by Ellen Kaplan
Very well: we’re therefore looking for ancestors in which each pair is relatively prime. It would have been natural for a Pythagorean to think at this point in terms of parity: whether a number is odd or even (since this distinction awoke so many hidden harmonies for them). They would therefore rephrase our latest conclusion by saying that a, b, and c can’t all be even, nor could any pair of them be—hence all three must be odd, or two will have to be odd and one even. This is the sort of dividing and conquering that number theoryt delights in. It was how Socrates caught the sophist in his nets,1 and how we, along with those ancient mathematicians, will catch the infinite variety of Pythagorean triples in a little box.
It can’t be that a, b, and c are all odd, since an odd number is of the form 2s + 1, and so an odd squared is odd. But this means that if a, b, and c were all odd, a2 + b2 = c2 would also be composed of all odd numbers. Yet an odd plus an odd is even, so we would have a2 + b2 (an even) equal to c2 (an odd), which is impossible. Hence two of a, b, c are odd and one is even. Which?
Were c even, say c = 2m, then c2 = 4m2, and 4 would be a divisor of c2. But since a and b have to be odd, we can write a = 2p + 1 and b = 2q + 1 for some p and q. That makes a2 + b2 = (2p + 1)2 + (2q + 1)2 = 4(p2 + p + q2 + q) + 2. Hence 4 can’t be a divisor of a2 + b2 (since it is a factor of the first but not the second of the terms that make it up), which contradicts the fact that it divides 4m2 = c2, which these add up to. So c can’t be even, and since we must have c odd, that means either a or b must be even. Does it matter which? No. Let’s take a to be odd and b even.
Pause to appreciate how far we’ve come. Invoking no more than parity, we’ve learned something important about a, b, and c—their genetic makeup, if you will. For what it’s worth, since c and a are both odd, c + a and c − a must both be even.
Let’s make use of this little knowledge by saying in a different way the only other thing we know: a2 + b2 = c2 means that b2 = c2 − a2, which in turn equals (c + a)(c − a). Yes, this confirms that b is even (since even times even is even), but it will turn out to do much more. For we claim that the only common factor of (c + a) and (c − a) is 2. While we have more or less followed our noses up to now, this latest claim isn’t at all obvious.
Here’s the ingenious proof (it’s like watching a major league shortstop turn a double play on a bad hop that would likely skip past a fielder in the minors). If c + a and c − a had in common a factor other than 2, call it d, then c + a = 2df and c − a = 2dg, for some whole numbers f and g. Dividing by 2, we would then have:
Adding these two together we get
This means that d is a factor of c.
Now let’s subtract our second term from the first:
This means that d is a factor of a. But that is impossible: c and a can’t have a common factor—they are relatively prime.
Hence when we divide (c + a) and (c − a) by 2, the results are relatively prime.
Let’s assess our gains: a and c are odd, b, (c + a) and (c − a) are even
It may seem that this dog has been chasing its own tail, but in fact he has just dug up the bone. Watch. Replacing the right-hand side of equation (3) above,
Adding the two equations in (4) gives us
while subtracting the second from the first in (4) tells us that
What does this mean for our quest? If you choose any two relatively prime whole numbers, u and v—one even, one odd (and let’s assume u is greater than v)—then you will automatically have made a Pythagorean triple:
Does a2 + b2 = c2, for these values? It’s the work of a minute to check: and yes indeed, this compact little kit yields all and only the primitive ancestors of the whole Pythagorean family. If you wish to dot your i’s and cross your t’s by confirming that when a, b, and c are so chosen they really are relatively prime, see the appendix.
