Chapter 6.3 — Exercise 6.2 — Prime Factorisation Method

Pythagorean Triplets

Three positive integers a, b, c form a Pythagorean triplet if a² + b² = c². The classic example: 3² + 4² = 9 + 16 = 25 = 5², so (3, 4, 5) is a triplet. Another: 5² + 12² = 25 + 144 = 169 = 13², giving (5, 12, 13).

A triplet with no common factor among all three numbers is called a primitive triplet. Multiplying every number in a triplet by the same factor always produces another valid triplet: (3,4,5) scaled by 2, 3, and 10 gives (6,8,10), (9,12,15), and (30,40,50) — all satisfy a²+b²=c² just as the original does, since squaring the common scale factor affects every term identically.

The Square Root, Defined

If a square number is written as the product of two equal factors, that repeated factor is the square root of the number, written √x. Since 16 = 4×4, √16 = 4; since 121 = 11×11, √121 = 11.

The Prime Factorisation Method, Step by Step

Break the number down into prime factors, group identical factors in pairs, and take one factor from each pair.

• 784 = 2×2×2×2×7×7 = (2×2×7)² → √784 = 28.
• 4096 = 2 multiplied by itself twelve times = (2⁶)² → √4096 = 2⁶ = 64.
• 7056 = 2×2×2×2×3×3×7×7 = (2×2×3×7)² → √7056 = 84.

Whenever every prime factor pairs up completely with no leftovers, the number is a perfect square, and its root is simply the product of one factor from each pair.

What to Do When a Factor Is Left Unpaired

3645 = 3×3×3×3×3×5 — five factors of 3 (an odd count, so one 3 is left unpaired) and a single unpaired 5. Multiplying by 5 pairs up the last factor and clears the imbalance, making the result a perfect square. Similarly, 2400 = 2×2×2×2×2×3×5×5 has an unpaired 2 and an unpaired 3; multiplying by 2×3 = 6 gives a perfect square, whose root is 2×2×2×3×5 = 120. The reverse operation also works: 7776 = 2×2×2×2×2×3×3×3×3×3 has an unpaired 2 and an unpaired 3, so dividing by 2×3 = 6 removes exactly the unpaired factors and leaves a perfect square behind.

Word Problems Built Around x² = Given Number

Several questions describe a situation where an unknown x appears twice as a factor of itself, translating directly into x² = (given number):

• 1521 trees planted with as many rows as trees per row: x² = 1521 = 3×3×13×13, so x = 3×13 = 39 rows of 39 trees each.
• ₹2601 collected in fees, with the fee per student equal to the number of students: x² = 2601 = 3×3×17×17, so x = 3×17 = 51 students.
• 7921 soldiers arranged with as many soldiers per row as rows: x² = 7921 = 89×89, so x = 89 rows.

A related but slightly different setup: the product of two numbers is 1296, and one number is 16 times the other. Writing the smaller number as x, then 16x × x = 1296, so 16x² = 1296, giving x² = 81 and x = 9 — so the numbers are 9 and 144.

A Two-Stage Word Problem

A square field of area 5184 m² has side √5184 = 72 m (found via prime factorisation: 5184 = 2⁶×3⁴, giving 2³×3² = 72), so its perimeter is 4×72 = 288 m. A rectangle with the same perimeter and length twice its breadth: if breadth = y, then 2(2y+y) = 288, so 6y = 288, giving y = 48 m (breadth) and 2y = 96 m (length). The rectangle's area is 96×48 = 4608 sq. m — noticeably different from the square's own area, even though the two shapes share a perimeter.

Why Pythagorean Triplets Scale So Cleanly

Scaling every number in a triplet by the same factor k always produces another valid triplet, and it's worth seeing why algebraically, not just by example. If a² + b² = c², multiplying every term by k² gives (ka)² + (kb)² = (kc)² — the equation still balances exactly, since k² factors out of every term identically. This is precisely why (3,4,5) scaled by 2, 3, and 10 all remain valid triplets: the underlying equation a²+b²=c² is unaffected by uniformly scaling every length involved, which is also the reason similar right triangles of any size all satisfy the same relationship.

Reading a Word Problem for Its Hidden x²

Several word problems in this exercise share a disguised structure: some quantity is described as being equal to another quantity that's multiplied by itself, even though the word "square" or "x²" never appears in the question. "As many trees in a row as there are rows," "fee per student equal to the number of students," and "as many soldiers per row as rows" are all the same sentence pattern wearing different words — each one directly translates to (unknown)² = (given total). Learning to recognise this pattern in unfamiliar wording, rather than needing the word "square" spelled out, is the real transferable skill these problems are teaching.

Unpaired Factors and Wrong Equations

• Forgetting to pair factors before concluding a number isn't a perfect square. Always finish the full prime factorisation first — an apparent leftover factor partway through might still pair up later.
• Multiplying by the wrong combination of unpaired factors. If more than one prime is left unpaired, the smallest number that fixes all of them is the product of each missing factor, not just one of them.
• Setting up the wrong equation in word problems. "As many rows as trees per row" means rows = trees-per-row, giving x², not x times a different unknown — read the condition carefully before writing the equation.

The Square-and-Rectangle Problem, Revisited

The final worked problem is worth a second look because it chains two different topics together: a square root finds the square's side, and then the resulting perimeter becomes the input to a completely separate ratio-based problem (the 2:1 length-to-breadth rectangle). Neither half is difficult alone, but recognising that the square root's job ends once the perimeter is found — and that everything after that is ordinary linear-equation reasoning about the rectangle — is what keeps the two stages from blurring together into a single, more confusing calculation than either one needs to be on its own — treat the square root as finished business the moment the perimeter is in hand, and start the rectangle problem fresh from there, as its own self-contained calculation with a clean numeric starting point of its own.

Where Factorisation Starts to Struggle

Prime factorisation works well for numbers with small, manageable prime factors, but it gets unwieldy for very large numbers — which is exactly the gap Exercise 6.3 fills with a method that doesn't require factorising at all. For the perfect-square recognition skills this exercise builds on, revisit Exercise 6.1.