Pythagorean Triples? (Solution #1, Part 2)

In Part 1, we have begun discussing primitive Pythagorean triples, and thought a little bit about them. Now, we want to try to characterize all primitive triples. That is now our goal.

Limiting The Possibilities

Suppose we are given a primitive triple (a,b,c). Recall that this means that the three positive whole numbers a, b, and c share no common factor and satisfy the equation a2 + b2 = c2. Our goal now is to think about these conditions, and to learn as much as possible about the three numbers. There are a lot of things that you can try (and very few people if any would come up with what I present here on their first try) but this line of thought is one thing that, with enough time, people thought to try.

For instance, it is an important fact that a perfect square, when divided by 4, has remainder either 0 (if the number is even) or 1 (if the number is odd). The reader can verify this for themselves. Using this information, keeping in mind that a2 + b2 and c2 must have the same remainder when divided by 4, we can discover that c must be odd, and exactly one of a, b is odd. Since a and b are interchangeable, we will suppose that a is odd and b even.

By manipulating the Pythagorean equation, subtracting b2 from both sides and factoring “the difference of squares” c2a2 = (ca)(c+a), we can conclude that b2 = (ca)(c+a). We now do not know about the factors of the two terms in (ca)(c+a), but we will use a trick with fractions to get around this. Dividing both sides by b(ca), we get a new equation

\dfrac{m}{n} = \dfrac{c+a}{b} = \dfrac{b}{c-a},

where we choose m/n to be the fraction in lowest terms. If you ‘flip’ the first and third terms in the equality above, you obtain the new equality

\dfrac{n}{m} = \dfrac{c-a}{b}.

We can add/subtract these equations together to see that

\dfrac{m}{n} + \dfrac{n}{m} = \dfrac{c+a}{b} + \dfrac{c-a}{b} = 2\dfrac{c}{b}


\dfrac{m}{n} - \dfrac{n}{m} = \dfrac{c+a}{b} - \dfrac{c-a}{b} = 2\dfrac{a}{b}.

Combining the fractions on the left of each equation using the common denominator mn and then dividing both sides by 2 gives us equations

\dfrac{c}{b} = \dfrac{m^2 + n^2}{2mn}, \ \ \ \dfrac{a}{b} = \dfrac{m^2 - n^2}{2mn}.

Since a, b, and c have no common factors, the fractions a/b and c/b are already in lowest terms. Because of this, if we can ensure that our new fractions are also in lowest terms, then the tops are equal and the bottoms are equal, and we would then be able to conclude that

a = m^2 - n^2, \ \ \ b = 2mn, \ \ \ c = m^2 + n^2.

So, we now have a pathway to part of our answer! All we have to do now is to set up boundaries for the values of m and n within which we know that each of the three values m2n2, 2mn, and m2 + n2 will have no common factors. As it turns out, we can actually force this to happen. To see this, suppose a prime number p happens to be a factor of all three of these expressions, it must be a factor of (m2n2) + (m2 + n2) = 2m2 and of (m2 + n2) – (m2n2) = 2n2, and because of this must also be a factor of both m and n. But, go back a few paragraphs, when we defined m and n. We defined the fraction m/n to be in lowest terms, but we have now claimed that m and n have a common factor. As it turns out, this makes no sense. By assuming we could find a common factor, we contradicted ourselves. This is a mathematical trick called a proof by contradiction (more on this another time), but for now we need only say that this means we can take for granted that our m and n can in fact be chosen in such a way that these three values in fact share no common divisor.

We’ve now done a good deal of exploration, we have actually solved half of the problem! We have established that every primitive triple (a,b,c) can be associated with these three numbers, m2n2, 2mn, and m2 + n2 exactly when they have no common factor. I will leave as a problem for a curious reader the following:

Claim: The numbers m2n2, 2mn, and m2 + n2 have no common factor if, and only if, m and n share no common factor and exactly one of them is even (and the other is odd).

This reduces every primitive triple to an easy-to-produce formula… but we actually aren’t quite done. We can ask now the reverse question of what we just did. Instead of starting with a, b, and c, what if we start with m and n? Does our new formula always work? Or only sometimes? In fact, it will always work. To see this, we want to add together the squares of the two smaller numbers and see if it is equal to the square of the larger number. In order to do this, recall briefly the “foiling” method that gives us the formula (x+y)^2 = x^2 + 2xy + y^2. Using this, we can see that

(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 = m^4 + 2m^2n^2 + n^4.

So these are the same! The Pythagorean equation is satisfied. Our new construction can now move both ways. We can start with (a,b,c) and find the numbers m and n, or we can start with m and n and find (a,b,c). What this does now is establish the following complete answer to our original question.

Theorem: The three numbers a, b, and c can form a primitive Pythagorean triple if, and only if, these three numbers can be expressed (in some order) by the numbers m2n2, 2mn, and m2 + n2, where the numbers m and n have no common factors, are not both odd numbers, and where m is larger than n.

Our question has been answered, which is quite a wonderful thing, but this approach requires a lot of guesswork and toying around to discover. There is a second, more beautiful and simple solution that gives us the same answer and I think gives a lot more insight about where all of this is coming from. This second solution will come in a post of its own.

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