If $adI'm working on a long excersice, and I'm stuck in one particular step...
Given some fixed integers $a,b,c,d \in \mathbb{N}$ such that $ad<bc$, which value $k\in\mathbb{N}$ should I take to ensure the existence of an integer $n\in\mathbb{N}$ that verifies the following? 
$$ abd^2 k^2 < n^2 < b^2cd k^2$$
In other words, I want to construct a $k$ dependant at most of $a,b,c$ and $d$ such that the above interval sandwich any perfect square.
I verify it computationally with several examples, and it seems to work. Even more, the existence of such $n$  makes sense because these two facts:
First, you can see that the  factor $(bdk)^2$ behaves like an homotethic transformation over the interval $(ad,bc)$, and therefore you can strech it as much as you want choosing large values of $k$.
On the other hand, the frequency in which perfect squares occur is quite often, because that sequence is just the sum of consecutive odd integers:
$ n^2= \sum_{j=1}^n (2j-1)$
Any ideas please?

Edit #1: I want to thank all the users who answered and commented. You gave me a deeper insight of this problem, but unfortunately I have omitted an important detail that makes me unable to use the solutions proposed so far. The thing is, this problem is the only incomplete step I left in my own version of proving the existence of the real numbers. Therefor, I can't make any reference to its properties, or to square roots. This is the reason I stated everything in terms of integers. I need the construction to stay within rational numbers at most.
 A: Expanding the comments I made earlier:

If $x, y$ and $z$ are real numbers such that $0 < z < y - x$, then there exists an integer $n$ such that $$x < nz < y.$$

PROOF: If an integer $n$ is such that $nz \geqslant y$, then $(n - 1)z \geqslant y - z > x$. It follows that if $n$ is the least integer such that $nz > x$ - a unique such integer does exist, by standard properties of the reals and the integers - we cannot have $nz \geqslant y$. The conclusion follows. $\square$
We are given $ad < bc$, therefore $\sqrt{ad} < \sqrt{bc}$, therefore $\sqrt{bd}\cdot\sqrt{ad} < \sqrt{bd}\cdot\sqrt{bc}$.
Let $k$ be any integer such that $$\frac{1}{k} < \sqrt{bd}\cdot(\sqrt{bc} - \sqrt{ad}).$$
Again, such an integer does exist, by standard properties of the reals and the integers.
By the proposition highlighted above, there exists an integer $n$ such that
$$\sqrt{bd}\cdot\sqrt{ad} < \frac{n}{k} < \sqrt{bd}\cdot\sqrt{bc}.$$
This condition simplifies to $abd^2k^2 < n^2 < b^2cdk^2$, as required.
To get a 'simple' value of $k$ that suffices, rewrite its defining condition as
$$k > \frac{\sqrt{bc} + \sqrt{ad}}{\sqrt{bd}\cdot(bc - ad)}.$$
Because $ad < bc$, and $a, b, c, d$ are integers, we have $bc - ad \geqslant 1$, so a sufficient condition for $k$ is
$$k > \frac{2\sqrt{bc}}{\sqrt{bd}} = 2\sqrt{\frac{c}{d}}.$$
We must assume $d \geqslant 1$ (or the problem has no solution), so it suffices to take $k > 2\sqrt{c}$. But $ad < bc$, therefore $c \geqslant 1$, therefore $c \geqslant \sqrt{c}$. So we may take $k$ to be any value greater than $2c$, such as $2c + 1$.
Edit:
The strict inequalities involving $k$ at the end could actually be replaced by non-strict ones: $k \geqslant 2\sqrt{c/d}$, $k \geqslant 2\sqrt{c}$, $k \geqslant 2c$.

Here is a proof using only the properties of the natural numbers.
(The small use made of rational arithmetic can probably be easily avoided.)
Let positive integers $m, r$ be such that there is no positive integer $n$ satisfying the inequalities $$r(m - 1) < n^2 < rm.$$ Then there exists a positive integer $q$ such that $$(q - 1)^2 \leqslant r(m - 1) < rm \leqslant q^2.$$
Consequently, $r \leqslant 2q - 1$, and
$$\frac{m - 1}{m} \geqslant \frac{(q - 1)^2}{q^2}.$$
We can rewrite the latter inequality as $$m \geqslant \frac{q^2}{2q - 1},$$
which then gives $$4m \geqslant 2q + 1 +  \frac{1}{2q - 1} > r + 2.$$
This proves:

If positive integers $m, r$ satisfy the inequality $r \geqslant 4m - 2$, then there exists a positive integer $n$ satisfying the inequalities: $$r(m - 1) < n^2 < rm.$$

In this result, take $m = bc$, and $r = 4bc^2d = 4cdm \geqslant 4m$.
The conclusion is that there exists a positive integer $n$ such that
$$4bc^2d(bc - 1) < n^2 < 4b^2c^3d.$$
Write $k = 2c$. Then we have:
$$abd^2k^2 = 4abc^2d^2 = 4bc^2d(ad) \leqslant
4bc^2d(bc - 1) < n^2 < 4b^2c^3d = b^2cdk^2,$$
as required.
A: When we subtract the square root of two integer and we get $\leq 1$ this tells us that at most, these integers are consecutive squares, so to ensure that there is a square between $a d b d k^2$ and $b c b d k^2$ and since the latter is bigger because $a d < b c$ we need to make sure that $\sqrt{b c b d k^2} - \sqrt{a d b d k^2} >1$ which is just $(\sqrt{b^2 c d } k - \sqrt{a b d^2 }k)  >1$  which is just $(\sqrt{c d} b - \sqrt{a b}d )k >1$ divide by $\sqrt{c d} b - \sqrt{a b}d >0$ which we know its true then we get that $k > \frac{1}{b\sqrt{c d}  - d\sqrt{a b} }$
For example : $a=5,b=4,c=9,d=7$ we get from the inequality that $k > \frac{1}{b\sqrt{c d}  - d\sqrt{a b} }= \frac{1}{4\sqrt{9*7}-7\sqrt{5*4}} \approx 2.252$
Because $k$ is integer then $k \geq 3$, will give infinitely many solutions for these fixed $a,b,c,d$.
For explicit formula, then how about $ k = \lceil \frac{1}{b\sqrt{c d}  - d\sqrt{a b} } \rceil+1$, adding one because the inequality have the notion $>$ and not $\geq$.
Generalized explicit formula is $ k = \lceil \frac{1}{b\sqrt{c d}  - d\sqrt{a b} } \rceil+m$ such that $m \in \mathbb{N} \geq 1$.
