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Chapter 2 · The Baudhāyana-Pythagoras Theorem

Baudhāyana-Pythagoras triples, and how to make infinitely many

यह वीडियो हिंदी में भी · Watch in Hindi

Integer triples, and beyond11 min

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11 min.

Also recorded in Hindi.Englishहिन्दी

3, 4, 5 — nine plus sixteen is twenty-five, so three whole numbers make a right triangle. There are infinitely many more, and a recipe for them.

The idea

The section builds two machines for making integer triples, and they answer two different questions. Scaling — multiply every term of a working triple by the same whole number — works because the k² factors straight out of a² + b², and it delivers infinitely many triples. But every triple it delivers is a magnified copy of one you already had, so it produces no new shape of triangle at all. Getting genuinely new triples needs a second idea from a completely different corner of arithmetic: adding the first n odd numbers gives n², so (n − 1)² plus the nth odd number is n². Whenever that nth odd number happens itself to be a square, you have a square plus a square making a square. So the honest answer to "how many triples are there?" is: infinitely many, twice over, and for two quite different reasons.

What you should be able to do

  • Test whether a given integer triple satisfies a² + b² = c²
  • Name the triples the chapter names, and give the four names the chapter offers for them
  • List every triple whose entries are all 20 or less, and say which of them are new shapes and which are magnifications
  • Prove that (ka, kb, kc) is a triple whenever (a, b, c) is, by taking k² outside the sum
  • Explain why that proof establishes infinitely many triples and no new triangles
  • Define a primitive triple, and decide primitivity for a given triple
  • Derive (n − 1)² + (2n − 1) = n² from the odd-numbers fact, and also by expanding
  • Generate triples by choosing an odd number that is itself a square, and identify which n it is
  • Argue that every triple this second machine produces is primitive, and give a primitive triple it cannot reach

Words to know

TermDefinition in one lineFirst introduced
Baudhāyana triplestriples of whole numbers that are the sides of a right triangleprinted in bold in this chapter (Part II, §2.5, p.48)
Baudhāyana-Pythagoras triplesthe same, under the chapter's joint nameprinted in bold in this chapter (Part II, §2.5, p.48)
Pythagorean triplesthe same, under the name in widest international useprinted in bold in this chapter (Part II, §2.5, p.48)
right-angled triangle triplesthe fourth name the chapter offers for the same objectsprinted in bold in this chapter (Part II, §2.5, p.48)
scaled versionwhat you get by multiplying every term of a triple by one whole numberprinted in bold in this chapter (Part II, §2.5, p.49)
primitive Baudhāyana triplea triple whose three terms share no factor greater than 1printed in bold in this chapter (Part II, §2.5, p.49)
conjecturea general statement put forward on the evidence of examples, before proofprinted in bold in this chapter (Part II, §2.5, p.48)
common factora number dividing every term of the tripleprinted in this chapter (Part II, §2.5, p.49)
square numbera whole number that is another whole number times itselfprinted in this chapter (Part II, §2.5, p.49)
odd squarean odd number that is also a square, which is what the second machine feeds onprinted in this chapter (Part II, §2.5, p.50)
nth odd numberthe odd number in position n, which the chapter writes as 2n − 1printed in this chapter (Part II, §2.5, p.49)
Verse 1.13the Śulba-Sūtra verse in which Baudhāyana lists the triplesprinted in this chapter (Part II, §2.5, p.48)
shape of a trianglethe thing a scaled triple leaves unchanged — same angles, different sizean added phrasing; the chapter distinguishes scaled from primitive without ever saying what scaling preserves

