Is it true that $f(x,y)=\frac{x^2+y^2}{xy-t}$ has only finitely many distinct positive integer values with $x$, $y$ positive integers? 
Prove or disprove that if $t$ is a positive integer, $$f(x,y)=\dfrac{x^2+y^2}{xy-t},$$ then $f(x,y)$ has only finitely many distinct positive integer values with $x,y$ positive integers. In other words, there exist $k\in\mathbb N$ such that if $n\gt k$ then $f(x,y)=n$ has no positive integer solutions.

This problem is a generalization of this famous problem.


*

*Below is the list of the set of $f(x,y)$ with $t\le 10$ (may be incomplete):


{t,{f(x,y)}}=
{1,{5}}
{2,{4,10}}
{3,{3,4,8,13,17}}
{4,{5,26}}
{5,{13,25,37}}
{6,{6,10,50}}
{7,{5,8,9,20,29,41,65}}
{8,{4,10,18,34,82}}
{9,{5,29,61,101}}
{10,{20,122}}
Thanks in advance!
 A: Let f(x,y) be any integer with t also an integer and find that will x and y necessarily be integers.
let$ f(x,y)=z$ ,
$z(xy)-zt=x^2+y^2$
Let $zt$ be another integer $w$,
$w=(z+2)(xy)-(x+y)^2$
Now the sum of $(z+2)xy$ and $-(x+y)^2$ to be integers both the terms should be separately integers.
Now you can say that let $xy=A$
Where $A$ is integer and $(x+y)^2=B$
now here $B$ will necessarily a perfect square other wise it will not satisfy that integer subtracted from integer is an integer.
So, now solve it and you'll find x and y are sum or subtraction of integers . Hence they are integers.
A: October 14, 2015. This is with $$ \frac{x^2 + y^2}{xy - t} = q > 0, $$
which I believe to be the intent of the question. 
THEOREM: $$ \color{red}{ q \leq (t+1)^2 + 1 }  $$
I got some help from Gerry Myerson on MO to finish the thing. 
 https://mathoverflow.net/questions/220834/optimal-bound-in-diophantine-representation-question/220844#220844 
As far as rapid computer computations, for a fixed $t,$ we can demand $1 \leq x \leq 4 t.$  For each $x,$ we can then demand $1 \leq y \leq x$ along with the very helpful $x y \leq 4 t.$ Having found an integer quotient $q,$ we then keep only those solutions with $2x \leq qy$ and $2y \leq qx.$
In particular, for $t=1$ we find $q=5,$ then for $t=2$ we find $q=4,10.$ In both cases we have $q \leq (t+1)^2 + 1.$ We continue with $t \geq 3.$
With $t \geq 3, $ we also have $t^2 \geq 3t > 3t - 1.$
We are able to demand $xy \leq 4t$ by taking a Hurwitz Grundlösung, that is $2x \leq qy$ and $2y \leq qx.$ Define $k = xy - t \geq 1.$
Now, $xy \leq 4t,$ then $k = xy - t \leq 3t,$ then $k-1 \leq 3t - 1.$ Reverse, $3t-1 \geq k-1.$ Since $t^2 > 3t - 1,$ we reach
$$ t^2 > k-1. $$ 
Next, $k \geq 1,$ so $(k-1) \geq 0.$ We therefore might get equality in
$$ (k-1)t^2 \geq (k-1)^2, $$
but only when $k=1.$
$$  0 \geq t^2 - k t^2 + k^2 - 2 k + 1,  $$
$$  k t^2 + 2 k \geq t^2 + k^2 + 1.  $$ Divide by $k,$
$$ t^2 + 2 \geq \frac{t^2}{k} + k + \frac{1}{k}.  $$ Add $2t,$
$$ t^2 +2t + 2 \geq \frac{t^2}{k} + 2 t + k + \frac{1}{k},  $$
with equality only when $k=1.$ Reverse,
$$  \frac{t^2}{k} + 2 t + k + \frac{1}{k} \leq  t^2 +2t + 2  $$
with equality only when $k=1.$
