Question

Prove that 64 is a factor of ${\mathbf{9}}^{\mathbf{n}}\mathbf{-}\mathbf{8}\mathbf{n}\mathbf{-}\mathbf{1}$for every positive integer n.

Short answer

4 is a factor of $7^n-3^n$for every positive integern.

Step-by-step solution

Step 1

Consider the following expression,

$P\left(n\right)=9^n-8n-1$

Step 2

Substitute 1 for n in the equation $P\left(n\right)=9^n-8n-1$as follows;

$$\begin{gathered} P\left(n\right)=9^1-8\left(1\right)-1 \\ =9-9 \\ =0 \end{gathered}$$

Since, 0 is divisible by 64 this implies that at n = 1 , it is true.

Substitute 2 for n in the equation $P\left(n\right)=9^n-8n-1$as follows;

$$\begin{gathered} P\left(2\right)=9^2-8\left(2\right)-1 \\ =81-16-1 \\ =64 \end{gathered}$$

Since, 64 is divisible by 64 this implies that at n = 2 , it is true.

Step 3

Substitute k for n in the equation $P\left(n\right)=9^n-8n-1$as follows;

$P\left(k\right)=9^k-8k-1$

Suppose thatP(K) is divisible by 64m where m is an integer.

So,

$$\begin{gathered} 9^k-8k-1=64m \\ {9}^k=64m+8k+1 \end{gathered}$$

Step 4

Substitute k+1 for n in the equation $P\left(n\right)=9^n-8n-1$ as follows;

$$\begin{gathered} P(k+1)=9^{k+1}-8(k+1)-1 \\ =9^k.9-8k-8-1 \\ =9^k.9-8k-9 \end{gathered}$$

Substitute $9^k=64m+8k+1$for in the equation $P(k+1)=9^k.9-8k-9$as follows;

$P(k+1)=(64m+8k+1).9-8k-9$

Hence, P(k+1) is divisible by 64 ifP(k) is divisible by 64.