If x and y are different integers , and if $2005 +x =y^2 ; 2005+y =x^2 $ then find xy...

Question

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.

Answer 1

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

Answer 2

Hint: $$2005 + x- (2005+y)=y^2 - x^2 \quad \Longrightarrow \quad (x-y)(x+y+1) = 0.$$

Answer 3

$2005 +x =y^2 .....(i) ; 2005+y =x^2 ......(ii)$ $(i)-(ii)\Leftrightarrow (x-y).(x+y+1)=0$ $\Rightarrow x=y ; x+y=-1$ $+x=y$, now we solve $x,y$ then $xy=...$ $+x+y=-1 \Rightarrow xy=-2004$