Here is Gerry's best bit, this would not have occurred to me. Here we are back to considering all solutions $(x,y)$ and all $k=xy-t.$ Draw the graph of the quarter circle $x^2 + y^2 = k q.$ As $x,y \geq 1,$ there are boundary points at $(1, \sqrt{kq-1})$ and   $( \sqrt{kq-1},1).$ The hyperbola $xy =  \sqrt{kq-1}$ passes through both points, but in between stays within the quarter circle. It follows by convexity (or Lagrange multipliers again) that, along the circular arc,
$$ \color{blue}{ xy \geq  \sqrt{kq-1}}. $$ But, of course, $x^2 + y^2 = k q = qxy - t q$ is equivalent to our original equation $x^2 - q x y + y^2 = -tq.$
We have
$$  -tq = x^2 - q x y + y^2 = (x^2 + y^2 ) - q x y = k q - q x y \leq kq - q  \sqrt{kq-1}, $$ or
$$  -tq  \leq kq - q  \sqrt{kq-1}, $$
 $$  -t  \leq k -   \sqrt{kq-1}, $$
 $$    \sqrt{kq-1} \leq t + k, $$
$$ kq -1 \leq t^2 + 2k t  + k^2,  $$
$$ kq  \leq t^2 + 2 kt  + k^2 + 1,  $$ divide by $k,$
$$  q \leq \frac{t^2}{k} + 2 t + k + \frac{1}{k}. $$
For $t \geq 3$ and a solution with $xy < 4t,$ we showed
$$  \frac{t^2}{k} + 2 t + k + \frac{1}{k} \leq  t^2 +2t + 2  $$
with equality only when $k=1.$ For all solutions, Gerry showed
$$  q \leq \frac{t^2}{k} + 2 t + k + \frac{1}{k}. $$
Put these together, we get
$$ q \leq  t^2 +2t + 2  $$
with equality only when $k=1,$ that is $xy = t+1.$
ADDENDUM, October 15. Here is another way to get Gerry's main observation, with $k = xy - t,$ that $xy \geq \sqrt{kq-1}.$ We have $x,y \geq 1$ and
$kq =x^2 + y^2 .$ So $kq \geq x^2 + 1$ and $kq -(x^2 + 1) \geq 0.$ We also have $x^2 - 1 \geq 0.$ Multiply,
$$ (x^2 - 1) kq - (x^4 - 1) \geq 0. $$
Next, $y^2 = kq - x^2,$ so $x^2 y^2 = kq x^2 - x^4.$ That is
$$ x^2 y^2 = (kq-1) + (x^2 - 1)kq - (x^4 - 1).  $$ However,
$$ (x^2 - 1) kq - (x^4 - 1) \geq 0, $$ so
$$ x^2 y^2 \geq kq - 1,   $$
$$ \color{blue}{ xy \geq  \sqrt{kq-1}}. $$
A: For some cases decisions may be infinitely many. You can use this formula and select the required ratios. That root was rational.
Though it is necessary to bring the decisions some pretty simple solutions:
the equation:  $aX^2+bXY+cY^2=f$
If the root of the whole:  $\sqrt{\frac{f}{a+b+c}}$
Then use the solution of Pell's equation:  $p^2-(b^2-4ac)s^2=1$
Solutions can be written:
$Y=((4a+2b)ps\pm(p^2+(b^2-4ac)s^2))\sqrt{\frac{f}{a+b+c}}$
$X=(-(4c+2b)ps\pm(p^2+(b^2-4ac)s^2))\sqrt{\frac{f}{a+b+c}}$
A: October 7, 2015. This is with $$ \frac{x^2 + y^2}{xy - t} > 0, $$
which I believe to be the intent of the question. 
I proved finiteness, with an explicit bound that is not that bad.
This works. Note that the original question requires $xy> t.$ Otherwise we could have listed $x=1,y=1,t=2$ to get
$(x^2 + y^2)/ (xy-t) = -2.$ This was not done. So we are keeping $xy>t>0,$ in
$$ \frac{x^2 + y^2}{xy-t} = q. $$ 
We have the arc of the hyperbola
$$ x^2 - q x y + y^2 = -tq $$ in the first quadrant $x,y > 0$ that lies in the sector of the first quadrant defined by
$$ 2 x \leq q y $$ and
$$ 2 y \leq q x. $$ Note that the points of intersection of the two boundary lines with the hyperbola branch give the two points with the minimum values of $x$ and of $y.$
As noted in the other answer, if there are any integer solutions $(x,y)$ with $q$ also an integer, then there is at least one solution between the indicated Hurwitz lines. 
