# Why Prove a Theorem Twice?

I have written before on the problem and solution of the problem of “Pythagorean Triples”. The problem, based on the Pythagorean theorem $a^2 + b^2 = c^2$ for right triangles, asks for all possible solutions $(a,b,c)$ to this equation which have all of $a,b,c$ whole numbers. Not just any right triangle works – for instance if $a = b = 1$ then $c = \sqrt{2}$, which is definitely not a whole number. To see a longer statement of the problem, check out this post. You can also check out Part 1 and Part 2 to the first solution of the problem. If you don’t want to read those, don’t worry. I’ll summarize them all below soon.

I am now coming back to write about this problem again. Even though we have solved it, this is a wonderful opportunity to put on display a principle that all mathematicians think about as they are doing their work. It is not always good enough just to solve a problem. Sometimes, it can be extremely important to solve the same problem in more than one way. This may sound a bit strange – why would you spend your time searching out a second solution when you already know the answer? The reason, I think, would be parallel to reading a book or watching a movie multiple times. When you revisit a favorite book, you are likely to see things you didn’t see before. You will find a new, fresh perspective on some character or some event in the book, and because of that new perspective you will understand the book better.

Mathematics is just like that. When you revisit the same problem from different angles, you can learn a lot more about the problem and about the angles you come at the problem from. The Pythagorean problem is a wonderful example of this. If you’ve read the first solution, it might strike you as odd that geometry had absolutely no part of the solution to this problem. The solution, which I will give a sketch of below, was entirely algebraic! Isn’t that strange? Shouldn’t there be a solution to the problem that involves geometry, since this is a geometry problem?

These are all quite worthwhile questions, and mathematicians explore these sorts of ideas all the time. We want to know how different ideas interact – how both algebra and geometry can solve the same problem. We also explore what that means for both algebra and for geometry – maybe there are other algebra problems that geometry could help us solve, and maybe there are geometry problem that algebra can help us solve. There is now a famous history of interaction between the two in mathematics – pretty much every single problem modern mathematicians care about is multi-disciplinary. In other words, if we find a particular geometry problem interesting, it is probably because it also has connections to algebra and other areas of math. Same with algebra problems – we can almost always connect them to lots of other ideas. This is a central aspect of the mathematician’s mind: we don’t merely want to solve problems, we want to connect problems. My goal in this blog post is to explore two totally different solutions to the same problem – the problem of so-called Pythagorean triples. Since I have previously written on the first solution, I will give an outline of that solution and then explain what sorts of “insights” a mathematician would see in that solution. I will then go on to put forward a completely different solution, one inspired from geometry rather than algebra. I will then explore how we learn very different sorts of information from the geometry angle rather than the algebra angle.

The Problem

If a right triangle has side lengths $a,b,c$ with $c$ the length of the longest side, then the famous Pythagorean theorem tells us that the algebraic equation $a^2 + b^2 = c^2$ must be true. This is covered in geometry class in school and is famous for its surprising simplicity and great usefulness.

When you start thinking of the equation $a^2 + b^2 = c^2$ as an algebra equation, you can ask many interesting questions about it. One question is to solve for $c$ if you already have $a$ and $b$. If you try some random examples of whole numbers $a$ and $b$, you will begin to notice a pattern that $c$ tends to be a rather ugly number. In particular, it is pretty difficult to find examples where $c$ turns out to be another whole number. The “smallest” example of this happening is $3^2 + 4^2 = 5^2$. The “next smallest” is $5^2 + 12^2 = 13^2$. There appear not to be very many of these special solutions.

This is interesting to a mathematical explorer – why are there so few of these? And can we find all of them? This is the heart of the problem we wish to consider. We will now try to just give some shorthand language for this problem. We will now call a triple of numbers $(a,b,c)$ a Pythagorean triple if all three of $a,b,c$ are whole numbers and $a^2 + b^2 = c^2$. In other words, $(a,b,c)$ is a Pythagorean triple if we can find a right triangle with whole number side lengths $a, b, c$. Our goal for the rest of the post is to explain how you can locate every possible Pythagorean triple in two different ways.

