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Pythagorean triples

Number theory · level 7, lesson 5 of 6

A question to start

The right triangle with sides 3, 4 and 5 is famous because all three sides are whole numbers: \(3^2 + 4^2 = 9 + 16 = 25 = 5^2\). So is 5, 12, 13: \(25 + 144 = 169\). And 8, 15, 17. And 20, 21, 29.

Look at these and see what you notice. In each one, a leg is even. A leg is a multiple of 3. And something is a multiple of 5: the 5 in 3, 4, 5; the 5 in 5, 12, 13; the 15 in 8, 15, 17; the 20 in 20, 21, 29. Is that luck, or must it always happen? And is there a recipe that makes every such triangle? Both answers are known, and this lesson proves them.

The idea

Triples and primitive triples

A Pythagorean triple is three whole numbers \(a, b, c\) (all at least 1) with

\[a^2 + b^2 = c^2.\]

By Pythagoras' theorem (see Geometry level 6) these are exactly the right triangles with whole-number sides; \(c\) is the hypotenuse, the longest side.

4 3 5 9 + 16 = 25

Doubling every side of a triple gives another: 6, 8, 10. So does tripling: 9, 12, 15. These are the same triangle, only bigger. A triple whose three numbers have no common factor (other than 1) is called primitive. Every triple is a whole-number multiple of a primitive one: divide all three by their greatest common divisor \(g\) (the equation still holds, since \(a^2 + b^2 = c^2\) divided by \(g^2\) is \((a/g)^2 + (b/g)^2 = (c/g)^2\)). So the real question is: find all primitive triples.

In a primitive triple, any two of the numbers are already coprime. If a prime \(p\) divided \(a\) and \(b\), it would divide \(a^2 + b^2 = c^2\), so (Euclid's lemma) it would divide \(c\), and the triple would not be primitive. The same works for any other pair.

One leg is always even

Squares mod 4 can only be 0 or 1: an even number \(2k\) has square \(4k^2 \equiv 0\), and an odd number \(2k + 1\) has square \(4k^2 + 4k + 1 \equiv 1 \pmod 4\).

  • Both legs even? Then 2 divides \(a\) and \(b\), so the triple is not primitive.
  • Both legs odd? Then \(a^2 + b^2 \equiv 1 + 1 = 2 \pmod 4\). But \(c^2\) is 0 or 1 mod 4, never 2. Impossible.

So in a primitive triple exactly one leg is even and the other is odd, and then \(c^2 \equiv 0 + 1 = 1\), so \(c\) is odd. From now on we call the odd leg \(a\) and the even leg \(b\).

The recipe

All primitive triples. Choose whole numbers \(m > n \ge 1\) that are coprime and not both odd. Then

\[a = m^2 - n^2, \qquad b = 2mn, \qquad c = m^2 + n^2\]

is a primitive triple. Every primitive triple arises this way, exactly once.

First, the formula always gives a triple. Expand:

\[(m^2 - n^2)^2 + (2mn)^2 = m^4 - 2m^2n^2 + n^4 + 4m^2n^2 = m^4 + 2m^2n^2 + n^4 = (m^2 + n^2)^2.\]

Here are the first few:

\(m\) \(n\) \(a = m^2 - n^2\) \(b = 2mn\) \(c = m^2 + n^2\)
2 1 3 4 5
3 2 5 12 13
4 1 15 8 17
4 3 7 24 25
5 2 21 20 29
5 4 9 40 41
6 1 35 12 37
6 5 11 60 61

The conditions on \(m\) and \(n\) matter. With \(m = 3\), \(n = 1\) (both odd) you get 8, 6, 10: a triple, but not primitive. With \(m = 4\), \(n = 2\) (not coprime) you get 12, 16, 20, also not primitive.

Why every primitive triple comes from the recipe

Start with a primitive triple, \(a\) odd, \(b\) even, \(c\) odd. Move \(a^2\) across and factor a difference of squares:

\[b^2 = c^2 - a^2 = (c - a)(c + a).\]

Both \(c - a\) and \(c + a\) are even (odd minus odd, odd plus odd). So set

\[u = \frac{c + a}{2}, \qquad v = \frac{c - a}{2}, \qquad \text{so} \qquad \left(\frac b2\right)^2 = u v.\]

\(u\) and \(v\) are coprime. A common factor of \(u\) and \(v\) divides \(u + v = c\) and \(u - v = a\), and \(a\) and \(c\) are coprime. So the only common factor is 1.