This outcome is up there with a fugue from The Well-Tempered Clavier or the dome of St. Paul’s. By pointedly looking away from the visible world, we have seen not only the two families of gnomonic triples discovered in Chapter Two, but triples you wouldn’t have guessed, in a genealogy limitlessly broad. Here, for example, are the choices of u and v for some members of our first (single gnomon) family. If u = 2 and v = 1, we get (3, 4, 5); u = 3 and v = 2 produces (5, 12, 13). From the second (two-gnomon) family, let u = 4 and v = 1 to get (8, 15, 17). Do you recall Neugebauer’s triple (4601, 4800, 6649)? It comes from u = 75 and v = 32. Choose your own monstrously large u and v fitting the requirements: you will have just made a Pythagorean triple most likely never imagined before.
Should you conclude that so much doggedness, wedded to such inspiration, is beyond your wildest dreams, remember that what you’ve followed here is the tidied remainder of who knows how many lively conversations—and conversations among practitioners as devoted to their craft as are cooks to theirs. Is a pinch of cardamom, a drop of vanilla, a hint of lemon zest needed to make the dish just so? And here: is it factoring, or thinking in terms of parity, or of primes, that would perfectly season our thinking? Construing then, as now, evolves past constructing, and accounts for how mathematics creates what it later reads as discovered.u This is why Einstein could say: “I hold it true that pure thought can grasp reality, as the ancients dreamed.”2
What we’e seen, in the previous chapter and this, has been not only proofs but a short history of proving. Let’s deepen that history by comparing this Greek way of finding all Pythagorean triples to its modern counterpart.v Just as tautologies let us look below the surface by saying the same thing in two different ways, so contrasted proofs of the same theorem tell us a great deal about the nature of shifted points of view. They might even raise in your mind the question of whether a theorem remains the same, when differently proved. Context alters content.w
Since we’re looking for all triples (a, b, c) of natural numbers such that a2 + b2 = c2, and since c, being a natural number, can’t be zero, we could divide by it and rewrite:
Think of this equation as saying that we’re looking for all pairs of rationals, , which solve the equation
Rather than fleeing from geometry to number, as we did above, let’s call geometry to our aid: this is the equation of a circle with radius 1,
centered at the origin of the Cartesian plane. What we’re asking for is all points on this circle whose coordinates, (x, y), are both rational.
We can quickly come up with four candidates: (1, 0), (0, 1), (-1, 0), and (0, −1). Are there others—and if so, how shall we find any or all of them? Half the art of invention lies in choosing a fruitful viewpoint—the other, in seeing something useful from it. While we know ourselves too little yet to be sure where such insights live, it often helps to ring a few doorbells. Geometry, circles, points . . . what about lines? Let’s anchor a straight line at one of these points—say (-1, 0)—and let it sweep its way across the circle:
Second, paired, insight: let’s consider this line only when its slope is a rational number m—for then (parlaying now between geometry and algebra) the equation of this line will be
and since (-1, 0) is on this line—i.e., x = −1, y = 0—we can solve for b and find b = m, so that our equation
or
We’ve set it up so that if x is a rational number, so is y; and if these points (x, y) lie both on the line and the circle—that is, where the line meets the circle a second time, after its intersection at (-1, 0)—then those points will be the ones we want: points with rational coordinates .
It remains only to find them—and this needs not art but craft: the craft of handling equations in general, and quadratic equations in particular. We want to solve our equations simultaneously, so taking out your dividers and multipliers from their velvet cases and substituting (2) in x2 + y2 = 1, you find
An easy way now to find all possible x (from which, by using equation (2), we’ll then get the paired possible y) is this. Since we know x = −1 is a solution, i.e., x + 1 = 0, let’s divide (3) by x + 1 to get the other root:
(If you don�
�t want to check this by doing the division, just multiply the quotient by the divisor to see if you get the dividend back.)
