Difference of Squares Identity — Factor 272 to Find One Clean Pair | Math Olympiad

Difference of squares identity, exponential Diophantine, olympiad number theory — rewrite as a²−b² = 272, factor 272 into all pairs, and only one pair yields powers of 6 and 2. 36^x − 4^y = 272. The insight is recognising 36 = 6² and 4 = 2². Rewrite: 6^(2x) − 2^(2y) = 272. Set a = 6^x and b = 2^y. The equation becomes a² − b² = 272. Apply the identity (a−b)(a+b) = 272 instantly. Now list all factor pairs of 272: (1,272), (2,136), (4,68), (8,34), (16,17). For each pair (p,q) , solve a−b = p and a+b = q to get a = (p+q)/2 and b = (q−p)/2. Both are integers only when p and q have the same parity. Check each: (1,272): odd, odd → a = 136.5, rejected (2,136): even, even → a = 69, b = 67. But 69 and 67 are not powers of 6 and 2 (4,68): even, even → a = 36, b = 32. Check: 36 = 6² ✓, 32 = 2⁵ ✓ (8,34): even, even → a = 21, b = 13. But 21 and 13 are not powers of 6 and 2 (16,17): odd, odd → a = 16.5, rejected Only one pair survives: a = 36 = 6² and b = 32 = 2⁵. So x = 2 and y = 5. The factorization of 272 is the entire problem. Once you list all pairs and check which consist of actual powers, the answer falls out. This is a complete proof — no other solution exists because all other factor pairs fail the power test. This factor-and-check technique applies to any exponential Diophantine a^(2x) − b^(2y) = k. Rewrite as a perfect square difference, factor k, test each pair. Completeness comes from exhausting all factorizations. 🔔 Daily olympiad problems — algebra, number theory, geometry, combinatorics. #OlympiadMath #DifferenceOfSquares #ExponentialDiophantine #FactorIdentity #NumberTheory #CompetitionMath #AMCPrep #AIМЕPrep #MathOlympiad #PowersOfIntegers #UniqueSolution #OlympiadNumberTheory