Prove that 32n+2 – 8n – 9 is divisible by 64 for any positive inte

1.	Prove that 32n+2 – 8n – 9 is divisible by 64 for any positive integer n.
Answer» Let T(n) be the statement: 3²ⁿ⁺² – 8n – 9 is divisible by 64. Basic Step: For n =1, 3^2×1+2 – 8 × 1 – 9 = 81 – 17 = 64 which is divisible by 64. ⇒ T(1) holds. Induction Step: Let T(k), k∈N hold, i.e., 3^2k+2 – 8k – 9 is divisible by 64. Then, T(k + 1) = 3^2(k+1) ^{+ 2} – 8(k + 1) – 9 = 3². 3^2k+2 – 8k – 17 = 9 (3^2k+2 – 8k – 9) + 64k + 64 = 9. T (k) + 64 (k + 1) ⇒ T(k + 1) is divisible by 64, whenever T(k) is divisible by 64. ⇒ T(n) is true for every natural number n.

Prove that 32n+2 – 8n – 9 is divisible by 64 for any positive integer n.

Answer»

Let T(n) be the statement: 3²ⁿ⁺² – 8n – 9 is divisible by 64.

Basic Step:

For n =1, 3^2×1+2 – 8 × 1 – 9 = 81 – 17 = 64 which is divisible by 64.

⇒ T(1) holds.

Induction Step:

Let T(k), k∈N hold, i.e.,

3^2k+2 – 8k – 9 is divisible by 64.

Then, T(k + 1) = 3^2(k+1) ^{+ 2} – 8(k + 1) – 9 = 3². 3^2k+2 – 8k – 17

= 9 (3^2k+2 – 8k – 9) + 64k + 64 = 9.

T (k) + 64 (k + 1)

⇒ T(k + 1) is divisible by 64, whenever T(k) is divisible by 64.

⇒ T(n) is true for every natural number n.