$, opposite parity:
  - $(7,24,25)$, $c=25 < 30$, primitive.
- $m=5$, $n=2$:
  - $(21,20,29)$, $c=29 < 30$, primitive.
- $m=5$, $n=4$:
  - $(9,40,41)$ has $c=41 \ge 30$, so ignore.
- $m=6$:
  - With $n=1$: $(35,12,37)$, $c=37 \ge 30$.
  - Larger $n$ give even bigger $c$.
So the primitive triples with $c < 30$ are:
- $(3,4,5)$
- $(5,12,13)$
- $(8,15,17)$
- $(7,24,25)$
- $(20,21,29)$
(You can reorder legs if desired.)
Prove that $c$ is always odd in a primitive triple.
Solution
Prove that $c$ is always odd in a primitive triple.
Let $(a,b,c)$ be primitive with $a^2 + b^2 = c^2$.
- From Exercise 5, exactly one of $a$ or $b$ is even.
  - So one is even, one is odd.
- Then:
  - Even square $\equiv 0 \pmod{4}$.
  - Odd square $\equiv 1 \pmod{4}$.
- So $a^2 + b^2 \equiv 0 + 1 = 1 \pmod{4}$.
- Hence $c^2 \equiv 1 \pmod{4}$.
- The only integers with square $\equiv 1 \pmod{4}$ are odd integers.
- Therefore, $c$ must be odd.
10. Challenge: Find a triple where the legs differ by $1$ (i.e., $b = a+1$).
We want integers $a$ and $c$ such that: $$a^2 + (a+1)^2 = c^2.$$
- Expand: $$a^2 + a^2 + 2a + 1 = c^2$$ $$2a^2 + 2a + 1 = c^2.$$
- Try small $a$:
  - $a=3$: $2\cdot 9 + 6 + 1 = 18 + 6 + 1 = 25 = 5^2$.
- So $a=3$, $b=4$, $c=5$ works.
Thus $(3,4,5)$ is a Pythagorean triple whose legs differ by $1$.
Challenge: Find a triple where the legs differ by $1$ (i.e., $b=a+1$).