The difference of two squares

Pattern 1. Look at these:

• 2(2² + 1²) = 3² + 1²
• 2(3² + 1²) = 4² + 2²
• 2(5² + 3²) = 8² + 2²
• 2(6² + 5²) = 11² + 1²

Twice a sum of two squares is again a sum of two squares.

Which two? Look at 2(5² + 6²) = (6 + 5)² + (6 − 5)²: the sum and the difference of the original pair.

The two identities you've just met explain it in two lines. Add them: (a + b)² + (a − b)² = (a² + 2ab + b²) + (a² − 2ab + b²). The +2ab and −2ab cancel, leaving 2a² + 2b². So 2(a² + b²) = (a + b)² + (a − b)², always.

Pattern 2. Now these:

• 9 × 9 − 1 × 1 = 10 × 8
• 8 × 8 − 6 × 6 = 14 × 2
• 7 × 7 − 2 × 2 = 9 × 5
• 10 × 10 − 4 × 4 = 14 × 6

Each time the answer is (sum) × (difference).

The guess is a² − b² = (a + b)(a − b). Check by distributing: (a + b)(a − b) = a² − ab + ba − b², and since ab and −ab cancel, we're left with a² − b². True.

Identity 1C: (a + b)(a − b) = a² − b². It turns an awkward product into a difference of two squares, which is often far easier:

• 98 × 102 = (100 − 2)(100 + 2) = 100² − 2² = 10000 − 4 = 9996;
• 45 × 55 = (50 − 5)(50 + 5) = 2500 − 25 = 2475.

It also tells you which product is bigger without working either out: 16 × 24 = 20² − 4² beats 14 × 26 = 20² − 6², because less is taken away.

Sridharacharya, around 750 CE, saw that it can be run backwards to square numbers. Rearrange it as a² = (a + b)(a − b) + b², and choose b to make the product easy.

For 31², take b = 1: 31² = 32 × 30 + 1² = 960 + 1 = 961. For 197², take b = 3 to land on a round 200: 197² = 200 × 194 + 3² = 38800 + 9 = 38809. A hard square becomes an easy product plus a tiny square.