If $\frac{1}{4} - b^2 n \in \mathbb Z$ where $b \in \mathbb Q$ and $n$ is square-free, then $b \in \frac{1}{2}\mathbb Z$. 
If $\frac{1}{4} - b^2 n \in \mathbb Z$ where $b \in \mathbb Q$ and $n$ is square-free, then $b \in \frac{1}{2}\mathbb Z$.

We must have $b \in \mathbb Q - \mathbb Z$ for otherwise $1-4b^2n \in 4\mathbb Z$ which is impossible.
Say, $b=\frac{c}{d}$.
Where can we go from here?
 A: You have that $\frac{1}{4}-b^2n\in \mathbb{Z}$ hence $4k=1-4b^2n$ and so $$1=4k+4b^2n=4k+(2b)^2n$$ from there we can conclude that $(2b)^2n\in \mathbb{Z}$ so in your notation $d^2|4c^2n$ but $(c,d)=1$ and $d\neq 1$ so $d^2|4n$ you know that $n=\prod _{i=1}^sp_i$ with $p_i$ prime and so the only choice for $d=2.$
A: Starting with $\frac{1}{4} - b^{2} n = z\in \mathbb Z$ and $b=\frac{c}{d}$
$$\frac{c^2}{d^2}n=\frac{1}{4}-z=\frac{1-4z}{4}$$
$$4c^2n=d^2(1-4z)$$ because 4 does not divide $(1-4z) $ has to divide $d^2$ and we get that 2 divides d.
so $b \in \frac{1}{2}\mathbb Q$
lets see if d=2
'd' can not divide 'c' so $d^2$ can not divide $c^2$, also $d^2$ can not divide n because it is square-free so $d^2$ divides 4 and finally we get the desired
$b \in \frac{1}{2}\mathbb Z$ because d =2.
A: We know that $\frac{1}{4} - b^{2} n = z\in \mathbb Z$ , $n\in \mathbb{N}-\{0\}$ square-free and $b=\frac{c}{d}$ where $c\in \mathbb{Z}$, $d\in \mathbb{N}-\{0\}$.
Now we are going to prove that $b \in \frac{1}{2}\mathbb Z$ without using the fact that $(c,d)=1$.
$$\frac{c^2}{d^2}n=\frac{1}{4}-z$$
$$4c^2n=d^2(1-4z)$$
$$\left(2c\right)^2n=d^2(1-4z)$$
$d\ne1$ indeed if $d=1$ then $4c^2n=1-4z$ and it leads to contradiction because $1$ is not a multiple of $4$.
If $p$ is any prime factor of $d$ and $p^r$ is the largest power of $p$ which divides $d$, then $p^{2r}$ divides $(2c)^2n$ and $p^{2r-1}$ divides $(2c)^2$ because $n$ is square-free. It follows that $p^r$ divides $2c$ too.
Hence $d$ divides $2c$, so $\frac{2c}{d}=2b\in \mathbb{Z}$ and it means that $b \in \frac{1}{2}\mathbb Z$.
A: Presumably $\gcd(c,d) = 1$.
$\frac 14 - \frac  {c^2}{d^2}n$ is an integer.  Multiply it by $4$ and
$1- \frac 1{d^2}4c^2n$ is an integer and $d^2|4c^2n$.  As $c$ and $d$ are relatively prime then $d^2|4n$.  If $p$ is an odd prime factor of $d$ then $p\ne \mid 4$ so $p|n$ so $p^2|n$ but $n$ is square free.
So $d$ has no odd prime factors and $d$ is a power of $2$. And $n$ has at most on power of $2$ as a factor.  At most.  So $d^2|8$.  $4^2\not \mid 8$ so $d = 2$  (because, as you pointed out $b\not \in \mathbb Z$ so $d\ne 1$).
So $b = \frac c2;c\in \mathbb Z$.
