Is $\frac{4n^2+4n+1}{8}$ an integer for any $n\in \mathbb{N}$?

Question

I've been thinking the following: $8|4n^2$ for some $n$ and $8|4n$ for some $n$, which would imply that there are $q_1,q_2\in \mathbb{Z}$ such that $4n^2=8q_1$ and $4n=8q_2$, the only solution for $q_1$ would be $0$, but there are plenty of non-zero solutions to $q_2$. Here's one of my problems:

• One of them has only $0$ as a solution and the other various solutions. I'm not sure if there is a solution for $4n^2+4n=8q_3$.

Which I guess that would help me in my final quest, knowing if $\frac{4n^2+4n+1}{8}$ has integer solutions such that $\frac{4n^2+4n+1}{8}$ is also an integer.

Another fairly naïve heuristic I used was the following:

• Looking at $\frac{4n^2+4n+1}{8}$ we see that $4n^2$ is even, $4n$ is even and $8$ is even. So independently of the $n$ chosen, $4n^2+4n+1$ is going to be odd because of the sum with $1$. And taking $8q$ for $q\in \mathbb{N}$, it's always going to be even, so I'd have to have $4n^2+4n+1=8q$ and I've just argued that one side is always odd and the other is always even. I guess this means that the answer to my question is negative.

I am still not sure if this guarantees what I'm looking for.

Answer 1

Hint: $4n^2+4n+1$ is always odd.

Answer 2

HINT:

$$4n^2+4n=8\frac{n(n+1)}2\equiv0\pmod8$$

Answer 3

a very basic thing to note is that the denominator is divisible by 2 while the numerator is an odd square. hence 2 doesn't divide the denominator as n is a natural number. so the given expression cannot possibly be an integer.