## Pythagorean Triples

I recently began writing some lectures for an online course — I’ll talk more about the nature of the course in next week’s post.  The broad topic is geometry, of course a favorite — and the specific topic for this unit is Triangles.

You can’t talk about triangles without talking about the Pythagorean Theorem.  Part of my job is also to compose problems for the lectures as well as for quizzes and exams, and to my surprise, I came up with a few interesting ones.  So I thought I’d share them with you.  I am always a fan of sharing mathematics as it happens!

The questions I wrote are based on the following parameterization of Pythagorean Triples:  Given positive integers $p$ and $q,$ then

$(q^2-p^2,2pq,q^2+p^2)$

is a Pythagorean Triple.  This parameterization generates all primitive Pythagorean Triples — that is, triples whose sides share no common factor.  But it is not possible to get $(9,12,15),$ for example, using this parameterization.  Of course $(9,12,15)$ is just three times the triple $(3,4,5);$ therefore, if you can generate all primitive Pythagorean Triples, you can take multiples of them to generate all Pythagorean Triples.

I thought of my first problem walking down the sidewalk going to lunch the other day.  The simplest Pythagorean Triple, $(3,4,5),$ has side lengths which are in arithmetic progression.  What other Pythagorean Triples have this property?

The simplest way to approach this is to parameterize such a triple by $(a,a+d,a+2d),$ where $a>0$ is the smallest integer in the arithmetic progression and $d>0$ is the common difference.  Since the triangle is a right triangle, we must have

$a^2+(a+d)^2=(a+2d)^2,$

which we may rewrite as

$a^2-2ad-3d^2=0.$

Now this factors:

$(a+d)(a-3d)=0,$

resulting in solutions $a=-d$ or $a=3d.$  We did assume that $d>0,$ so we eliminate the solution $a=-d.$  Note that this would generate the triple $(a,0,-a),$ and in fact $a^2+0^2=(-a)^2.$ But one side length is zero and another is negative, so no triangle is possible with these side lengths.

What about the solution $a=3d$?  Here, we get

$(a,a+d,a+2d)=(3d,4d,5d),$

which you can observe is just a multiple $d$ of the primitive Pythagorean Triple $(3,4,5).$

The conclusion?  The only Pythagorean Triples possible whose side lengths are in arithmetic progression are multiples of the $(3,4,5)$ right triangle.

I really didn’t know the answer would come out so nicely — but since the algebra involved was fairly straightforward, I thought I could include this as a non-routine example of an application of the Pythagorean Theorem at the high school level.

The previous problem was part of a lecture.  The next problem was written as a possible exam question for teachers; once I realized I had more than one interesting problem, I thought there would be enough for a blog post….

I was just looking for interesting patterns in Pythagorean Triples, and noticed that with the $(6,8,10)$ triangle, the area and perimeter were both $24.$  A coincidence?  Were there other triangles with this property?

Of course there had to be finitely many — as the side lengths get larger, the area gets larger faster than the perimeter, as the area is essentially a quadratic function, while the perimeter is essentially a linear function.  So how many others are there?  Make a mental note of your guess before reading further….

We begin by parameterizing by

$(k(q^2-p^2),2kpq,k(q^2+p^2));$

the factor of $k$ is necessary since the two-variable version generates all primitive Pythagorean Triples, but not necessarily all Pythagorean Triples.

Setting the perimeter and area equal to each other results in

$\dfrac12k(q^2-p^2)\cdot2kpq=k(q^2-p^2+2pq+q^2+p^2),$

Cancelling out factors of $k,$ $q,$ and $p+q$ results in

$kp(q-p)=2.$

This equation clearly has just three solutions, since one of the factors must be $2$ and the other two factors must be $1.$

None is particular difficult; let’s take them one at a time.  When $k=2,$ then $p=q-p=1,$ so that $q=2.$  Substituting back into the parameterization, we obtain the Pythagorean Triple $2(3,4,5),$ which is the triple $(6,8,10).$

When $p=2,$ then $k=q-p=1,$ so that $q=3.$ This generates a new Pythagorean Triple, $(5,12,13).$

Finally, when $q-p=2,$ then $k=p=1$ and $q=3,$ so that the Pythagorean Triple $(8,6,10)$ is generated.  Of course this is just a duplicate of the first solution.

Surprised that there was just one more solution?  I was!  It was such a nice, straightforward solution, that I couldn’t help but include it.

There was a third problem which I liked, but the algebra was a little too intense — there was a nice geometrical solution, but it required ideas learned later on in the course.  So here it is if you want a challenge:  suppose you are given two right triangles, and you know that their perimeters and areas are the same.  Prove that they are congruent.

I think you might enjoy solving this purely algebraically.  I did like it so much, though, that I included a simpler version in one of my lectures:  suppose you are given two right triangles, and you know that their hypotenuses are both of length $8$ and that their perimeters are equal.  Prove that the triangles are congruent.

To be honest, I never knew I’d find problem solving with the Pythagorean Theorem so interesting.  It’s nice to know that there is always more geometry to learn!  Even with something as apparently simple as the venerable Pythagorean Theorem….