Two coprime numbers whose product is a square are each squares. Look at the prime factorisation (level 6: it is unique). In a square, every prime appears an even number of times. Since \(u\) and \(v\) share no prime, each prime of \(uv\) comes wholly from \(u\) or wholly from \(v\), so each prime appears an even number of times in \(u\), or in \(v\). So \(u = m^2\) and \(v = n^2\) for some whole numbers \(m > n \ge 1\).

Now undo: \(c = u + v = m^2 + n^2\), \(a = u - v = m^2 - n^2\), and \((b/2)^2 = m^2 n^2\), so \(b = 2mn\). Since \(u = m^2\) and \(v = n^2\) are coprime, so are \(m\) and \(n\); and if both were odd, \(c = m^2 + n^2\) would be even. ∎

A picture: points on a circle

Divide \(a^2 + b^2 = c^2\) by \(c^2\): \(\left(\tfrac ac\right)^2 + \left(\tfrac bc\right)^2 = 1\). So each triple gives a point \((x, y)\) with fraction coordinates on the circle of radius 1. Here is a geometric way to find all of them. Draw a line through the point \((-1, 0)\) with a fraction slope \(t\). It meets the circle at one more point:

\[x = \frac{1 - t^2}{1 + t^2}, \qquad y = \frac{2t}{1 + t^2}.\]

With \(t = \tfrac nm\) this is \(x = \dfrac{m^2 - n^2}{m^2 + n^2}\), \(y = \dfrac{2mn}{m^2 + n^2}\): the recipe again.

(−1, 0) (3/5, 4/5) slope 1/2

With slope \(\tfrac12\) the line hits \(\left(\tfrac35, \tfrac45\right)\): the 3, 4, 5 triangle. Every fraction slope between 0 and 1 gives a point, and every point with fraction coordinates in that quarter comes from some slope (the slope of the line from \((-1, 0)\) to a point with fraction coordinates is a fraction). This is the same answer, found by geometry instead of factoring.

Hidden factors 3, 4 and 5

The recipe explains the patterns in the opening question. In every primitive triple:

  • A leg is a multiple of 4. \(b = 2mn\), and one of \(m\), \(n\) is even, so \(b\) is a multiple of 4.
  • A leg is a multiple of 3. Squares mod 3 are 0 or 1. If neither \(m\) nor \(n\) is a multiple of 3, then \(m^2 \equiv n^2 \equiv 1\), so \(a = m^2 - n^2 \equiv 0 \pmod 3\). Otherwise \(b = 2mn\) is a multiple of 3.
  • Some side is a multiple of 5. Squares mod 5 are 0, 1 or 4. If 5 divides \(m\) or \(n\), it divides \(b\). If not, \(m^2\) and \(n^2\) are each 1 or 4 mod 5. If they are equal, \(a = m^2 - n^2 \equiv 0\); if one is 1 and the other 4, \(c = m^2 + n^2 \equiv 5 \equiv 0\).

So \(abc\) is always a multiple of \(3 \times 4 \times 5 = 60\), and the product of the two legs (twice the area) is always a multiple of 12. These facts carry over to every triple, primitive or not, since multiplying all sides by a whole number keeps divisibility.

A clay tablet from ancient Babylon, now called Plimpton 322 (from about 1800 BCE), lists numbers closely tied to Pythagorean triples, including large ones. The recipe itself appears in Euclid's Elements, Book X.

Worked examples

1. From \(m\) and \(n\). \(m = 7\), \(n = 4\) (coprime, not both odd): \(a = 49 - 16 = 33\), \(b = 56\), \(c = 49 + 16 = 65\). Check: \(33^2 + 56^2 = 1089 + 3136 = 4225 = 65^2\). ✓

2. Back to \(m\) and \(n\). Which \(m\), \(n\) give 20, 21, 29? The odd leg is 21 and the hypotenuse 29, so \(u = (29 + 21)/2 = 25 = 5^2\) and \(v = (29 - 21)/2 = 4 = 2^2\). So \(m = 5\), \(n = 2\), and \(b = 2 \times 5 \times 2 = 20\). ✓

3. All triples with a leg of 12. Primitive ones need 12 as the even leg \(2mn\), so \(mn = 6\): \((m, n) = (6, 1)\) gives 35, 12, 37; \((3, 2)\) gives 5, 12, 13. Non-primitive ones are multiples of smaller primitive triples with a leg dividing 12: from 3, 4, 5 we get \(\times 3\): 9, 12, 15 and \(\times 4\): 12, 16, 20. All together: (5, 12, 13), (9, 12, 15), (12, 16, 20), (12, 35, 37).

