How Do You Find Every Right Triangle with Whole Sides?

Spend a few minutes with the hunt above. Try these, in order.

• a.Find all four triangles with a short side of 12.
• b.Find 6, 8, 10 and 9, 12, 15. How are they related to 3, 4, 5?
• c.Find a triangle on your list that is not a bigger copy of a smaller one.
• d.Look at 3, 4, 5, then 5, 12, 13, then 7, 24, 25. What do the long side and the middle side always do? Predict the next one, starting from 9, and check it.

A right triangle with whole-number sides is a Pythagorean triangle, and its three numbers a Pythagorean triple. A triple whose numbers have no common factor, like 3, 4, 5, is a primitive triple; every other triple is a primitive one multiplied up. So the real hunt is for the primitive ones.

Here is the idea that finds every triangle with a given side. Write the rule with the known side a on its own:

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

Picture it. $c^2$ is a square of side c. Take out a square of side b from one corner and what is left is an L of width c − b. So the rule says: the $a^2$ small squares of a square of side a can be rearranged into an L that fits round a square of side b.

The L’s area is easy to count. Cut it at the corner and lay its two arms end to end: a strip c − b wide and c + b long. So

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

which is the difference of two squares from Grade 9’s factorization, now with a job to do. For 5, 12, 13: $5^2 = 25 = (13 - 12)(13 + 12) = 1 \times 25$. The L is one square wide.

So to find every triangle with a short side of 12, write 144 = d × e with d = c − b the smaller factor and e = c + b the larger. Then c and b are the half-sum and half-difference:

$$c = \frac{e + d}{2}, \qquad b = \frac{e - d}{2}$$

For these to be whole, d and e must both be even or both odd.

This is how Bhaskara did it in the Lilavati (1150 AD). In Colebrooke’s 1817 translation: “a side is put. Its square, divided by an arbitrary number, is set down in two places: and the arbitrary number being added and subtracted, and the sum and difference halved, the results are the hypotenuse and upright.” His example is the side 12: with the arbitrary number 2 he gets 35, 37, then 16, 20 and 9, 15. His commentator Ganesa says why it works: “the difference of the squares is equal to the product of the sum and difference of the roots.”

The L finds the triangles with one side you know. To make every triangle, start from two numbers instead. Take any whole numbers p > q, and make

$$a = p^2 - q^2, \qquad b = 2pq, \qquad c = p^2 + q^2$$

With p = 2, q = 1: 3, 4, 5. With p = 3, q = 2: 5, 12, 13. With p = 5, q = 2: 21, 20, 29, the triangle from the card on this topic. The table shows more.

	q = 1	q = 2	q = 3	q = 4	q = 5
p = 2	3, 4, 5
p = 3	8, 6, 10	5, 12, 13
p = 4	15, 8, 17	12, 16, 20	7, 24, 25
p = 5	24, 10, 26	21, 20, 29	16, 30, 34	9, 40, 41
p = 6	35, 12, 37	32, 24, 40	27, 36, 45	20, 48, 52	11, 60, 61

Why must it always work? Not because the table’s fifteen cases check out: because the rule is an identity, true for every p and q. Expand both sides.

$$(p^2 - q^2)^2 + (2pq)^2 = p^4 - 2p^2q^2 + q^4 + 4p^2q^2 = p^4 + 2p^2q^2 + q^4 = (p^2 + q^2)^2$$

The $-2p^2q^2$ from the first square and the $4p^2q^2$ from the second leave $+2p^2q^2$, exactly what $(p^2 + q^2)^2$ has in its middle. Since it holds for all numbers, any p > q gives a right triangle with whole sides, and there are infinitely many p and q, so infinitely many triangles that aren’t copies of each other. This recipe is usually called Euclid’s formula, after the Elements, about 300 BC.

The L is hiding inside it too: $c - b = p^2 + q^2 - 2pq = (p - q)^2$ and $c + b = (p + q)^2$. The L’s width and length are both squares.

The table makes infinitely many triangles. But does it make all of them? Perhaps some strange primitive triple with sides in the millions isn’t in it. Here is why that can’t happen. Take any primitive triple a, b, c.

