GANITA PRAKASH · GRADE 8 · PART II · CHAPTER 2
GENERATE — Turn an odd square into a triple
Think first. Use AI purposefully. Verify always.
DISCOVER
Start with a sketch and a prediction.
The difference between consecutive squares is odd. When is that difference itself a square?
Try on paper first. Use the model to test your idea before accepting the explanation.
Understand
Since (n−1) + (2n−1) = , an odd square = 2n−1 yields (n−1,m,n). Use odd integers m ≥ 3 so every side is positive. These generated triples are primitive because n−1 and n are consecutive.
Key Ideas
(n − 1) + (2n − 1) =
(8,15,17) is primitive but not generated: neither leg is 16, one less than its hypotenuse.
Visualise
Generate from an odd square
= 9; n = = 5.
+ =
Triple: (4, 3, 5). The consecutive entries 4 and 5 have gcd 1, so the triple is primitive.
This does not produce (8,15,17): neither leg is one less than 17.
Method
Choose an odd m ≥ 3. Calculate n = . Use legs n−1 and m with hypotenuse n. Verify the equality and do not claim the method generates every primitive triple.
Worked Example
Given
Odd square 49 = .
To Find
The triple generated from this odd square
Property
2n−1 = 49 and (n−1) + 49 = .
Solution
- n = 25 and n−1 = 24.
- + = 576 + 49 = 625 = .
Result
The primitive triple is (24,7,25).
AIMS Method
- A — Attempt independently
Sketch and predict before using a formula. - I — Investigate the idea
Separate given facts from visual assumptions. - M — Make connections
Connect area, length, algebra and geometric constructions. - S — Scrutinise your reasoning
Check each hypothesis and justify the conclusion.
Think first. Use AI purposefully. Verify always.
Focused Practice
Does m = 1 yield a positive-integer triple?
Check & Feedback
Name the known quantities and target, establish the theorem conditions and check your result in the original relationship.
Textbook connection: Ganita Prakash · Grade 8 · Part II · Chapter 2 — The Baudhāyana–Pythagoras Theorem · printed page 50