Baudhāyana-Pythagoras triples, and how to make infinitely many

What to assume they know

• Combining two different squares: a² + b² = c² — the theorem a² + b² = c² and what it says about a right triangle's sides
• Squaring whole numbers up to about 40, and recognising squares on sight
• Expanding (k a)² as k² a², and taking a common factor out of a sum
• Highest common factor of a set of whole numbers, and what having no common factor above 1 means
• That adding the first n odd numbers gives n² — established in earlier work on square numbers
• Writing the nth term of the odd numbers as 2n − 1, and expanding (n − 1)²
• Checking a proposed general statement by substituting a letter, not by testing more cases

What they 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

Where it usually goes wrong

• "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.

Questions to check understanding

• Decide whether a given triple satisfies the relation, showing the three squares
• List all triples with entries below a stated bound, and mark the primitives
• Given a primitive triple, produce a stated number of scaled versions
• Decide whether a given triple is primitive, and if not, reduce it
• Prove that scaling a triple gives a triple, using k
• Derive (n − 1)² + (2n − 1) = n² and use it on a supplied odd square
• Given an odd square, find its position among the odd numbers and write down the triple
• Explain why the odd-square machine never produces a non-primitive triple
• Give a primitive triple that the odd-square machine cannot produce, and say how you know
• True or false, with justification: every triple is primitive or a scaling of one
• Produce rectangles whose sides and diagonals are all whole numbers (Part II, p.53, no.6) — the same content asked as a rectangle rather than a triangle

Examples worth working on the board

Values marked derived are added here; the chapter answers only some of its own questions here.

• Baudhāyana's list (Part II, §2.5, p.48). From Śulba-Sūtra Verse 1.13: (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25), (12, 35, 37) and (15, 36, 39). Six triples. Derived, and worth showing as a sorting exercise: five of them are primitive; (15, 36, 39) is three times (5, 12, 13).
• The chapter's own working list up to 20 (Part II, §2.5, p.48): (3, 4, 5), (6, 8, 10), (9, 12, 15), (12, 16, 20). The chapter is careful — it says the full list contains these, not that it is these — and it uses them to draw out the scaling pattern. Derived, and this is the answer to the Math Talk question that opens the page: the complete list of triples with all three entries 20 or under is those four together with (5, 12, 13) and (8, 15, 17), six in all. Three of the six are primitive: (3, 4, 5), (5, 12, 13), (8, 15, 17). That answers the chapter's later question on Part II p.49 as well.
• The scaling questions (Part II, §2.5, p.48). Is (30, 40, 50) a triple? Is (300, 400, 500)? Derived: yes and yes, ten times and a hundred times the first.
• The conjecture and its check, worked on the page (Part II, §2.5, p.48). Conjecture: (3k, 4k, 5k) is a triple for every positive whole k. Check: (3k)² + (4k)² = 9k² + 16k² = 25k², which is (5k)². The chapter concludes that there are infinitely many triples.
• The general scaling proof, worked on the page (Part II, §2.5, p.49). (ka)² = k²a² and (kb)² = k²b², so the sum is k²a² + k²b², which factors as k²(a² + b²) = k²c² = (kc)². The chapter calls (ka, kb, kc) a scaled version.
• Primitivity (Part II, §2.5, p.49). A triple with no common factor above 1 is primitive. The chapter gives (3, 4, 5) as primitive and (9, 12, 15) as not. Its questions: is (5, 12, 13) primitive; what are the other primitives at 20 or below; produce five scaled versions of each primitive and say whether those are primitive; and, for a non-primitive triple with common factor f, is (a/f, b/f, c/f) a triple — checked on (9, 12, 15), with a justification wanted. Derived: yes, (5, 12, 13) is primitive; scaled versions are not primitive for any k > 1, since k is then a common factor — while k = 1 returns the primitive unchanged, which is the case the chapter's question does not intend; and dividing through by f does give a triple, by the scaling proof run backwards — (9, 12, 15) divided by 3 is (3, 4, 5).
• The odd-numbers identity (Part II, §2.5, pp.49–50). The chapter prints 1 = 1², 1 + 3 = 2², 1 + 3 + 5 = 3², states that adding the first n odd numbers gives n², gives the nth odd number as 2n − 1, and underlines the first (n − 1) odd numbers inside the full sum to reach (n − 1)² + (2n − 1) = n². It notes the same equation drops out of expanding (n − 1)² and adding 2n − 1.
• Machine run 1, worked on the page (Part II, §2.5, p.50). 9 is an odd square, and it is the 5th odd number because 9 = 2 × 5 − 1. So 1 + 3 + 5 + 7 + 9 = 5², which is 4² + 3² = 5². Via the equation with n = 5: (5 − 1)² + 9 = 5².
• Machine run 2, worked on the page (Part II, §2.5, p.50). 25 is an odd square and the 13th odd number, since 25 = 2 × 13 − 1. So the odd numbers up to 25 add to 13², which is 12² + 5² = 13². Via n = 13: (13 − 1)² + 25 = 13². The page sets the two derivations in a two-column tinted box, side by side.
• The next runs, for the exercise (Part II, p.50, Figure it Out no.1, which asks for five more). Derived, with the odd square first and the resulting triple after: 49 is the 25th odd number, giving (7, 24, 25); 81 is the 41st, giving (9, 40, 41); 121 is the 61st, giving (11, 60, 61); 169 is the 85th, giving (13, 84, 85); 225 is the 113th, giving (15, 112, 113). One of these five, (7, 24, 25), is on Baudhāyana's own list at Part II p.48 — the other four are not — and (9, 40, 41) is the triple hiding in one of the chapter's own exercise figures on Part II p.53, the triangle marked 40 and 41.
• The primitivity of everything this machine makes (Part II, p.50, Figure it Out no.2, whose hint observes that one of the two shorter sides always falls one short of the hypotenuse). Derived, and it is a clean argument for video: the larger leg and the hypotenuse are consecutive whole numbers, so any common factor of the triple divides their difference, which is 1. So every triple from this machine is primitive, and the answer to no.2 is no.
• What the machine misses (Part II, p.50, Figure it Out no.3, marked Math Talk). Derived: since the machine only ever produces triples whose larger leg and hypotenuse differ by 1, it cannot reach (8, 15, 17), where the gap is 2, nor (12, 35, 37), nor (20, 21, 29). Two of those three are on Baudhāyana's own list, which makes the point sharply: his list is not the output of this one machine.
• The chapter's own closing claim (Part II, p.53, Figure it Out no.5). A true-or-false item asserting that every triple is either primitive or a scaled version of a primitive one. Derived: true — divide any triple through by the highest common factor of its three terms and what remains is primitive, by the argument the chapter set up on Part II p.49.
• The SUMMARY line (Part II, p.54). It gives (3, 4, 5), (6, 8, 10) and (5, 12, 13) as examples and records that infinitely many can be built.