The other root is therefore the solution of
This means that for every rational number m we get a solution to (x, y) in rationals:
We’ve managed to show that if m is rational, then we have a solution in rationals to (1). But are these all the rational solutions? That is, if (r, s) is a solution in rationals to (1), will there be an m that produces it? Well, if (r, s) is such a solution, then the line through it and (-1, 0) will have slope
which is rational. Our modern process has indeed yielded every rational solution to (1).x
How very different the results of the Greek and modern approaches look. Has the nature of number, or of the Pythagorean relation, changed over the millennia, or need we only compare? Let’s see. If we write our rational m as and substitute this in (6), we have
One difference, though, is that in these liberal times our procedure has given us all, not just the ancestral, Pythagorean triples. It even includes a triangle with sides and hypotenuse 1—not a generalization even our most adventurous travels came upon. To restrict ourselves to the ancestors we would need to put the old restraints back on u and v: that they be relatively prime whole numbers, one even, one odd, and with u > v.
We’ve been describing the modern counterpart of the way the Greeks handled Pythagorean triples—but “modern” is a floating index. The Sumerians must have felt as modern as we do (recall, for example, that the Hypermodern school of thought in chess flourished a century ago). Since there never seems to be a last word in mathematics, would it surprise you to hear of this leap past the steps we’ve just taken: c is the hypotenuse of an ancestral Pythagorean triple (a, b, c) if and only if it is a product of primes, each of which leaves a remainder of 1 when divided by 4?3
The dynamic character of contemporary approaches may well have struck you. What has become of the Greek conviction that things mathematical are immovably part of what is? A great many mathematicians still consider themselves Platonists (a name that covers many variations), but an altered view of mathematical activity—from constructing to construing, as we’ve put it—allows them to shift and make shifts in the middle world of language: processes at whose limit those immutable structures lie. The fecund tension is no longer between geometry and algebra, but play within and between them generates insightful ways of re-saying and so re-seeing the eternal same, as mind planes over the procreating world.
You catch a flicker of this evolution in the movement from Loomis’s old strictures against trigonometric proofs of the Pythagorean Theorem, to the trigonometry-like work we’ve just done together in finding Pythagorean triples. The One remains, the terms it is that change.
APPENDIX
The fine print guaranteeing that our hard-won a, b, and c are indeed relatively prime reads like this:
Since a and c are both odd, 2 couldn’t be a common factor. But say they had an odd common factor, p, so that a = pm and c = pk. Then since
This means p would be a factor of both 2u2 and 2v2, hence of u2 and v2—and therefore of u and v.
But these, we said, were chosen to be relatively prime—so, indeed, a and c have no common factor.
Had a and b a common odd factor q, since b = 2uv, q would have to be a factor of u or v—say u. But a = u2 − v2 = (u + v)(u − v), hence q must divide one of these terms. But since it divides u, it would also therefore have to divide v, which is impossible, since u and v are relatively prime.
CHAPTER EIGHT
Living at the Limit
We’ve traveled very far since leaving a bold conjecture stranded, half clothed, in Chapter Six: might the square on the ‘body-diagonal’ of an n-dimensional box be the sum of the squares on the n sides of the box? We know this is true for n = 2 (for this is the Pythagorean Theorem), and saw it was true for n = 3.
Being experienced voyagers, we are confident that induction will easily carry us home.y For assuming the conjecture true in k dimensions, the square on the body-diagonal of a k + 1 dimensional box will (in the plane made by the new side in dimension k + 1 and the previous body-diagonal in dimension k) be the square of the new side plus the square on that previous diagonal, by the Pythagorean Theorem; and by the inductive assumption, this square is itself the sum of the squares on the previous k sides. When Descartes, as you saw, wrote 1, 2, 3, 4 in his secret journal, he might have whispered in his mind: “. . . ”—where the dots bear infinitely more weight than the sturdy integers that precede them.
FROM HERE TO ETERNITY
Two further journeys open from this: one will change our viewpoint, the other immeasurably broaden it. Both begin by acknowledging that the n-dimensional box is a natural framework for the Pythagorean Theorem, with the hypotenuse as a diagonal strut. But aren’t we perversely making nested matryoshka dolls? Why picture such a box in n-dimensional space, when that space could itself be all the framework we require? For it is a ‘box’ open in every direction from the neat corner at the origin where its n axes meet, each at right angles to the rest.
