Show that for every integer n1, there are integers x y1 such that
$$
\frac { x^2 + y^2 -1}{ x y} =n

$$

Question

Show that for every integer n>1, there are integers x > y>1 such that
$$
\frac { x^2 + y^2 -1}{ x y} =n

$$

anonymous · Answer

Are there more conditions on x and y?  Because you could always take x = n and y = 1.  That would work for any integer n.

anonymous · Answer

bah i take that back. y must be greater than 1.

kinggeorge · Answer

I'm back for a moment to propose that using $x=n$ and $y=n^2-1$ has been working for the examples I've looked at so far. 

Now I'm actually off to bed :P

anonymous · Answer

yeah, thats the solution:$$\frac{x^2+y^2-1}{xy}=\frac{n^2+(n^2-1)^2-1}{n(n^2-1)}=\frac{n^2+n^4-2n^2+1-1}{n(n^2-1)}$$$$=\frac{n^4-n^2}{n(n^2-1)}=\frac{n^2(n^2-1)}{n(n^2-1)}=n.$$ Nice insight :)

anonymous · Answer

Maybe just switch x and y so that the x > y > 1 statement holds.