The Key Simplifying Idea

Both of the methods involve a key simplifying idea that makes the problem easier. To see what we mean here, consider the two solutions $3^2 + 4^2 = 5^2$ and $6^2 + 8^2 = 10^2$. Both of there are genuine solutions. However, notice that if you work out all these numbers, you can divide the second equation by 4 on both sides to obtain the first. In other words, if you multiply the equation $3^2 + 4^2 = 5^2$ by four on both sides, you just get the equation $6^2 + 8^2 = 10^2$. So, there is a sense in which these are really the “same solution”. On the other hand, there isn’t anything like that to relate $3^2 + 4^2 = 5^2$ to the solution $5^2 + 12^2 = 13^2$. These are really different answers.

To make the problem easier on ourselves, we should take advantage of this observation that some solutions are “the same”. We have just observed that the Pythagorean triples $(3,4,5)$ and $(6,8,10)$ should be viewed as essentially the same solution. This is because if you multiply all the numbers in $(3,4,5)$ by $2$, you get $(6,8,10)$. This gives us the idea that we can multiply or divide a solution $(a,b,c)$ by anything we want to get another solution. We will implement this idea in two slightly different ways in the two methods.

The Algebraic Method

Using the key simplifying idea, we’d like to remove any common factors between numbers. So, we don’t care about the triples $(na, nb, nc)$, since those are really just the same thing as $(a,b,c)$. Therefore, we will assume for the rest of this method that $a, b, c$ have no common factors between them.

We can immediately make use of this fact using a key observation about the value of $x^2$ for even versus odd values of $x$. If $x$ is even, notice that that $x^2$ must be a multiple of 4 (you get two copies of 2). If $x$ is odd, then both $x-1$ and $x+1$ are even, and so $x^2-1 = (x-1)(x+1)$ is a multiple of 4 as well. To put it another way, $x^2$ is one more than a multiple of 4 if $x$ is odd, and $x^2$ is a multiple of 4 when $x$ is even. This actually enables us to learn something about Pythagorean triples $(a,b,c)$. If $a,b$ were both odd, then $a^2 + b^2$ would be two more than a multiple of 4. But $c^2$ isn’t allowed to be two more than a multiple of 2 because of what we just discussed. Since we’ve already assumed that $a,b,c$ don’t share factors, $a,b$ can’t both be even either.

Therefore, we’ve reached an interesting conclusion: one of $a,b$ must be odd and the other must be even. This tells us a fairly significant amount of information already – and this is to me the major “moral of the story” to this approach to solving the problem. If you look at other mathematical problems that involve solving some kind of equation with powers in it, one thing you’ll very frequently see is that people will consider whether the variables are odd or even and very often you can make some significant progress just by making these considerations.

The rest of this solution involves some manipulation of the original equation. You can subtract $a^2$ from both sides and factor to show that if $(a,b,c)$ is a Pythagorean triple, then $(c-a)(c+a) = c^2 - a^2 = b^2$. Since $b^2 = (c-a)(c+a)$, if we divide both sides by $b(c-a)$ we can see that

$\dfrac{b}{c-a} = \dfrac{c+a}{b}$.

These fractions may or may not be in “lowest terms”, so let’s use the fraction $\dfrac{m}{n}$ to be the “lowest-terms” form of both of the fractions above, so

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

Using $\dfrac{n}{m} = \dfrac{c-a}{b}$ and doing some additional algebra which I will leave to the curious reader, we can show that

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

This has now solved our problem – the triples all look like

$(a,b,c) = (m^2 - n^2, 2mn, m^2 + n^2)$.

We haven’t learned anything particularly enlightening from all this algebra, other than perhaps the idea that factorizing polynomials is very useful whenever it is possible. While this solution is a completely correct solution to the problem, it is not my favorite solution. My favorite is the geometric method, which I now demonstrate below.

The Geometric Method

Recall that we began the algebraic method by using the key idea to simplify to the case where there aren’t any factors in common between $a,b,c$. For the geometric method, we are going to start differently. We are going to temporarily allow ourselves fractional solutions to the equation in order to “get rid of” $c$ altogether. What I mean by that is that if $a^2 + b^2 = c^2$, then

$\bigg( \dfrac{a}{c} \bigg)^2 + \bigg( \dfrac{b}{c^2} \bigg)^2 = \dfrac{a^2 + b^2}{c^2} = \dfrac{c^2}{c^2} = 1$.