One of us, at thirteen, got the shock of a lifetime on reading in Sir Arthur Eddington’s little book, The Expanding Universe, his explanation of why the galaxies were all receding from ours. Think of the galaxies, he wrote, as ink drops spattered on a balloon, which you then inflate. Each drop sees the others as moving away from it, but in fact it is the space itself which is expanding. So that’s how adults thought of things! Well, perhaps not—but it is the view geometers come to: what they study isn’t so much shapes in space as the space of shapes. What we’ll do now is rethink the Pythagorean Theorem as a property of n-dimensional space.
For this we’ll need such things as angles with their measures, and lines along with their lengths, and the squares of these lengths; none of these imported from elsewhere, but (in the spirit of trying to reach the innermost doll) intrinsic. To that end, we won’t think of space as a grid of points with coordinates, but as a vector space: a set of line segments each with a direction and length, most naturally pictured as arrows with their tails at that common origin and their heads at what before were just those isolated points.
How easily now a limpid picture begins to coalesce. That origin, as with everything else in this space, is itself a vector—whose length is 0. Out of it, like quills upon the fretful porpentine, grows a set of special arrows, each at right angles to the others, one for each of the space’s n dimensions—and again, for the simplicity that uniformity brings, let each of these be of length 1. These special vectors are the axes for the space, and the set of them is aptly called a basis. In two dimensions therefore the basis consists of the two vectors {(1, 0), (0, 1)}; in three, {(1, 0, 0), (0, 1, 0), (0, 0, 1)}—and so on. As a shorthand for the n axes in n-dimensional space, these vectors tend to be called e1, e2, . . . , en.
In order to stretch or shrink these axes to an infinity of other vectors in their directions, we provide ourselves with a kit of scalars: all the real numbers, thought of as multipliers.z To make a vector 17 units long in the e1 direction, for example, just multiply each coordinate of e1 by 17: 17e1 = (17, 0, 0, . . . , 0).
And what about all the vectors that don’t point along any axis but are at angles to them? We can form these with one of the last bits of machinery we’ll need. For we want to be able to add vectors to one another, as well as multiply each by scalars. But how do you add two arrows? As so often happens, physics gives us the clue: if you have two component forces acting on a particle, it squirts out between them in a quite specific way. Picture this in two dimensions: the resultant of the component forces (vectors) X and Y is their sum, X + Y, understood as the long diagonal of the parallelogram these components generate:
To speak of this as a ‘sum’ isn’t as arbitrary as it may at first seem, because the coordinates of the vector X + Y are indeed the sums of the respective coordinates of X and Y: (x1 + y1, x2 + y2).
So the vector that is the sum of X = (1, 0) and Y = (
0, 1) is X + Y = (1, 1):
and (staying in two dimensions for ease in visualizing), you could write a vector like (8, −1) as (2, 2) + (6, −3). Or at an even further remove, since (2, 2) is 2(1, 1) and (6, −3) is 3(2, −1), why not write (8, −1) = 2(1, 1) + 3(2, −1).
In this way we can manufacture any vector we like from the sum of others, suitably scaled.
The only work left to do before rephrasing the Pythagorean Theorem in the language of vector spaces is to understand how to calculate a vector’s length, and what signals that two vectors are perpendicular to each other. We’ll let the Pythagorean Theorem itself lead us. The vector (4, 0) has length ; the vector (0, 3) has length , and the vector (4, 3) = (4, 0) + (0, 3) has length : that is,
the square root of the sum of each of its coordinates squared. This accords with our common intuition of distance, and is what the rectangular box showed us: The length of a vector (x, y) in two dimensions is , and of a vector (x1, . . . , xn) in n dimensions is .
So, fancying up the absolute value sign commonly used for distance, we write the length of a vector X = (x1, . . . , xn) as
To go along with this elevated notation, the custom is to rechristen this more abstract length ‘norm’.