Chapter 9.5, Problem 19E

Solution

To prove : using mathematical induction to prove that statement $3$ is a factor of $n^3+2n$ holds true for all positive integers.

Given information :

The statement $n^3+2n$ .

Formula used :

Follow the steps to prove that the statement is true for any positive integer.

Anchor step: Prove that any statement $P_n$ is true for $n=1$ .

Inductive hypothesis: Assume $P_n$ is true for $n=k$ .

Inductive Step: Prove that $P_n$ is true for $n=k+1$ .

Proof :

Conjecture:

Take $P_n$ to be statement $n^3+2n$ .

Anchor step:

Prove that $P_n$ is true for $n=1$ .

For $n=1$ ,

  $\begin{array}{l}{1}^1+2\cdot 1\stackrel{?}{=}3\\ 1+2\stackrel{?}{=}3\\ 3=3\end{array}$

Inductive hypothesis:

Assume $P_n$ is true for $n=k$ . Replace $n$ with $k$ in $P_n$ .

  $P_k:k^3+2k$ is divisible by $3$ .

Inductive Step:

Prove that $P_n$ is true for $n=k+1$ . This means it is needed to be proved that $P_{k+1}$ must be true.

To prove $P_{k+1}:{\left(k+1\right)}^3+2\left(k+1\right)$ is also divisible by $3$ .

Proceed with the inductive hypothesis.

  $P_k:k^3+2k$ is divisible by $3$ .

So,

  $k\left(k^2+2\right)$ is divisible by $3$ .

This implies that $k$ is divisible by $3$ or $k^2+2$ is divisible by $3$ .

Case 1: $k$ is divisible by $3$ .

Then,

  $\begin{array}{c}{\left(k+1\right)}^3+2\left(k+1\right)=k^3+3k^2+5k+3\\ =k\left(k^2+3k+5\right)+3\end{array}$

As $k$ is divisible by $3$ , so $k\left(k^2+3k+5\right)+3$ is also divisible by $3$ .

In the way, $k^3+3k^2+5k+3$ is also divisible by $3$ .

Case 2: $k^2+2$ is divisible by $3$ .

Then,

  $\begin{array}{c}{\left(k+1\right)}^3+2\left(k+1\right)=k^3+3k^2+5k+3\\ =\left(k^3+2k\right)+\left(3k^2+3k+3\right)\\ =k\left(k^2+2\right)+3\left(k^2+k+1\right)\end{array}$

As $k\left(k^2+2\right)$ and $3\left(k^2+k+1\right)$ are divisible by $3$ so $k\left(k^2+2\right)+3\left(k^2+k+1\right)$ is also divisible by $3$ .

In the way, $k^3+3k^2+5k+3$ is also divisible by $3$ .

In both cases, the statement is same as the statement $P_{k+1}$ , so $P_{k+1}$ is true.

So, if $P_n$ is true for $n=k$ , then it is also true for $n=k+1$ .

Conclusion:

As $P_n$ is true for $n=1$ and $P_k$ further implies $P_{k+1}$ , this means $P_n$ is true for $n=2$ , $n=3$ and so on.

By mathematical induction, $3$ is a factor of $n^3+2n$ is true for all positive integers $n$ .