If $n$ is any integer, which of the following options is an odd number?
(a) $n^{2}-n+2$
(b) $n^{2}+n+2$
(c) $n^{2}+n+5$
(d) $n^{2}+5$
(e) $n^{3}+5$

## Solutions to List 2

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Let's recall that:

- $n$ and $n^{2}$ have the same parity: $(\text{even})^{2}=$ even and (odd $)^{2}=$ odd;
- the sum or difference of numbers with the same parity is an even number: (even $\pm$ even=even and odd $\pm$ odd=even).

Solution 1: Observe that $n^{2}+n$ and $n^{2}-n$ are the sum and difference of two numbers that always have the same parity, so these numbers will always be even. Therefore, $n^{2}+n+5=\left(n^{2}+n\right)+5$ is the sum of an even number and an odd number, so it will always be odd for any integer value of $n$. The correct option is (c).

Solution 2: Observe that $n^{2}+n=n(n+1)$ and $n^{2}-n=n(n-1)$ are products of two consecutive numbers, so these numbers are even. Therefore, $n^{2}+n+5=\left(n^{2}+n\right)+5$ is the sum of an even number and an odd number, so this number is odd for any integer value of $n$.

Observe that (a) and (b) are always even numbers for any value of $n$, while (d) and (e) can be even or odd, depending on whether $n$ is odd or even.