Euclid's Formula for Pythagorean Triples

Euclid's formula generates Pythagorean triples from any two positive whole numbers m > n: a = m² − n², b = 2mn and c = m² + n². With m = 2 and n = 1 it gives 3, 4, 5. When m and n share no factor and are not both odd, the triple is primitive, and every primitive triple appears exactly once.

How to use Euclid's formula

1. Pick two whole numbers m and n with m > n > 0.
2. Compute m² − n², 2mn and m² + n².
3. Order the legs if you like: m² − n² can come out bigger than 2mn, as in (m, n) = (4, 1), which gives 15, 8, 17.

Why the formula always works

Expand both sides. The cross terms −2m²n² and +4m²n² combine into +2m²n², which is exactly the middle term of (m² + n²)²:

This is an identity, true for every pair of numbers, so whole numbers m > n > 0 always give a genuine triple with three positive sides. That makes the formula ideal for writing your own practice problems.

Derivation 1: factoring a difference of squares

Where could such a formula come from? Start from a² + b² = c² and move a² across, then factor the difference of two squares:

We want the product (c − a)(c + a) to be a perfect square. The simplest way to arrange that is to make each factor twice a square:

Then (c − a)(c + a) = 4m²n² = (2mn)², so b = 2mn. Adding the two equations and halving gives c = m² + n²; subtracting and halving gives a = m² − n². That is Euclid's formula.

The choice "twice a square" looks like a lucky guess, but it is forced. For a primitive triple with a odd, c + a over 2 and c − a over 2 share no factor and multiply to a square, so each must be a square. The full proof is on the primitive Pythagorean triples page.

Derivation 2: rational points on the unit circle

Divide a² + b² = c² by c²:

So every triple gives a point (a over c, b over c) on the unit circle x² + y² = 1 (the circle of radius 1 centered at the origin) whose coordinates are fractions. Conversely, a point with fraction coordinates, written over a common denominator as (p over r, q over r), gives p² + q² = r². Finding triples is the same problem as finding rational points on the circle.

Here is the trick. We already know one rational point, (−1, 0). Draw a line through it with slope t, so y = t(x + 1), and see where else it meets the circle. Substituting:

One solution is x = −1, the point we started from. The two solutions of a quadratic add up to −2t² over 1 + t², so the other one is

If the slope t is a fraction, these coordinates are fractions too. And if a point (x, y) on the circle has fraction coordinates, the line joining it to (−1, 0) has the fraction slope y over x + 1. So rational points and rational slopes match up one to one. Writing the slope as t = n over m and multiplying top and bottom by m²:

The numerators and the denominator are exactly Euclid's a, b and c.

Look at the slope 1 over 3. It gives (8 over 10, 6 over 10), which cancels to (4 over 5, 3 over 5): the 3-4-5 triangle again, with its legs swapped. That is why pairs with m and n both odd add nothing new. The same picture explains a trigonometry formula: if the point sits at angle θ, the line makes angle θ over 2, so t = tanθ over 2 and cosθ = 1 − t² over 1 + t², sinθ = 2t over 1 + t². Check them with the Pythagorean identity.

When is the triple primitive?

• If m and n share a factor d, then d² divides all three numbers. Example: (6, 3) gives 27-36-45, which is 9 × (3-4-5).
• If m and n are both odd, all three numbers are even. Halving gives the triple from M = m + n over 2, N = m − n over 2 with the legs swapped. Example: (3, 1) gives 8-6-10, and halving gives 4-3-5, the triple from (2, 1).
• If both are even, they share the factor 2, so the first rule already rules them out.

What about non-primitive triples? Put a multiplier in front: a = k(m² − n²), b = 2kmn, c = k(m² + n²). With k included, the formula produces every Pythagorean triple (up to the order of the legs). Without it, some are missed: 9-12-15 cannot be written as (m² − n², 2mn, m² + n²) for any whole numbers m and n.

Table of triples for m up to 8

All 28 pairs with 1 ≤ n < m ≤ 8. Exactly 15 of them give primitive triples; the rest are multiples, and the last column names the primitive triple each one is a multiple of.

For bigger values, or to list every triple containing a particular number, use the Pythagorean triples generator, or browse the full list of Pythagorean triples.

Other formulas for generating triples

Pythagoras' formula: start from an odd number

For any odd number a = 2k + 1, the numbers b = 2k² + 2k and c = 2k² + 2k + 1 complete a triple. Equivalently, b = a² − 1 over 2 and c = b + 1. This is Euclid's formula with m = k + 1 and n = k, and it produces 3-4-5, 5-12-13, 7-24-25, 9-40-41, 11-60-61 and every other triple whose hypotenuse is one more than a leg.

The picture shows the idea. Square numbers grow by odd numbers: k² + (2k + 1) = (k + 1)². Whenever the odd number 2k + 1 happens to be a perfect square, say a², you get k² + a² = (k + 1)², a triple.

Plato's formula: start from an even number

For an even number a = 2k, take b = k² − 1 and c = k² + 1. This is Euclid's formula with m = k and n = 1. With k = 4 it gives 8-15-17 and with k = 6 it gives 12-35-37. When k is odd all three numbers come out even (k = 3 gives 6-8-10), so it is primitive only for even k. Proclus (5th century CE), in his commentary on Euclid's proof of the theorem, credits the odd-number method to Pythagoras and the even-number method to Plato.

Fibonacci's method: adding odd numbers

In his Book of Squares (Liber Quadratorum, 1225), Leonardo of Pisa, known as Fibonacci, used the fact that the first k odd numbers add up to k². Choose an odd square, such as 81 = 9². It is the odd number 2k + 1 with k = 40. The 40 odd numbers before it add up to 40², and including 81 makes 41 odd numbers, which add up to 41²:

It is the same family as Pythagoras' formula, seen through sums of odd numbers.

Dickson's method: one formula for every triple

Leonard Eugene Dickson described a method that finds all triples, primitive or not. Choose an even number r and split r² over 2 into two factors, r² over 2 = st. Then (r + s,  r + t,  r + s + t) is a Pythagorean triple, because expanding shows (r + s)² + (r + t)² − (r + s + t)² = r² − 2st = 0.

Comparing the methods

There is one more generator with a very different flavour: three matrices that grow every primitive triple, exactly once, from 3-4-5. See the tree of primitive Pythagorean triples.

Worked examples

Where the formula comes from

The formula is named after Euclid because it appears in his Elements (about 300 BCE), in Book X, as Lemma 1 attached to Proposition 29: "to find two square numbers such that their sum is also square". Euclid's version is stated with lengths rather than algebra, but it is the same construction. Much earlier, the Babylonian tablet Plimpton 322 (about 1800 BCE) records the sides of large right triangles such as 119-120-169 (its first row gives 119 and 169), which suggests its scribes had a reliable method; scholars still debate exactly what it was.

The circle method is close to how Diophantus of Alexandria (probably 3rd century CE) solved Problem 8 of Book II of his Arithmetica, "divide a given square into two squares". For the square 16 he used a line and found 16 = 256 over 25 + 144 over 25. Fermat later wrote his famous note about Fermat's Last Theorem in the margin beside this very problem.