If we use $x,y$ as a sort of shortcut for $\dfrac{a}{c}$, $\dfrac{b}{c}$, then our Pythagorean equation is now $x^2 + y^2 = 1$, and the question is now to ask what the fractional solutions are to this equation. The solution $(3,4,5)$ now translates into the solution $(3/5)^2 + (4/5)^2 = 1$.

Why is this useful? One way this makes things easier is that now there are only two variables instead of three. It is slightly harder to deal with fractions instead of just whole numbers, but there is a second advantage that makes the difficulty of dealing with fractions well worth it. The important fact to know here is that $x^2 + y^2 = 1$ is a famous equation in geometry – this is the graph of a circle! This is awesome because now we are allowed to use everything we know about circles to help us out.

The big idea now is to visualize solutions as being points on this circle. To make things a bit easier, we can pick a sort of reference point $(1,0)$ on the circle, which corresponds to a rather silly solution $1^2 + 0^2 = 1^2$. Now, as has already been discussed, any Pythagorean triple $(a,b,c)$ corresponds to a solution $(a/c)^2 + (b/c)^2 = 1$, which we could graph as a point $(a/c, b/c)$ on our circle graph. This is especially nice because the correspondence goes both ways – if we have a point $(x,y)$ on the graph of the circle where both $x,y$ are some kinds of fractions, then $(x,y)$ has to basically look like $(a/c, b/c)$, which leads us to a Pythagorean triple $(a,b,c)$.

To summarize what we’ve just set up, the whole number solutions $(a,b,c)$ to the equation $a^2 + b^2 = c^2$ are really the same thing as pairs of fractions $(x,y)$ which satisfy $x^2 + y^2 = 1$. So, if we can identify every pair of fractions on the circle, we can identify all Pythagorean triples! This is one of the deep insights we get from this second method of proof – we can now see much more clearly how geometry plays a role in finding Pythagorean triples. But… how can we make use of this geometry to actually find all the solutions?

This is the next big idea. Imagine we’ve identified some point $(x,y)$ on the circle $x^2 + y^2 = 1$ but we don’t know yet whether or not $x$ and $y$ are fractions. How could be use geometry to figure that out? The answer, as it turns out, is found in another geometric object: the line. I mentioned earlier that we can find the “obvious point” $(1,0)$ on the circle. What happens if we connect $(x,y)$ to $(1,0)$ by a line? Let’s call that line $L$. What is the equation for $L$? Using the “rise over run” formula for the slope of a line, we can find the slope of this line is

$\dfrac{y - 0}{x - 1} = \dfrac{y}{x-1}$.

Notice that if $(x,y)$ are fractions, so is the slope of the line $L$. It would be very nice if we could “go the other way” too – that is, could it be true that if I graph a line $L$ going through the point $(1,0)$ and intersect that line with the circle, will that new point $(x,y)$ have fractional entries? If so, then we would have shown that “lines through $(1,0)$ with fraction slopes” are the same thing as the points $(x,y)$ with fractional entries on the circle. If this is so, then we have solved the Pythagorean triples problem in the following steps:

1. Write down all lines $L$ going through $(1,0)$ with slope $m/n$.
2. Translate each line $L$ into the point $(x,y)$ with fractional entries.
3. Translate each $(x,y)$ into a Pythagorean triple $(a,b,c)$ by the formulas $x = a/c$ and $y = b/c$.

We’ve actually already shown how to do Step 3. So, all we need to do now is Step 1 and Step 2. This will complete the method.

Step 1 is fairly straightforward. Pick a slope $m/n$. Then using the ‘point-slope’ formula for a line, the line $L$ going through the point $(1,0)$ with slope $m/n$ has equation

$y = \dfrac{m}{n}(x - 1)$.

Simplifying this equation, the line $L$ has the equation

$y = \dfrac{m}{n} x - \dfrac{m}{n}$.

