Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 1 Exercise 1.11

Solution:

### Part a

If $n^2+5n+1$ is even, then

\begin{equation}\label{1-11-1}

(n+1)^2+5(n+1)+1=(n^2+5n+1)+2n+6

\end{equation} is also even. Therefore $P_{n+1}$ is true whenever $P_n$ is true.

### Part b

There is no integer $n$ such that $n^2+5n+1$ is even. It can be seen easily by looking at $P_1:1^2+5\cdot 1+1$ is even which is clearly wrong. In factor, using \eqref{1-11-1}, one can show that $n^2+5n+1$ is odd by induction. You just need to show it like Part a.