<s>[INST] If x and y are different integers , and if $2005 +x =y^2 ; 2005+y =x^2 $ then find xy... Problem : 
If $2005 +x =y^2 ; 2005+y =x^2$ then find xy...
My approach : 
Let $2005 +x =y^2 .....(i) ; 2005+y =x^2 ......(ii) $
Now from (i) we get : 
$ y = \sqrt{x + 2005}$ 
Now putting this value of y in (ii) we get : 
$ \Rightarrow 2005 +\sqrt{x + 2005} =x^2$ 
$ \Rightarrow \sqrt{x + 2005} =x^2 -2005 $ 
Now squaring both sides we  get : 
$\Rightarrow (\sqrt{x + 2005})^2 =(x^2 - 2005)^2$ 
Is there any other way I can solve this problem please suggest... thanks.
 [/INST] I am new here so don't know how to type math equation but I am providing the solution.
2005+x=y^2  ..(1)
 2005+y=x^2 ..(2)
After subtracting 2nd equation from first
x-y=y^2-x^2
x-y=(x-y)(x+y)
(x-y)+(x-y)(x+y)=0     (we change the sign x-y to y-x here)
(x-y)(x+y+1)=o
because as mentioned x and y are different integers so x-y!=0
In this case
x+y+1=0  -> x+y=-1 ...(3)
Now adding equation (1) and second we got
4010+x+y=x^2+y^2
because x+y=-1
so 4010-1=x^2+y^2
4009=x^2+y^2
because x^2+y^=(x+y)^2-2xy
as (x+y)^2=x^2+y^2+2xy
so 4009=(x+y)^2-2xy
and 4009=(-1)^2-2xy
and 2xy=1-4009
xy=-2004
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