# Problem #3

**Question**: say whether the following is true or false and support your answer by a proof: For any integer $n$, the number $n^2 + n + 1$ is odd.

**Answer**: it's true.

**Claim**: $(\forall n \in {Z})[n^2 + n + 1$ is odd]

**Proof**: let's prove it directly. By the definition of odd number, it can be presented as $k+1$, where $k$ is an even number. In our case $k=n^2+n$, so if $k$ is always even, $k+1=n^2+n+1$ is always odd.

If $k=n^2+n$, then, by algebra, $k=n(n+1)$.

By the properties of even numbers, the product of two integers is even, when one of factors is even.

Suppose $n$ is even, then $k=n(n+1)$ is even.

Suppose $n$ is odd, then $n+1$ is even and $k=n(n+1)$ is even.

Thus, $k=n(n+1)=n^2+n$ is always even, therefore $k+1=n^2+n+1$ is always odd and for any integer $n$, the number $n^2 + n + 1$ is odd. $\blacksquare$