Question

Prove that 5 is a factor of${\mathbf{2}}^{\mathbf{4}\mathbf{n}\mathbf{-}\mathbf{2}}\mathbf{+}\mathbf{1}$for every positive integer n.

Short answer

5 is a factor of $2^{4n-2}+1$for every positive integer n.

Step-by-step solution

Step 1

Consider the following expression,

$P\left(n\right)=2^{4n-2}+1$

Step 2

Substitute 1 for n in the equation $P\left(n\right)=2^{4n-2}+1$as follows;

$$\begin{gathered} P\left(1\right)=2^{4\left(1\right)-2}+1 \\ =2^2+1 \\ =5 \end{gathered}$$

Since, 5 is divisible by 5 this implies that at n = 1 , it is true.

Substitute 2 for n in the equation $P\left(n\right)=2^{4n-2}+1$as follows;

$$\begin{gathered} P\left(2\right)=2^{4\left(2\right)-2}+1 \\ =2^6+1 \\ =65 \end{gathered}$$

Since, 65 is divisible by 5 this implies that at n = 2 , it is true.

Step 3

Substitute k for n in the equation $P\left(n\right)=2^{4n-2}+1$as follows;

$P\left(n\right)=2^{4n-2}+1$

Suppose thatP(K) is divisible by 5m where m is an integer.

So,

$$\begin{gathered} 2^{4n-2}+1=5m \\ {2}^{4n-2}=5m-1 \end{gathered}$$

Step 4

Substitute k+1 for n in the equation $P\left(n\right)=2^{4n-2}+1$ as follows;

localid="1660545718465" $$\begin{gathered} P(k+1)=2^{4(k+1)-2}+1 \\ =2^{4k+4-2}+1 \\ =2^{4k-2}·2^4+1 \end{gathered}$$

Substitute $2^{4k-2}=5m-1$for in the equation localid="1660545727038" $P(k+1)=2^{4k-2}·2^4+1$as follows;

localid="1660545785111" role="math" $$\begin{gathered} P(k+1)=(5m-1)·2^4+1 \\ =5m·2^4-1·2^4+1 \\ =5m·2^4-16+1 \\ =5(m·2^4-3) \end{gathered}$$

Hence, $P(k+1)$is divisible by 5 if $P\left(k\right)$is divisible by 5.