Where people slip up

  • "Infinitely many triples means infinitely many different right triangles with whole-number sides." True, but not because of the scaling proof, and this is the section's sharpest point. Scaling gives infinitely many triples that are all the same three or four triangles blown up. The odd-square machine is what supplies genuinely new triangles. An explanation that runs only the scaling proof has answered the easier question and left the interesting one standing.
  • "(6, 8, 10) is a new triple." It is (3, 4, 5) doubled. Draw the two triangles on top of each other, scaled, so the student sees one shape.
  • "Testing (30, 40, 50) and (300, 400, 500) proves the conjecture." It does not, and the chapter is explicit about this: it substitutes k and checks algebraically. Examples motivate a conjecture; only the letter settles it.
  • "Any three numbers in the ratio 3 : 4 : 5 form a triple." Only whole-number multiples give whole-number triples; 1.5, 2, 2.5 satisfies the relation but is not a triple of integers, which is what the section is about.
  • "Primitive means small." It means no shared factor. (8, 15, 17) is primitive and (9, 12, 15) is not, though they are the same size.
  • "Every odd number gives a triple." Only an odd number that is itself a square. 9, 25, 49, 81 work; 7, 11, 15 do not. Students will try the first few odd numbers and conclude the method is broken.
  • "The odd-square machine finds them all." It does not — it only reaches triples whose larger leg and hypotenuse are consecutive. Section 12 exists to say so, and the chapter's own Math Talk item asks the student to notice it. Baudhāyana's list contains counter-instances.
  • "So we have a method that generates all triples." The chapter says mathematicians have found one, and does not give it.
Transcript1,446 words

Three, four, five. Nine plus sixteen is twenty-five, so those three whole numbers are the sides of a right triangle. Three whole numbers that do that are worth a name, and they have one: a triple. Both words matter. The relation has to hold, and all three have to be whole numbers. Halve them and you get one point five, two, two point five. That still satisfies the relation exactly.

But it is not a triple, because two of the three are no longer whole numbers. It is the same triangle, drawn at half the size. What was lost was not the shape. It was the whole numbers. A side of no length, or a negative one, satisfies it too, and is no triangle at all. Here is a list of six of them, written down long ago. Three, four, five. Five, twelve, thirteen. Eight, fifteen, seventeen. Seven, twenty-four, twenty-five. Twelve, thirty-five, thirty-seven. And fifteen, thirty-six, thirty-nine.

Every one of the six checks out, which is worth doing rather than assuming. But one of the six is not like the others. Fifteen, thirty-six, thirty-nine. Divide all three by three and you get five, twelve, thirteen, which is already on the list. So it is not a new triangle. It is one of the others, drawn three times bigger. Six triples. Five triangles. Hold on to that gap, because the whole video is about it.

Now let us take nobody's word for it, and find every triple whose three numbers are twenty or less. Not remember them. Search for them, one combination at a time. There are six. Three, four, five. Five, twelve, thirteen. Six, eight, ten. Eight, fifteen, seventeen. Nine, twelve, fifteen. Twelve, sixteen, twenty. And look at what turns up. Six, eight, ten is three, four, five doubled. Nine, twelve, fifteen is it tripled. Twelve, sixteen, twenty is it quadrupled.

So six triples again, and this time only three triangles. Half of what we found was one triangle in different sizes. That pattern is a machine worth turning the handle on. Take three, four, five and multiply every term by ten. Thirty, forty, fifty. Nine hundred plus sixteen hundred is two thousand five hundred. It works. Multiply by a hundred. Three hundred, four hundred, five hundred. It works again. At this point most of us would say: it always works. And most of us would be right.

But saying it always works, on the evidence of two cases, is a guess with good manners. That has a name: a conjecture. Turning a conjecture into a fact takes something the two examples cannot supply. The wrong way to settle it is to try more numbers. Try a thousand. Try a million. You will be right every time and you will never be finished. The right way is to stop using numbers.

Take three k, four k, five k, where k stands for any whole number at all. Three k squared is nine k squared. Four k squared is sixteen k squared. Add them: twenty-five k squared. And five k squared is also twenty-five k squared. The two sides are the same expression. That is not evidence any more. That is settled, for every k at once, including the ones nobody will ever write down.

And notice that three, four and five did no work in that argument. Run it again with letters where the numbers were. Take any triple a, b, c. Scale all three by k. k a squared is k squared times a squared. k b squared is k squared times b squared. Add them and pull the k squared out in front: k squared, times a squared plus b squared. But a squared plus b squared is c squared, because a, b, c was a triple to begin with. So the total is k squared c squared, which is k c squared.

