Chapter 9.4, Problem 32E

To determine

To proof: The expression $n^4-n+4$ is a factor of 2 with help of mathematical induction for integers $n\ge 1$ .

Explanation of Solution

Given information:

The expression $n^4-n+4$ .

Formula used:

The steps to prove a statement by mathematical induction are as follows:

Step 1. Show that the statement is true for $n=1$ .

Step 2. Assume that the statement is true for $n=k$ .

Step 3. Prove that the statement is true for $n=k+1$ with help of step 2.

Proof:

Consider expression $n^4-n+4$ .

Denote the expression by $S_n$ .

Recall that the steps to prove a statement by mathematical induction are as follows:

Step 1. Show that the statement is true for $n=1$ .

Step 2. Assume that the statement is true for $n=k$ .

Step 3. Prove that the statement is true for $n=k+1$ with help of step 2.

Substitute $n=1$ in the expression,

  $\begin{array}{c}{1}^4-1+4=1-1+4\\ =4\end{array}$

The above equation holds true as 2 is a factor of 4.

Therefore, the statement is true for $n=1$ so $S_1$ is true.

Now, assume that the statement if true for $n=k$ so, $k^4-k+4$ is a factor of 2.

Let $k^4-k+4=2a$ where a is any number.

Therefore, the statement holds true for $n=k$ so $S_k$ is true.

Now, prove it for $n=k+1$ . Write the expression for $n=k+1$ .

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

Substitute the value of $S_k$ ,

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

Where $\lambda =\left(a+2k^2+2k^3+2k\right)$

Therefore, it is concluded that 2 is a factor of ${\left(k+1\right)}^4-\left(k+1\right)+4$ where $\left(a+2k^2+2k^3+2k\right)$ is any polynomial.

Hence, by the principle of mathematical induction the statement $n^4-n+4$ is a factor of 2 is true for all integers $n\ge 1$ .