Chapter 9.4, Problem 34E

To determine

To prove that a factor of $2^{2n+1}+1$ is 3 for all integers $n\ge 1$ by using mathematical induction.

Answer to Problem 34E

The property that a factor of $2^{2n+1}+1$ is 3 for all integers $n\ge 1$ is valid.

Explanation of Solution

Given:

Let $a_n=2^{2n+1}+1$ .

Concept used:

The Principle of Mathematical Induction states that for a statement $P_n$ involving the positive integer n . If

1. $P_1$ is true, and

2. for every positive integer k , the truth of $P_k$ implies the truth of $P_{k+1}$ then the statement $P_n$ must be true for all positive integers n .

Calculation:

First to show that the property is valid for $n=1$ . When $n=1$ , the property is valid because:

  $\begin{array}{c}{2}^{2\left(1\right)+1}+1=2^3+1\\ =8+1\\ =9\end{array}$

And, clearly 3 is a factor of 9.

Let us assume that for $n=k$ , a factor of $a_k=2^{2k+1}+1$ is 3, we need to show that for $n=k+1$ , a factor of $a_{k+1}=2^{2k+1}+1$ is 3.

For $n=k+1$ , the expression will be:

  $\begin{array}{c}{a}_{k+1}=2^{2\left(k+1\right)+1}+1\\ =2^{2k+2+1}+1\end{array}$

Now add and subtract 4 on the LHS of the equation and get,

  $\begin{array}{c}{a}_{k+1}=2^{2k+1}{2}^2+1+4-4\\ =4\left(2^{2k+1}+1\right)-3\end{array}$

By assumption we know, that a factor of $a_k=2^{2k+1}+1$ is 3, and 3 is a factor of 3 (Trivial), therefore 3 will be a factor of $4\left(2^{2k+1}+1\right)-3$ .

Therefore, by combining all the results, we can conclude by the mathematical induction that the property is valid for all $n\ge 1$ .