Manipulating Integer Inequalities. Suppose a, b, c and d are positive integers satisfying:
\begin{align}
& {{a}^{2}}<2{{b}^{2}}<2{{c}^{2}}<{{d}^{2}} \
&  \
\end{align}
We can take square-roots and re-arrange the terms to get:
$(d-a)>\sqrt{2}(c-b)$
And then, we could write:
${{(d-a)}^{2}}>2{{(c-b)}^{2}}$
The question is, can we get from the first stated inequality to the last using only integer arithmetic?
 A: Here is my attempt at solving your problem. Let us assume that we start with certain values for the four integers $a$, $b$, $c$ and $d$. And these values satisfy the prescribed inequalities: $$a^2 < 2b^2 < 2c^2<d^2$$ It is well known that the square root of $2$ can be approximated with high accuracy by rational numbers, of the form $M/N$. My claim is that we can always find integers $M$ and $N$, in such a way that their ratio is just slightly larger than the square root of $2$, while on the other hand the following inequalities are satisfied:
$$a^2 < (M/N)^2b^2 < (M/N)^2c^2 < d^2$$
For example the values $M = 66922$ and $N=47321$, corresponding to $(M/N)^2 = 2.0000000009$ will work in the vast majority of cases. But the point is that even better approximations can always be obtained. Now the formula above leads us directly to the following relation between integers (in fact they are squares):
$$N^2a^2 < M^2b^2 < M^2c^2 < N^2d^2$$
Taking the square root from each term leads to:
$$Na < Mb < Mc < Nd$$
Re-arranging terms yields another integer relation:
$$N(d-a)>M(c-b)$$
Squaring yields $$(d-a)^2 > (M/N)^2(c-b)^2$$
Since we have chosen $(M/N)^2 > 2$ we find:
$$(d-a)^2 > 2(c-b)^2$$
A: Suppose we have four positive integers $a, b, c, d$ which satisfy:
${{a}^{2}}<2{{b}^{2}}<2{{c}^{2}}<{{d}^{2}}.$
We want to show that:
${{(d-a)}^{2}}>2{{(c-b)}^{2}}.$
But – We are not allowed to introduce anything non-integer into our calculations!
First, I need to go on a short excursion:
Consider the two expressions:
$\begin{align}
  & p=(3d+4c) \\ 
 & q=(2d+3c) \\ 
\end{align}$
It’s easy to see that $2{{q}^{2}}<{{p}^{2}}$:
$2{{q}^{2}}-{{p}^{2}}=(8{{d}^{2}}+24cd+18{{c}^{2}})-(9{{d}^{2}}+24cd+16{{c}^{2}})=2{{c}^{2}}-{{d}^{2}}<0.$
It’s also easy to see that $pc<qd$:
$pc-qd=(3cd+4{{c}^{2}})-(2{{d}^{2}}+3cd)=4{{c}^{2}}-2{{d}^{2}}<0.$
Now, we can multiply our starting inequalities through by ${{q}^{2}}$   :
${{q}^{2}}{{a}^{2}}<2{{q}^{2}}{{b}^{2}}<{{p}^{2}}{{b}^{2}}<{{p}^{2}}{{c}^{2}}<{{q}^{2}}{{d}^{2}}$
Taking square roots, and re-arranging:
$q(d-a)>p(c-b)$
Squaring both sides:
${{q}^{2}}{{(d-a)}^{2}}>{{p}^{2}}{{(c-b)}^{2}}>2{{q}^{2}}{{(c-b)}^{2}}$
Cancelling the q’s:
${{(d-a)}^{2}}>2{{(c-b)}^{2}}$