Next, we always have $q \geq 3.$ In $ x^2 - q x y + y^2 = -tq ,$ if $q=1$ the quadratic form on the left hand side is positive definite and can never equal the right hand side, which is negative. If $q=2$ the quadratic form on the left hand side is positive semi-definite ($(x-y)^2$) and can never equal the right hand side, which is negative. 
The key to finiteness was simply the size of $xy/t.$ We know already that $xy > t,$ that is $xy/t > 1.$ By Lagrange multipliers, the smallest value occurs when $x = y,$ at which point
$$ \frac{xy}{t} = \frac{q}{q-2} = 1 +  \frac{2}{q-2}. $$
Once again by Lagrange multipliers, the largest value of $xy/t$ within the Hurwitz region occurs at the boundary point where one of the lines meets the hyperbola.  One of them is at
$$ y = \left( \frac{2}{q} \right) x. $$ Plugging this into $ x^2 - q x y + y^2 = -tq $ gives a nice value for $x^2,$ then $ y^2 = \left( \frac{4}{q^2} \right) x^2 $ gives a nice value for $y^2.$ These turn out to be
$$  x^2 = \frac{q^3 t}{q^2 - 4}, \; \; \; y^2 = \frac{4 q t}{q^2 - 4}.  $$
Together 
$$  x^2 y^2 =  \frac{4 q^4 t^2}{(q^2 - 4)^2},  $$ and
$$  x y =  \frac{2 q^2 t}{q^2 - 4},  $$ or
$$  \frac{x y}{t} =  \frac{2 q^2 }{q^2 - 4} =   \frac{2 q^2 - 8 }{q^2 - 4} +  \frac{8 }{q^2 - 4} = 2  +  \frac{8 }{q^2 - 4} .  $$
This gives the maximum. Since $q \geq 3,$ 
$$  \frac{x y}{t} \leq  2  +  \frac{8 }{3^2 - 4} = \frac{18}{5} = 3.6 .  $$
Here we finally return to integers. We have $x \geq 1,$ which tells us that a Hurwitz fundamental solution always has
$$ y \leq  \frac{18}{5} t. $$ Once again, Lagrange multipliers tell us that $x^2 + y^2$ is maximized at the boundary point $x=1$ on the curve $xy= 18t/5,$ so
$$ x^2 + y^2 \leq 1 + \frac{324}{25} t^2.  $$ However, $xy - t \geq 1,$ meaning $q \leq x^2 + y^2.$ We then get finiteness from
$$  q  \leq 1 + \frac{324}{25} t^2.  $$
Computations, as above, suggest the stronger $q \leq t^2 + 2 t + 2.$ 
As far as rapid computer computations, for a fixed $t,$ we can demand $1 \leq x \leq 4 t$ because $18/5 < 4.$ For each $x,$ we can then demand $1 \leq y \leq x$ along with the very helpful $x y \leq 4 t.$ Having found an integer quotient $q,$ we then keep only those solutions with $2x \leq qy$ and $2y \leq qx.$
To repeat the good part: if there is any solution $(x,y)$ then there is at least one fundamental solution, that is with 
$$  \color{blue}{ 2x \leq qy}$$ and $$  \color{blue}{ 2y \leq qx}.$$
For such a fundamental solution, we have
$$ \color{blue}{ 1 + \frac{2}{q-2} \leq \frac{xy}{t} \leq  2 + \frac{8}{q^2-4} }. $$
Since $x^2 - qxy + y^2$ is positive (semi)-definite when $q = 1,2,$ we know that $q \geq 3$ always. Thus $q^2 - 4 \geq 5.$ As $\frac{8}{5} \leq 2,$ we get
$$  \color{blue}{xy \leq 4t}.  $$
Here is a graph for $t=1, q=5,$ showing the region where fundamental solutions must lie:

