Finding x in an Olympiad simultaneous equation I have been practicing for a an upcoming intermediate math olympiad and I came across the following question:

Let $x$ and $y$ be positive integers that satisfy the equations $$\begin{cases} xy = 2048 \\ \frac{x}{y}-\frac{y}{x}= 7.875.\end{cases}$$ Find x.    

My approach
$$xy=2048$$
$$x=\frac{2048}{y}$$
$$\frac{\frac{2048}{y}}{y} - \frac{y}{\frac{2048}{y}} = 7.875$$
$$\frac{2048}{y^2} -\frac{y^2}{2048} = 7.875$$
$$2048^2 -y^4 =7.875\cdot 2048 y^2$$
If we compare the the first equation (x/y) - (y/x) = 7.875 to the euqtion that we just found ((sqrt. 2048)/y) - (y/(sqrt. 2048)) = 7.875 , x is in the same place so, x must be equal to the sqrt. of 2048. 
Is this solution correct or are there any better solutions? 
Thank you :) 
 A: The best way to solve math olympiad questions is the fast way - A long solution is just as bad as none, given the time constraints. Therefore, I would recommend the following:
First, notice that $2048 = 2^{11}$. With the stipulation that $x$ and $y$ are integers, this means that they must both be integer powers of two, and so are $\frac xy$ and $\frac yx$.
Next, notice that $\frac yx$ must be less than one, since $\frac xy$ is greater than one, since the difference is greater than one.
Thus, the only candidate for $\frac xy$ is 8. Thus, since there is guaranteed to be a solution, $\frac xy = 8 = 2^3$. Thus, $x = 2^7$ and $y = 2^4$. 
A: HINT...$\frac xy-\frac yx=7.875=8-\frac 18$
Therefore either $\frac xy=8$ or $-\frac 18$
A: Your solution is incorrect because it was stated that x is a positive integer, and $\sqrt{2048}$ is not an integer, therefore it can't be a solution.
Hints to solve the problem:
$$2048 = 2^{11}$$
$$ 7.875 = 8 - \frac{1}{8} $$
(Spoiler alert) My solution:
Because $2048 = xy = 2^{11}$, we can say that $x = 2^a$ and $y = 2^b$, where $a+b=11$ (equation 1).
Now, we see that: 
$$\frac{2^a}{2^b} - \frac{2^b}{2^a} = 7.875$$
$$\frac{2^a}{2^b} - \frac{2^b}{2^a} = 8 - 1/8$$
Therefore:
$$\frac{2^a}{2^b} = 8$$ 
Which can be rewritten as:
$$2^{a-b} = 2^3$$
$$a-b = 3 (Equation 2)$$
Now, adding equation 1 and 2:
$$2a = 14$$
$$a = 7$$
Therefore, $x = 2^{7} = 128$
A: Consider the second equation and define $z=\frac xy$. So, you have $$z-\frac 1z=7.875$$ which is a quadratic in $z$; its roots are $z=-\frac 18$ (to be discarded) and $z=8$ (to keep). So $xy=2048$ and $\frac xy=8$.
I am sure that you can take from here.
A: When solving quiz problems (and to some extent even real-world problems), always ask yourself if the problem is easier than it seems. In this case, 2048 is easily recognizable as a power of 2: $2^{11} = 2048$. Because $x$ and $y$ are integers, they have to be powers of two as well since $xy = 2048$. Moreover if $x = 2^n$, then $y = 2^{11-n}$. That's only a few combinations to try (the equation is symmetric on $x$ and $y$ except for the sign of the result), so you immediately get the solution $x = 2^7, y = 2^4$.
In this case the problem specifies that x and y are positive integers, so this approach was guaranteed to work. But it would have been worth exploring even if it was just a hunch.
A: Clearly, one can just use the naive approach and solve the quadratic in $y$.
As others noted, being given that $x,y$ are integers immediately tells you that they are powers of two.
So we have:
$x = 2^a, y = 2^b$
${x\over y} - {y\over x} = {63 \over 8} $
$\implies {{x^2 - y^2}\over 2^{11}}= {(2^6-1)\over {2^3}}$
$\implies {{2^{2a} - 2^{2b}}}= 2^8{(2^6-1)} = 2^{14} - 2^8$
At which point $a,b$ fall out by inspection.
Moral of the story: When the given equations have some kind of pattern, keep it in the working. 
In this case, the powers of 2 are very important. The "$xy = 2048$" is very obviously so. The "$7.875$" is sufficiently odd as to encourage me to immediately think that it is going to be worthwhile writing as the more "normal" $63\over 8$.
A: How about this one :
Let, $x/y = A$
$\implies y/x = 1/A$
Then it's going to be like this 
$$A-1/A = 7.875$$ 
And the answer would be $A=8$
Then you got these solutions : $x/y=8$ & $x.y=2048$
ready to go. Easier and faster $x=128$ & $y=16$
