Monday Math: Pythagorean Triples, cont.

Posted by jrosenhouse on January 24, 2011
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Time to finish what we started last week.

We saw that if a, b, c was a primitive Pythagorean triple, then at least one of a and b is even and one is odd. Let us declare, then, that we will use a to denote the odd length and b to denote the even one. By rearranging the Pythagorean equation and factoring we get:


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

 

Let’s try this out for a few specific triples:


\[
3^2=(5+4)(5-4)=(9)(1)
\]


\[
5^2=(13+12)(13-12)=(25)(1)
\]


\[
15^2=(17+8)(17-8)=(25)(9)
\]


\[
45^2=(53+28)(53-28)=(81)(25)
\]

 

In each case we find that c+b and c-b are both perfect squares. Even more important, they are relatively prime. Let’s see if we can prove that.

We begin by noting that, since b is even and c is odd, we have that c+b and c-b are both odd.

Now suppose that d is some number that divides both of them. We want to show that d has to be one. Since d divides both c+b and c-b, we must also have that d divides their sum and their difference. That is, d divides:


\[
(c+b)+(c-b)=2c \phantom{xxx} \textrm{and} \phantom{xxx} (c+b)-(c-b)=2b
\]

 

Recall, though, that we are discussing primitive Pythagorean triples. That means that b and c are relatively prime. It follows that d must divide 2, meaning that d=1 or d=2.

Having already established that c+b and c-b are both odd, we know that d is not two. We conclude that d=1 as desired.

This is significant for the following reason: If the product of two relatively prime numbers is a perfect square, then the numbers must themselves be perfect squares. I won’t stop to prove that here, though it is straightforward if you think in terms of prime factorizations.

We can now write


\[
c+b=s^2 \phantom{xxx} \textrm{and} \phantom{xxx} c-b=t^2,
\]

 

where s and t are odd, relatively prime integers. But now we can solve for b and c to obtain


\[
c= \frac{s^2+t^2}{2} \phantom{xxxx} \textrm{and} \phantom{xxxx} b=\frac{s^2-t^2}{2},
\]

 

from which it follows that


\[
a=st.
\]

 

And we’re done! So long as s and t are odd and relatively prime, these formulas will give you a primitive Pythagorean triple. Moreover, all such triples can be produced in this way. For example, choosing s=3 and t=1 gives you (3,4,5) and choosing s=5 and t=1 gives you (5,12,13).

 

As was pointed out in the comments to last week’s post, we could also take a purely geometric approach to this problem. Notice that if (a,b,c) is a Pythagorean triple, then we have


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

 

meaning that any such triple corresponds to a rational point on the curve


\[
x^2+y^2=1
\]

 

That’s the equation for the unit circle. Imagine now that we draw a line, with rational slope m, through the point (-1,0). If you remember your elementary algebra you will see that the equation for this line is


\[
y=m(x+1)
\]

 

This line will hit the unit circle in exactly one other point, as shown below:



 

It is now a straightforward algebraic exercise to solve our equations simultaneously to work out the coordinates of the second point of intersection, which will be a rational point on the circle. When that is done, you quickly recover our formulas for producing Pythagorean triples. Very nice!

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Comments

  1. #1 Jon Wharf
    January 24, 2011

    Nice. I like the way you can clearly only get a prime leg when the even leg is within 1 of the hypoteneuse (a split-the-square triple).

    Here’s my different derivation of a two parameter formula:

    For a primitive pythagorean triple (p,q,r), p^2 + q^2 = r^2
    Set p=r-m, q=r-n
    (r-m)^2 + (r-n)^2 = r^2
    Then r^2 – 2(m+n)r + (m^2+n^2) = 0
    r = m + n + sqrt(2mn)
    => if m and n have a common factor then r also has that factor and (p,q,r) is not primitive
    => m,n coprime
    sqrt(2mn) is integer => m, n are an odd square and half an even square.
    Set m odd, => m=x^2 and n=y^2/2
    For any coprime x odd and y even, (p,q,r)
    p = y^2/2 + xy
    q = x^2 + xy
    r = x^2 + y^2/2 + xy
    is a primitive pythagorean triple. I think t=x and s=x+y is the mapping to your solution.

    Incidentally I was wrong, last week, that x & y can be uniquely extracted from any natural number. So it remains a two-parameter formula.

  2. #2 joemac53
    January 24, 2011

    My elementary algebra students got a kick out of breaking up odd perfect squares to make a pythagorean triple. The odd prime r^2, in the form 2n +1, could be broken into n and n+1. Those two and r made the triple.

    Those freshman tortured the upperclassmen for a few days. I would teach them math parlor tricks, saying it would make them very popular at parties. They refused to believe me.

  3. #3 madarab
    January 24, 2011

    You can also note that every square is the sum of odd numbers Σ(n=0 to k) 2n+1=k^2 That means that every odd square (2k+1)^2 forms a primitive triple of the form [(2k+1), (2k^2+2k), (2k^2+2k+1)]. This generates an infinite number of primitive triples.

  4. #4 David Winfrey
    January 25, 2011

    Here’s a similar question; I’m not sure how closely it’s related, or whether it has a solution.

    For y = ax**2 – bx + c, where the coefficients are positive integers, is there any way to find the smallest positive integer x such that y is a square?

    This arises out of an attempt to find prime factors of large composites, so it may not be possible.

  5. #5 Jon Wharf
    January 25, 2011

    Also note that my method does not need to “know” in advance whether p,q and r are odd or even, or anything about patterns of squares.

    That’s probably why I found it :-) .

    The other thing I should have noted is that it tells us directly that the difference between the two leg lengths is constrained to be a difference between an odd square and half an even square (that is coprime to the odd square).

  6. #6 parisa
    February 5, 2011

    Hi
    I need to notaion to number theory please help me

  7. #7 bhbh
    y7y
    August 10, 2012

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