This is all of Step 1! Now we can move on to Step 2, where we try to find a point $(x,y)$ which is on the circle $x^2 + y^2 = 1$ and also on the line $y = \dfrac{m}{n} x - \dfrac{m}{n}$. Since we have two equations, we can use the method of substitution to solve them. By plugging in the equation for $y$ into the circle, we now need to try to solve the equation

$x^2 + \bigg( \dfrac{m}{n} x - \dfrac{m}{n} \bigg)^2 = 1$.

By foiling out the square term and making some simplifications, this equation reduces to

$\bigg( \dfrac{m^2 + n^2}{n^2} \right) x^2 - \dfrac{2m^2}{n^2}x + \dfrac{m^2 - n^2}{n^2} = 0$.

If we multiply both sides by $n^2$, then we find the quadratic equation

$(m^2 + n^2)x^2 - 2m^2 x + (m^2 - n^2) = 0$.

We can simply solve this using the quadratic formula. By plugging in all the relevant values, the quadratic formula gives us a solution

$x = \dfrac{2m^2 \pm \sqrt{(2m^2)^2 - 4(m^2+n^2)(m^2-n^2)}}{2(m^2+n^2)}$.

After some rather tedious simplifying,

$x = \dfrac{2m^2 \pm 2n^2}{2(m^2 + n^2)}$,

and so the two solutions are $x = 1$ and $x = \dfrac{m^2-n^2}{m^2 + n^2}$. We already knew the first solution – $x = 1$ was just the point $(1,0)$. But the second solution will lead to a Pythagorean triple. Following the work we’ve already laid out, we should now have a Pythagorean triple $(a,b,c)$ with

$x = \dfrac{a}{c} = \dfrac{m^2 - n^2}{m^2 + n^2}$.

When you work out the value of $b$, you get $b = 2mn$, which is the same solution we got by the algebraic method.

Concluding Remarks

We have just shown two different ways of solving the same problem – one algebraic, one geometric. I now want to claim that the geometric approach to solving the problem is far, far more important. But what could that possibly mean? Both methods solve the problem, so how can one be better or more important than the other?

I say this method is better because the “toolkit” we learn when we look at the problem from a bird’s eye view is much more useful. To see what I mean, think about the two solutions. The algebraic proof used some weird tricks that don’t really work unless you happen to be dealing with a circle equation. However, the geometric proof isn’t like that. The geometric proof uses lines as its main tool. Lines make perfectly good sense in all sorts of contexts. This ends up making this method better than the other method because, it turns out, this method is helpful for basically any algebraic formula you’d ever be interested in! In other words, for any algebraic formula you can graph, you can learn at least something about points on it by drawing lines that go through the graph. You don’t always wind up with a total solution to the problem that way, but you can always learn at least something. This makes the geometric approach more useful than the algebraic one, since you can’t always use that approach.

To show how much you really can learn, I’ll give an example of what mathematicians have discovered using this method. The circle belongs to a group of equations called conics in geometry – conics are things like circles, ovals, and parabolas (also a more difficult shape called hyperbolas). The equations for these things are similar to circles in that all the powers of terms are at most 2, just like in the circle equation $x^2 + y^2 = 1$ we used to study Pythagorean triples. The full equation that works for any conic at all is

$A x^2 + B xy + C y^2 + Dx + Ey + F = 0.$

This equation, as you can see pretty easily, is significantly more complicated than a circle. And yet, mathematicians have proven that if you can find even one point that solves this equation, then every single other point can be found by drawing lines just like we did above! In other words, the method we discovered for solving this circle equation actually works for way more equations than just circles.

(By the way, there actually are some conceptually useful tools that show up in the algebraic method, but these are far less obvious and the usefulness of these methods isn’t quite clear until you learn a lot more about equations. This is why I say the geometric method is better – the usefulness of this approach is much more clear in this case. Any graph can be analyzed using this line method, whereas it isn’t very clear which sorts of equations can be handled using the algebraic method I used here.)

I hope you can get an idea now of why mathematicians might sometimes try to prove things they already know in new ways. Sometimes, you can learn a lot of new ideas by tackling the same problem in different ways.