oh.. i hvn tried... just changed to feel confident.. now its my q..not urs :P
Let " x " and " y " be positive integers such that " xy " divides " x ^{2} + y ^{2} + 1 " . Prove that 
x ^{2} + y ^{2} + 1xy = 3 .
I will add some more if I see that people are taking interest in this .

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7 Answers
Changing the question . Find all k such that the diophantine equation x^{2}+y^{2}+1 = kxy has solution in positive integers.
xy divides x2+y2+1
so x divides x2+y2+1 and y divides x2+y2+1
xâ‰¡0 mod (x2+y2+1) and yâ‰¡0 mod (x2+y2+1)
this implies xâ‰¡0 mod (y2+1) and yâ‰¡0 mod (x2+1) dince its obvious that x2 is divisible by x and y2 is divisible by y .
adding these 2 gives us (x+y)â‰¡(x+y)2+22xy
this means (x+y) should divide 22xy
this means (x+y) should divide 2(1xy)
since x,y are positive integers, the above condition holds only when x=y=1
therefore (x2+y2+1)/xy=3
Sir , don't you think that it is not obvious , that line no. 4 implies line no. 5 ?
that's an application of Infinite decent...
Let us consider the case when xâ‰ y.WLOG x>y
Let x_{1}, y_{1}, n_{1} be a solution to the diophantine equation x^{2}+y^{2}+1 = nxy.
This gives us (y_{1}n_{1}x_{1})^{2}+ y_{1}^{2}+1 = (y_{1}n_{1}x_{1})y_{1}n_{1} hence (y_{1}n_{1}x_{1}),y_{1},n_{1} is also a solution..
we have y_{1}^{2}+1= x_{1}(y_{1}n_{1}x_{1}) > 0, implying (y_{1}n_{1}x_{1})>0
since x_{1}>y_{1} , x_{1}^{2}>y_{1}^{2}+1= x_{1}(y_{1}n_{1}x_{1}) we have x_{1} > (y_{1}n_{1}x_{1})
Therefore one can form a decreasing sequence of positive integers..but no such sequence exists!
so x=y. This gives x^{2}(n2) = 1 hence we must have n = 3
oops, yes ricky u r right.. mine wa smistake
and dont call me sir...
i am neither that old nor that worthy of the title