$n^2 + 7n + 1$ is odd Prove that for any integer $n$, the integer $n^2 + 7n + 1$ is odd.
I have  $n=2k+1$ for some $k\in Z$
I really do not how to do this problem. any help in understanding would be greatly appreciated.
 A: Hint: You should be able to fill in the the following blanks and justify each claim made. We start by noting that $$n^2+7n=n(n+7).$$ If $n$ is an integer, then either $n$ or $n+7$ must be _____ and the other must be _____. Hence, $n(n+7)$ must be _____, and so....
A: There are only two scenarios here: $n$ is odd or $n$ is even. That alone is enough to determine whether $n^2 + 7n + 1$ is odd or even.


*

*If $n$ is odd, then so is $n^2$, as well as $7n$. Then $n^2 + 7n$ is even, because two odd numbers add up to an even number. But then there's the $+ 1$, which makes the number odd again.

*If $n$ is even, then so is $n^2$, as well as $7n$. Two even numbers also add up to an even number. But again we have the $+ 1$, which then makes the number odd.


Try it with a few specific odd and even values of $n$. I'll do two for you:


*

*$n = 1$, then we have $1 + 7 + 1 = 9$, which is odd.

*$n = 2$, then we have $4 + 14 + 1 = 19$, which is also odd.

A: $n^2 + 7n + 1= n(n+1)+6n+1$
$n(n+1)=2\binom{n}{2}$ and so is even.
$6n+1=2(3n)+1$ and so is odd.
So the sum is odd.
A: You can distinguish two cases: $n=2k$ or $n=2k+1$ ($n$ even or odd, respectively) and substituting in $n^2+7n+1$ you should be able to rewrite the result as $2l+1$ for a certain $l$ in both cases.
A: Hint: if $n$ is odd, then is $n^2$ odd or even? what about $7n$? Ask the same questions if $n$ is even.
A: Option 1: plug in $n=2k+1$ into the form of the number you have, expand and show you get something of the form $2\cdot (some integer)+1$. Repeat by plugging in $n=2k$. Option 2: Do you know that the sum of three odd numbers is odd? Do you know that the square of an odd number is odd? Do you know that the product of two odd numbers is odd? What about even numbers? 
A: $$n^2+7n+1=n(n+7)+1.$$
Note that:


*

*if $n$ is even, $n+7$ is odd.

*if $n$ is odd, $n+7$ is even.


Thus $n(n+7)$ is the product of an odd and an even term, which is even. Thus $n(n+7)+1$ is an even term plus $1$, which is odd. 