1. One short side is even, the other odd. They can’t both be even, or all three would share the factor 2. They can’t both be odd either. Divide any square by 4 and look at the remainder: an even number squared, $(2k)^2 = 4k^2$, leaves 0; an odd number squared, $(2k + 1)^2 = 4k^2 + 4k + 1$, leaves 1. Two odd squares added leave 2, and no square leaves 2. So exactly one short side is even; call it b. Then c is odd.

2. Split the L. $b^2 = (c - a)(c + a)$, and c and a are both odd, so c − a and c + a are both even. Write c + a = 2m and c − a = 2n. Then $b^2 = 4mn$, so $mn = (b/2)^2$, a perfect square, with c = m + n and a = m − n.

3. m and n share no factor. Anything that divides both divides m + n = c and m − n = a too. And a and c share no prime: one that divided both would divide $c^2 - a^2 = b^2$, so b as well, and the triple wouldn’t be primitive.

4. So each of them is a square. Every whole number splits into primes in only one way. In a perfect square each prime comes an even number of times (36 = 2 × 2 × 3 × 3). Here m and n share no prime, so all the 2s of mn belong to one of them, all the 3s to one of them, and so on, each prime with its even count. So m is a square, $p^2$, and n is a square, $q^2$.

Put them back: $c = m + n = p^2 + q^2$, $a = m - n = p^2 - q^2$, and $b = 2\sqrt{mn} = 2pq$. Since m and n share no factor, neither do p and q; since $c = p^2 + q^2$ is odd, one of them is even. That is the recipe. Every primitive triple is in the table; every other triple is a primitive one times a whole number.

There is a picture of all of this at once. Divide $a^2 + b^2 = c^2$ by $c^2$:

$$\left(\frac{a}{c}\right)^2 + \left(\frac{b}{c}\right)^2 = 1$$

So the point (a/c,  b/c) lies on the circle $x^2 + y^2 = 1$ (Grade 10 Optional Math, the circle), and it is a point whose coordinates are both fractions. 3, 4, 5 is the point (3/5,  4/5). Finding Pythagorean triples is the same as finding the points on the circle with fractions for coordinates.

The way to find those points is a line. Start at the point (−1, 0) on the circle and draw a line with a fraction for its slope, q/p: up q for every p across. It meets the circle in one more point, and that point turns out to have fractions for coordinates too.

Why fractions? Write t = q/p for the slope, so the line is y = t(x + 1). Put it into $x^2 + y^2 = 1$, in the form $(x^2 - 1) + y^2 = 0$:

$$(x - 1)(x + 1) + t^2(x + 1)^2 = 0 \quad\Longrightarrow\quad (x + 1)\bigl[(x - 1) + t^2(x + 1)\bigr] = 0$$

The factor x + 1 is the point we started from, x = −1. The other point makes the bracket 0: $x(1 + t^2) = 1 - t^2$, so $x = (1 - t^2)/(1 + t^2)$, a fraction, with no square root anywhere. Put t = q/p and multiply top and bottom by $p^2$:

$$x = \frac{p^2 - q^2}{p^2 + q^2}, \qquad y = \frac{2pq}{p^2 + q^2}$$

the recipe again. And every point with fractions for coordinates is reached this way, since the line from (−1, 0) to it has a fraction for its slope. Diophantus of Alexandria, in the third century AD, used the same trick in numbers (a line through a point he already knew) to split a square into two squares in his Arithmetica (Book II, Problem 8): his example splits 16 into $(16/5)^2 + (12/5)^2$, which is 3, 4, 5 again, times 4/5.

The recipe answers questions that would be hopeless by hunting.

Which numbers can be a side? Every whole number from 3 up. An odd number n is in $n,\ (n^2 - 1)/2,\ (n^2 + 1)/2$, the family you found (p and q one apart). A multiple of 4, 4k, is in $4k,\ 4k^2 - 1,\ 4k^2 + 1$ (p = 2k, q = 1). The numbers left over, 6, 10, 14, 18…, are twice an odd number: the even side of a primitive triple is 2pq with one of p, q even, so a multiple of 4, and these numbers can only appear in copies, like 6, 8, 10 or 14, 48, 50.

