Question

If a and b are distinct integers, prove that a − b is a factor of aⁿ − bⁿ, whenever n is a positive integer. [Hint: write aⁿ = (a − b + b)ⁿ and expaand]

Answer 1

a = a − b + b
So, aⁿ = [a − b +b]ⁿ
= [(a − b) + b]ⁿ
= ⁿC₀ (a − b)ⁿ + ⁿC₁ (a − b)ⁿ⁻¹b¹ + ⁿC₂ (a − b)ⁿ⁻²b² + ....... + ⁿC_n−1 (a − b)bⁿ⁻¹ + ⁿC_n (bⁿ)
⇒ aⁿ − bⁿ = (a − b)_n + ⁿC₁ (a − b)ⁿ⁻¹b + ⁿC₂ (a − b)ⁿ⁻²b² + ....... + ⁿC_n−1 (a − b)bⁿ⁻¹
= (a − b) [(a − b)ⁿ⁻¹ + ⁿC₁ (a − b)ⁿ⁻²b + ⁿC₂ (a − b)ⁿ⁻³b² + ...... + ⁿC_n−1 bⁿ⁻¹]
= (a – b)[aⁿ integer]
⇒ aⁿ – bⁿ is divisible by (a – b)