Figure it Out — Generating Baudhāyana Triples — The Baudhāyana–Pythagoras Theorem

1. 14 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 50

   Find 5 more Baudhayana triples using the odd-square idea — that if the nth odd number 2n − 1 is itself a square, then (n − 1)² + (2n − 1) = n² gives a triple.

   Hint. Pick an odd number m, square it, then work out which odd number m² is.

   The recipe, stated once. Choose an odd number m and square it. Since m² is odd, it is the nth odd number for some n, and 2n − 1 = m² gives n = (m² + 1) ÷ 2. The identity (n − 1)² + (2n − 1) = n² then reads m² + (n − 1)² = n², so (m, n − 1, n) is a Baudhayana triple. The book has already done m = 3 and m = 5; carry on from m = 7.

   m = 7 → m² = 49, n = (49 + 1) ÷ 2 = 25 → (7, 24, 25) Check: 49 + 576 = 625 = 25² ✓

   m = 9 → m² = 81, n = (81 + 1) ÷ 2 = 41 → (9, 40, 41) Check: 81 + 1600 = 1681 = 41² ✓

   m = 11 → m² = 121, n = (121 + 1) ÷ 2 = 61 → (11, 60, 61) Check: 121 + 3600 = 3721 = 61² ✓

   m = 13 → m² = 169, n = (169 + 1) ÷ 2 = 85 → (13, 84, 85) Check: 169 + 7056 = 7225 = 85² ✓

   m = 15 → m² = 225, n = (225 + 1) ÷ 2 = 113 → (15, 112, 113) Check: 225 + 12544 = 12769 = 113² ✓

   Why it never fails. Every odd square is an odd number, so it always occupies some position in the list of odd numbers, and n always comes out a whole number because m² + 1 is even. Since there are infinitely many odd numbers, the recipe produces infinitely many triples. Notice also that the two larger numbers always differ by 1 — 24 and 25, 40 and 41, 60 and 61 — which is the clue the next question turns on.

   ✦ (7, 24, 25), (9, 40, 41), (11, 60, 61), (13, 84, 85) and (15, 112, 113).

2. 23 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 50

   Does this method yield non-primitive Baudhayana triples? [Hint: observe that among the triples generated, one of the smaller sidelengths is one less than the hypotenuse.]

   Hint. If a number divides two numbers, it also divides their difference.

   Spot the pattern first. Every triple this method produces has the form (m, n − 1, n): the larger leg and the hypotenuse are consecutive integers. Look back at the list — 24 and 25, 40 and 41, 60 and 61, 84 and 85, 112 and 113. Their difference is always 1.

   Turn that into an argument. Suppose some whole number d divides all three members of such a triple. Then d divides n − 1 and d divides n. A number that divides two numbers must also divide their difference, so d divides n − (n − 1) = 1. The only positive whole number dividing 1 is 1 itself, so d = 1.

   Conclusion. No common factor greater than 1 can exist, and therefore every triple produced by this method is primitive. The method cannot yield a non-primitive triple at all.

   A quick confirmation. Take (9, 40, 41): 9 is odd and 40 is even, so 2 is out; 41 is prime and does not divide 9 or 40; nothing survives. It is primitive, as promised.

   ✦ No. In every triple this method gives, the larger leg and the hypotenuse differ by 1, so any common factor would have to divide 1 — hence all such triples are primitive.

3. 33 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 50

   Are there primitive triples that cannot be obtained through this method? If yes, give examples.

   Hint. Use the fact just established — this method always produces a hypotenuse one more than a leg. So look for a primitive triple where that fails.

   Where to look. The previous question showed the method always produces triples in which the hypotenuse is exactly one more than one of the legs. So any primitive triple in which no leg is one less than the hypotenuse is out of the method's reach.

   A clear example — (8, 15, 17). It is a genuine triple: 64 + 225 = 289 = 17² ✓ It is primitive: 8, 15 and 17 share no factor above 1. But 17 − 15 = 2 and 17 − 8 = 9, and neither difference is 1. So (8, 15, 17) cannot come from this method.

   More examples, all primitive and all with the wrong differences: · (20, 21, 29) — differences 29 − 21 = 8 and 29 − 20 = 9. Check: 400 + 441 = 841 = 29² ✓ · (12, 35, 37) — differences 2 and 25. Check: 144 + 1225 = 1369 = 37² ✓ · (28, 45, 53) — differences 8 and 25. Check: 784 + 2025 = 2809 = 53² ✓ · (33, 56, 65) — differences 9 and 32. Check: 1089 + 3136 = 4225 = 65² ✓

   What this tells us. The odd-square method is a genuine triple factory — it never stops and it only makes primitives — but it is not a complete one. It reaches only the primitives with c − b = 1, and misses infinitely many others. A method that generates every primitive triple exists, but it belongs to a later class.

   ✦ Yes. Since this method always gives a hypotenuse one more than a leg, primitives such as (8, 15, 17), (20, 21, 29), (12, 35, 37), (28, 45, 53) and (33, 56, 65) can never be produced by it.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 8 Mathematics textbook Ganita Prakash Part 2, Reprint 2026-27 (hegp202.pdf), where this is Chapter 2 (pages 33-54) — the ninth chapter of the Class 8 course. Like the rest of Part 2, this PDF carries NO printed answer key, so every answer here was derived from first principles and then independently recomputed in Python before being written — including all the √2-style one-decimal bounds, the six figure triangles, the rhombus side, the odd-square triple generator, and the complete list of Baudhāyana triples with all numbers at most 20. TWO POINTS WHERE THE BOOK'S OWN TEXT NEEDS CARE ARE FLAGGED IN PLACE: (1) on page 48 the book lists four triples with numbers at most 20 and says the list 'contains' them — the complete list has six, since (5, 12, 13) and (8, 15, 17) also qualify and are not multiples of (3, 4, 5); (2) the six right triangles in Figure it Out Q2 on pages 52-53 are labelled only in the printed figure, so each one's right-angle position was read directly off the rendered PDF page before solving, and the reading is stated in the solution so a student can check it against the book.. Questions are referenced from the NCERT textbook for identification.