Weird Roots

Solution:

Let's call, for speeding things up, that

$\sqrt{x^2+2013}-x = \alpha$ .

It holds that:

$\alpha(\sqrt{y^2+2013}-y) = 2013 \Rightarrow \sqrt{y^2+2013}-y = \frac{2013}{\alpha}$

$\sqrt{y^2+2013} = y + \frac{2013}{ \alpha } \Rightarrow y^2 + 2013 = y^2 + \frac{4026y}{ \alpha } + (\frac{2013}{ \alpha }) ^2$

$1 = \frac{2y}{\alpha} + (\frac{2013}{\alpha^2}) \Rightarrow \frac {\alpha^2 - 2013}{2\alpha} = y$ .

Undoing the $\alpha$ substitution, we get:

$y = \frac {x^2+2013 - 2x\sqrt{x^2+2013} + x^2 - 2013}{2\sqrt{x^2+2013} - 2x} \Rightarrow y = \frac{-x(2\sqrt{x^2+2013}-2x)}{(2\sqrt{x^2+2013} - 2x)} \Rightarrow y = -x \Rightarrow$

$\boxed{x + y = 0.}$

Comments: What an amazing problem, Bogdan! Did you make it all by yourself?

$(\sqrt{x^{2}+2013}-x)(\sqrt{y^{2}+2013}-y)=2013$ ........(i)

$\implies \frac{(\sqrt{x^{2}+2013}-x)(\sqrt{x^{2}+2013}+x)(\sqrt{y^{2}+2013}-y)(\sqrt{y^{2}+2013}+y)}{(\sqrt{x^{2}+2013}+x)(\sqrt{y^{2}+2013}+y)}=2013$

$\implies \frac{(x^{2}+2013-x^{2})(y^{2}+2013-y^{2})}{(\sqrt{x^{2}+2013}+x)(\sqrt{y^{2}+2013}+y)}=2013$

$\implies \frac{2013 \times 2013}{(\sqrt{x^{2}+2013}+x)(\sqrt{y^{2}+2013}+y)}=2013$

$\implies (\sqrt{x^{2}+2013}+x)(\sqrt{y^{2}+2013}+y)=2013$ .......(ii)

From the equations (i) and (ii), we get-----

$(\sqrt{x^{2}+2013}-x)(\sqrt{y^{2}+2013}-y) = (\sqrt{x^{2}+2013}+x)(\sqrt{y^{2}+2013}+y)$

We see that $x=0$ and $y=0$ satisfy this equation, so $x=0,y=0$

Now, $x+y=0+0=\boxed{0}$

Notice that when x=-y ,the equation holds. Thus, x + y = 0.If anyone could prove that x= -y are the only solutions, please inform me in the comments.

$(\sqrt{x^{2}+2013} - x)(\sqrt{y^{2}+2013} - y) = 2013$

Multiplying by $(\sqrt{x^{2}+2013} + x)$ to both sides;

$(x^{2} + 2013 - x^{2})(\sqrt{y^{2}+2013} - y) = 2013(\sqrt{x^{2}+2013} + x)$

$2013(\sqrt{y^{2}+2013} - y) = 2013(\sqrt{x^{2}+2013} + x)$

Dividing by 2013 to both sides;

$(\sqrt{y^{2}+2013} - y) = (\sqrt{x^{2}+2013} + x)$

We can now see that the equation is true for every $x=-y$

So, $x+y = x + (-x) = \boxed{0}$

Let, √(x^2+2013)-x=p. So we get, x^2+2013=(p+x)^2. => x=(2013-p^2)/p. Similarly, we can get y=(2013-q^2)/q. From here, if we add x and y, we get x+y=[2013(p+q)-pq(p+q)]/pq. As p.q=2013, x+y=0.

since x and y can take any value..so assigning 0 to both

Divide both sides by (sqrt(x^2+2013)-x) and then multiply by (sqrt(y^2+2013)+y), you will get (y^2+2013-y^2)=((2013)(sqrt(y^2+2013)+y))/(sqrt(x^2+2013)-x). Combine like terms and divide both sides by 2013, you will get 1=(sqrt(y^2+2013)+y)/(sqrt(x^2+2013)-x). It follows that (sqrt(x^2+2013)-x)=(sqrt(y^2+2013)+y). Then substitute (sqrt(y^2+2013)+y) to the orginial equation. You will get y=0 and then x=0. The sum is 0.

let the term containing x be a& let the term containing y be b as 2013 is a prime number either a or b must be 2013 &1 solving we get -2012=2x 2012=2y therefore x+y =0

He doesn't say that[; x \neq y ;] so just make it as simple as possible and take [; (\sqrt {x^2+2013}-x)(\sqrt {y^2+2013}-y)=(\sqrt {x^2+2013}-x)(\sqrt {x^2+2013}-x)=2013 \rightarrow \sqrt{2013}=(\sqrt{x^2+2013}-x) \approx 44.9 ;] If we would take away the x^2 from our sqrt we would get the answer out of that part that we wanted, 44.9. the x^2 in [; \sqrt{x^2+2013}-x ;] is bigger than x because [; \sqrt{x^2+2013} \neq \sqrt{x^2}-\sqrt{2013} ;] so if we were to put an integer bigger than 1 as x we wouldn't get sqrt(2013) so [; (\sqrt{0^2+2013}-0)^2=2013 ;] so [; x=0 ;] and [; y=0 ;], [; x+y=0+0=0 ;] THE ANSWER IS [; 0 ;]!