12, 13) are both primitive triples. Can we find a formula for generating all primitive triples?

We first note that if (a, b, c) is a primitive triple, then any two of the numbers a, b, and c must be coprime, for if (for example) a and b have a common factor that’s greater than 1, then they must have a common prime factor p. It follows that p must divide , which equals c2, and so p (being prime) also divides c. This contradicts the fact that a, b, and c have no common factor. We can therefore assume that

But we can say more. If (a, b, c) is a primitive triple, then a and b cannot both be even, as we’ve just seen. Can they both be odd? As we saw in Chapter 2, every square has the form 4n or , and so if a and b are both odd, then a2 and b2 must have the form , and c2 must have the form , which is impossible. So one of a and b must be even and the other must be odd, and c2, and hence c, must also be odd. For definiteness, we’ll always take a to be odd and b to be even.

Now, because a and c are both odd, their sum and difference are both even and we can write and , for some integers u and v, with . So and .

Also, b is even and , and so

What can we say about u and v? If , where , then d divides and , and so divides both c and a, which cannot happen. So u and v are coprime. Also, because uv is a square and u and v are coprime, u and v must separately be squares, and so we can write and , for some integers x and y with . So

It can be checked that one of the numbers x and y is odd and the other is even, and that they are coprime. Tying all this together, we have our desired formula for primitive triples:

Primitive Pythagorean triples: If (a, b, c) is a primitive Pythagorean triple, then

where x and y are coprime integers, one odd and the other even, with .

For example, if and , then , , and , giving the triple (3, 4, 5). Likewise, if and , then , , and , giving the triple (5, 12, 13).

We can use this recipe to draw up a table of primitive Pythagorean triples, by listing all pairs of coprime integers x and y with one odd and the other even and , and calculating the corresponding values of a, b, and c. Table 4 lists all the primitive triples with no numbers exceeding 100. By extending this table as far as necessary, and then taking multiples, we can generate all the right-angled triangles with whole number sides.

Table 4. Primitive Pythagorean triples

We can also use our formula for primitive triples to answer such questions as:

How many primitive triples include the number 60?

Because 60 is even, we have , and so . Remembering that and that x and y are coprime with one of them even and the other odd, we find the following four possibilities:

A similar question is:

How many right-angled triangles with whole-number sides have a side of length 29?

Because 29 is prime, the lengths of the sides must form a primitive triple, so either

In the former case, and , and the triple is (21, 20, 29).

In the latter case, , so and . So and , and the triple is (29, 420, 421).

Sums of squares

Having just investigated the equation , we may ask a more general question that can be traced back to Diophantus:

Which numbers can be written as the sum of two perfect squares?

The first few examples are:

The numbers up to 20 that cannot be written as the sum of two squares are 3, 6, 7, 11, 12, 14, 15, and 19. Can we deduce a general pattern from this?

The first thing to remember is that any square has the form 4n or , and so the sum of two squares must have the form 4n, , or . So any number of the form cannot be a sum of two squares, and this rules out 3, 7, 11, 15, and 19. We also note that 6, 12, and 14, which are multiples of the forbidden numbers 3 and 7, cannot be written as the sum of two squares, whereas 9 and 18, which are multiples of 32, can be so written. By using these observations as guidelines, it turns out that we can completely describe which numbers can be written as the sum of two squares. The following result was stated by Fermat, and proved by Legendre in 1798:

Sum of two squares: A number can be written as the sum of two squares if and only if every prime factor that is congruent to 3 (mod 4) occurs to an even power.

For example, is the sum of two squares because 3 occurs to an even power, whereas and cannot be the sum of two squares because 3 occurs to an odd power in each case.

A useful result is that

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