4. A hypotenuse of 25. Primitive: \(m^2 + n^2 = 25\) with \(m > n\), so \((m, n) = (4, 3)\): 7, 24, 25. Non-primitive: \(5 \times (3, 4, 5) = 15, 20, 25\). These are the only two.

5. No triple has both legs odd: a numerical check. \(3^2 + 5^2 = 34\), \(5^2 + 7^2 = 74\), \(7^2 + 9^2 = 130\): each is 2 more than a multiple of 4, so none is a square. Squares are never 2 mod 4.

Try it

  1. Use the recipe with \((m, n) = (7, 2)\) and \((8, 3)\). Check each triple.
  2. Is 9, 40, 41 primitive? Find its \(m\) and \(n\).
  3. Find all Pythagorean triples with a leg of 15.
  4. Why does \((m, n) = (5, 3)\) not give a primitive triple? What does it give?
  5. Look at 3, 4, 5 and 5, 12, 13 and 8, 15, 17. In a primitive triple with even leg \(b\), is \(c - b\) always a perfect square? Prove your answer.
  6. Find a primitive triple whose hypotenuse is 65 in two different ways. (Hint: \(65 = 8^2 + 1^2 = 7^2 + 4^2\).)
  7. Prove that the hypotenuse of a primitive triple is never a multiple of 3.
Answers
  1. \((7, 2)\): \(45, 28, 53\); check \(2025 + 784 = 2809 = 53^2\). ✓ \((8, 3)\): \(55, 48, 73\); check \(3025 + 2304 = 5329 = 73^2\). ✓
  2. Yes: 9 and 40 share no factor. \(u = (41 + 9)/2 = 25\), \(v = (41 - 9)/2 = 16\), so \(m = 5\), \(n = 4\).
  3. 15 is odd, so in a primitive triple it would be \(m^2 - n^2 = (m - n)(m + n)\): \((m - n, m + n) = (1, 15)\) gives \(m = 8\), \(n = 7\): 15, 112, 113; and \((3, 5)\) gives \(m = 4\), \(n = 1\): 15, 8, 17. Non-primitive: \(3 \times\) (5, 12, 13) = 15, 36, 39 and \(5 \times\) (3, 4, 5) = 15, 20, 25. So (8, 15, 17), (15, 20, 25), (15, 36, 39), (15, 112, 113).
  4. Both are odd. It gives \(25 - 9 = 16\), \(30\), \(34\): the triple 16, 30, 34, which is \(2 \times\) (8, 15, 17).
  5. Yes. \(c - b = m^2 + n^2 - 2mn = (m - n)^2\), a perfect square. Examples: \(5 - 4 = 1\), \(13 - 12 = 1\), \(17 - 8 = 9\), \(29 - 20 = 9\). (Also \(c - a = 2n^2\), twice a square.)
  6. \((m, n) = (8, 1)\): 63, 16, 65. \((7, 4)\): 33, 56, 65. Both primitive. (The non-primitive ones with hypotenuse 65 are 25, 60, 65 and 39, 52, 65.)
  7. \(c = m^2 + n^2\). \(m\) and \(n\) are coprime, so not both multiples of 3. Squares mod 3 are 0 or 1, so \(m^2 + n^2 \equiv 0 + 1\), \(1 + 0\) or \(1 + 1\): that is 1, 1 or 2, never 0. So 3 never divides \(c\).

Common mistake

Forgetting the conditions on \(m\) and \(n\). The recipe with any \(m > n\) gives a triple, but only coprime \(m\), \(n\) of opposite parity give a primitive one, and only then is each primitive triple made once. With \((3, 1)\) you get \((8, 6, 10)\), a doubled 3, 4, 5.

Thinking the recipe gives every triple. It gives every primitive triple. 9, 12, 15 is not of the form \(m^2 - n^2, 2mn, m^2 + n^2\) for any whole \(m\), \(n\); it is 3 times a recipe triple.

Where it leads