One argument. Every triple, every whole number. There are infinitely many triples, and we have just proved it. So that question is answered, except it was not the question anyone was really asking. Take three, four, five and scale it by one, two, three, four, five and six. Six triples. Every one of them different. Every one of them genuinely a triple. Now divide each one back down by whatever its terms share. Every single one collapses to three, four, five.

Six triples. One triangle. The scaling machine makes infinitely many triples and never once a new shape. It cannot: multiplying every side by the same number is exactly what leaving the angles alone means. So the interesting question is still standing, and it is: how many different right triangles have whole-number sides? To ask that properly we need a word for a triple that is not a magnified copy of anything.

It is one whose three terms share no factor above one. Those are the primitive ones. Three, four, five shares nothing. Nine, twelve, fifteen shares three, so it is a copy. Careful here: primitive does not mean small. Eight, fifteen, seventeen is primitive, and it is bigger than nine, twelve, fifteen, which is not. And every triple is one or the other. Divide any triple by whatever its three terms share, and what is left is primitive, and still a triple.

Nine, twelve, fifteen divided by three is three, four, five. The scaling proof run backwards. So finding all the primitive ones would be finding all of them. That is the real question, and scaling has not touched it. The answer comes from somewhere that looks like it has nothing to do with triangles. Add up the odd numbers, starting from one, and see what you land on. One is one. One plus three is four. One plus three plus five is nine. Add seven as well and you get sixteen.

One, four, nine, sixteen. Every one of them a square. Add the first n odd numbers and you get n squared, every time. Now write it in two halves. The first n minus one already add to n minus one, squared. So one more odd number takes you from one square to the next. n minus one, squared, plus the nth odd number, equals n squared. Look at that sentence again with a triangle in mind.

Something squared, plus something, equals something else squared. It is nearly the relation we want. The only thing missing is that the middle term be a square too. So here is the whole machine: find an odd number that is also a square. Nine is odd, and nine is three squared. Which odd number is it? Nine is two times five, minus one, so it is the fifth. So the first five odd numbers add to five squared, and the first four add to four squared. Four squared plus nine is five squared.

And nine is three squared. Three, four, five falls straight out. Twenty-five is odd, and it is five squared. It is the thirteenth odd number, because twenty-five is two times thirteen minus one. So twelve squared plus twenty-five is thirteen squared. Five, twelve, thirteen. Feed it forty-nine and out comes seven, twenty-four, twenty-five. Eighty-one gives nine, forty, forty-one. A hundred and twenty-one gives eleven, sixty, sixty-one. A hundred and sixty-nine gives thirteen, eighty-four, eighty-five. Two hundred and twenty-five gives fifteen, one hundred and twelve, one hundred and thirteen.

Seven odd squares, seven triples, and this time seven different triangles. Not one is a copy of another. And every one of them is primitive, for a reason you can see in the numbers: the longer leg and the longest side are always one apart. Anything dividing all three would have to divide the difference between two of them, and that difference is one. Nothing above one divides one. One warning, though. Do not try this on every odd number.

Seven, eleven, fifteen are all odd, and none of them is a square, so none of them gives you anything. Between three and twenty-five, only two odd numbers work out of twelve. And the machine has a blind spot of its own. Every triple it makes has that gap of one. So it can never produce eight, fifteen, seventeen, where the gap is two. Nor twelve, thirty-five, thirty-seven. Nor twenty, twenty-one, twenty-nine, where the gap is eight.

Two of those three were on the list we started with, so that list was never one machine's output. So: infinitely many triples, twice over, for two completely different reasons. One machine copies what you have, forever. The other keeps finding shapes nobody had. And there are still triangles neither of them reaches. The counting is not finished, and that is the good news.

Where this fits

Taken from the notes each video was made from, not from the reading order — these are the ideas this one rests on and the ones that later rest on it.

Builds on

Comes up again in

The book

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