<s>[INST] $\frac{x^2+y^2}{x-y}|1995$. Find positive integer $x,y$ 
$x,y$ are positive integer . Find all $x,y$ such that 
  $\frac{x^2+y^2}{x-y}|1995$.

My answer: ${x^2+y^2}|1995$.
Therefore, $x^2+y^2$ can be $1$, $3$, $5$, $7$, $19$, $15$, $21$, $35$, $57$, $95$, $133$, $105$, $285$, $399$, $665$ or $1\,995$.
I observed that only $5$ can be represented as the sum of two square ( I haven't check the last two number because they are very big).
So, $(x,y)=(1,2),(2,1)$.
Am I right??
I believe there is a easier method.I am looking for that. Please help me.
 [/INST] Your first assertion that $x^2+y^2 \mid 1995$ doesn't follow from your hypotheses.  Note that $(6,3)$ and $(38,19)$ are also solutions.  Certainly $(x^2+y^2)/(x-y) >1995$ for $x>1995$, so there can be only finitely many solutions.  I believe that $(399,1197)$ is the largest.
I observe that if $a=(x^2+y^2)/(x-y)$ and that $(x,y)$ is a solution, then so is $(a-x,y)$.
Edit:  I also observe that every solution is of the form $(2y,y)$ or $(3y,y)$ and in all cases $a=5y$.
Another Edit.  Let $k$ be a divisor of $1995$.  Then you have $x^2+y^2=k(x-y).$  Put everything on one side and complete the square:
$$\left(x-\frac{k}{2}\right)^2+\left(y+\frac{k}{2}\right)^2 = \frac{k^2}{2}.$$
Multiply through by $4$:
$$(2x-k)^2 +(2y+k)^2 = 2k^2.$$
So find, in the usual way, all the ways of writing $2k^2$ as a sum of two squares.  Since many of the prime divisors are congruent to $3$ mod $4$, there are not many solutions.  Then solve for $x$ and $y$.  Do this for each $k$ and you have all solutions.  Here's one example.  Take $k=35$.  So we find all solutions to 
$$u^2+v^2 = 2\cdot35^2$$
and these are $(7,49)$ and $(35,35)$.  The second one leads to $x=0$, which is not positive.  The first one gives $2x-35 = 7$ and $2y+35 = 49$ which leads to $(x,y) = (21,7)$.  Then from my observation above $y=35-21 = 14$ gives a second solution.  Lather, rinse, repeat for all $k$.
</s>