If n is an integer, n >= 1, then show that 3^2^n – 1 is divisible by 2^n + 2.

Question

If n is an integer, n \(\geq\)1, then show that \(3^{2^n}\)– 1 is divisible by 2^{n + 2}.

Answer 1

Let T(n) be the statement:

\(3^{2^n}\) – 1 is divisible by 2^{n + 2} 

Basic Step: 

For n = 1, 

\(3^{2^1}\) – 1 = 8 and 2^{n + 2} = 8 

⇒ T (1) is true 

Induction Step: 

Assume T(k) to be true, i.e., 

T(k) = \(3^{2^k}\) – 1 is divisible by 2^{k + 2} 

= \(3^{2^k}\) – 1 = m. 2^{k + 2} when m∈N ...(i) 

= \(3^{2^k}\)= m. 2^{k + 2} + 1 

Now we need to prove that T(k + 1) holds true. 

∴ \(3^{2^{k+1}}\) –1 = \(3^{2^k.2}\) – 1 

= (m . 2^{k + 2} + 1)² – 1 (using (i)) 

= m² (2^k+2)² + 2m . 2^k+2 + 1 – 1 = 2^k+2 (m² . 2^k+2 + 2m) 

⇒ T(k + 1) = \(3^{2^{k+1}}\) – 1 is divisible by 2^k+2, whenever T(k) holds. 

Thus \(3^{2^n}\) – 1 is divisible by 2^{n + 2} for all integers n \(\geq\) 1.