Which numbers can be the long side? Among the primitive triples up to 100 the long sides are 5, 13, 17, 25, 29, 37, 41, 53, 61, 65, 73, 85, 89 and 97. Each leaves 1 when divided by 4, as the recipe says it must: $p^2 + q^2$ is an odd square (remainder 1) plus an even one (remainder 0). But 9, 21, 33, 45, 49, 57, 69, 77, 81 and 93 also leave 1, and none is ever the long side of a primitive triple. Their trouble is a prime factor (3, 7, 11, 19, 23 or 31) that leaves 3 when divided by 4. The full answer rests on a theorem Albert Girard noticed in 1625, Fermat stated in a letter at Christmas 1640, and Euler first proved (published 1752–1755): an odd prime is a sum of two squares exactly when it leaves 1 when divided by 4 (5 = 1 + 4, 13 = 4 + 9, 17 = 1 + 16). So a number is the long side of a primitive triple exactly when all its prime factors are like that. The proof is harder than anything on this page.

Can two triangles share a long side? Yes: 65 = 5 × 13 is the long side of two primitive triangles and two copies (Question a finds them). Each new prime of the right kind doubles the number of primitive triangles.

Short sides one apart. 3, 4, 5 has short sides one apart. So does 20, 21, 29, the triangle on this topic’s card, and so does 119, 120, 169. Their p, q are 2, 1, then 5, 2, then 12, 5: can you see how each pair comes from the one before? (Question e asks for the next triangle.) These triangles are nearly half-squares, so their long side divided by a short side creeps toward √2 = 1.41421…: 5/3 = 1.667, 29/20 = 1.45, 169/119 = 1.4202. They never reach it, because no triple has two equal short sides: that would need $c/a = \sqrt 2$ to be a fraction, which it isn’t.

Cubes. If squares split so easily into two squares, can a cube be split into two cubes, $x^3 + y^3 = z^3$? No whole numbers work. Some come agonizingly close: $6^3 + 8^3 = 728$, one less than $9^3 = 729$, and $9^3 + 10^3 = 1{,}729$, one more than $12^3$. (1,729 is the taxi number Ramanujan told G. H. Hardy was “a very interesting number”: the smallest that is a sum of two cubes in two ways, $1^3 + 12^3$ and $9^3 + 10^3$.) Around 1637 Pierre de Fermat wrote in the margin of his Arithmetica, next to Diophantus’ problem of splitting a square into two squares, that no cube, fourth power or any higher power splits into two of the same power, and that he had a proof the margin was too narrow to hold. It took until 1995 and Andrew Wiles to prove it. The story is in Can You Buy 100 Birds with 100 Coins?

The area is never a square. 3, 4, 5 has area 6; 5, 12, 13 has 30; 20, 21, 29 has 210. No Pythagorean triangle has an area that is a perfect square, and this one Fermat really did prove, by showing that any such triangle would lead to a smaller one, and that one to a smaller one still, for ever, which whole numbers can’t do. It is the only complete proof Fermat left, published in 1670 after his death, and it settles his margin note for fourth powers.

• Can You Prove √2 Is Not a Fraction? Why no right triangle with whole sides has two equal short sides, and how the Greeks proved it.
• How Did Bhaskara Solve Fermat’s Challenge 500 Years Early? $x^2 - 61y^2 = 1$ and the chakravala, the method behind the triangles with short sides one apart.
• What Don’t Mathematicians Know About Prime Numbers? Which primes are sums of two squares, and the questions about primes nobody can answer yet.
• Can You Buy 100 Birds with 100 Coins? Whole-number answers to equations with a straight line, and Fermat’s margin.
• How Many Ways Can You Prove Pythagoras’ Theorem? (Grades 6–8) The theorem itself, from Bhaskara’s five pieces to two schoolgirls’ trigonometry.
• How Does a Calculator Find a Square Root? (Computing, Going Further, Grades 9–12) For every long side that isn’t whole.
• Grade 10, Quadratic Equation: factorizing a difference of two squares, the L of this topic.