MQ42 --- Divisibility

5²ⁿ⁻¹ - 3²ⁿ⁻¹ - 2²ⁿ⁻¹
= 5²ⁿ⁻¹ - ( (5-2)²ⁿ⁻¹ + 2²ⁿ⁻¹ )
= 5²ⁿ⁻¹ - ( 5a - 2²ⁿ⁻¹ + 2²ⁿ⁻¹ ) ... (By binomial theorem , a is an integer)
= 5 ²ⁿ⁻¹ - 5a
≡ 0 (mod 5)
5²ⁿ⁻¹ - 3²ⁿ⁻¹ - 2²ⁿ⁻¹
= (3+2)²ⁿ⁻¹ - 2²ⁿ⁻¹ - 3²ⁿ⁻¹
= 3b + 2²ⁿ⁻¹ - 2²ⁿ⁻¹ - 3²ⁿ⁻¹ ... (b is an integer)
= 3b - 3²ⁿ⁻¹
≡ 0 (mod 3)
5²ⁿ⁻¹ - 3²ⁿ⁻¹ - 2²ⁿ⁻¹
= (2+3)²ⁿ⁻¹ - 3²ⁿ⁻¹ - 2²ⁿ⁻¹
= 2c + 3²ⁿ⁻¹ - 3²ⁿ⁻¹ - 2²ⁿ⁻¹ ... (c is an integer)
= 2c - 2²ⁿ⁻¹
≡ 0 (mod 2)
∴ 5²ⁿ⁻¹ - 3²ⁿ⁻¹ - 2²ⁿ⁻¹ mod (2 * 3 * 5 = 30) ≡ 0 for all n ∈ ℕ since (2 , 3 , 5) = 1.