Figures to have open

• A dot-square diagram for adding odd numbers: an L-shaped shell of dots added to a square of dots to make the next square, with the shell size 2n − 1 (Part II, §2.5, p.49, where the chapter gives the identity in symbols only). This is an addition made here and it is what makes section 10 land, because it shows why the odd numbers build squares rather than asserting it.
• The scaling family tree: primitives on one row with their multiples hanging below (Part II, §2.5, pp.48–49). Standard schematic; the chapter's Part II p.49 questions are exactly this picture.
• Two right triangles at true relative scale, 3-4-5 and 30-40-50, for section 8. Standard schematic.
• The chapter's two-column tinted box for the n = 13 run (Part II, §2.5, p.50), which shows the two routes to the same triple side by side. Redraw as a schematic.
• A table with columns for the odd square, its position n, and the resulting triple, filled for 9 and 25 and blank for 49, 81, 121, 169, 225. Not printed in the chapter; it is the shape the exercise wants.
• No photograph is needed. Checked on the printed pages for Part II pp.48, 49 and 50: §2.5 prints no diagram and no illustration anywhere in its three pages. Its only graphic elements are colour emphasis on some lines of working, and one tinted two-column box on Part II p.50.

Where this sits in the book

• NCERT Ganita Prakash Class 8, Part II, printed Chapter 2, "The Baudhāyana-Pythagoras Theorem", §2.5 "Right–Triangles Having Integer Sidelengths", Part II pp.48–50. The section's own Figure it Out, three items, is at Part II p.50.
• Baudhāyana's list is attributed in the chapter to Śulba-Sūtra Verse 1.13 (Part II p.48).
• Related later items: Part II p.53, Figure it Out no.5 (primitive or scaled) and no.6 (integer rectangles). The triple (9, 40, 41) is the answer to one of the six triangles in Part II p.53 no.2, and (12, 35, 37) to the rhombus in no.3.
• Backward dependency inside the chapter: the theorem itself, §2.4, Part II pp.41–47.
• Forward pointer: §2.6 at Part II pp.50–51 takes the abundance established here and asks the same question one power up.
• The chapter's SUMMARY at Part II p.54 carries three example triples and the infinitude claim.