Back to square one

Algebra Level 4

$\large \left( x+ \sqrt{x^2+\sqrt{2015}}\right)\left( y+ \sqrt{y^2+\sqrt{2015}}\right) = \sqrt{2015}$

If $x$ and $y$ are real numbers that satisfy the equation above, find the value of $S$ such that $S = x+y$ .

Hint : Multiply both sides by with $\left( \sqrt{x^2+\sqrt{2015}} - x\right)$ and $\left( \sqrt{y^2+\sqrt{2015}} - y\right)$ .

2 solutions

Leah Smith
Sep 16, 2015

Pulkit Gupta
Dec 4, 2015

Firstly, the equation given is clearly cyclic. Hence, x = y

Secondly, we observe that on substituting x for -(x) & y for -(y) yields the same equation as with positive x & y .

Hence the equation is cyclic even for reversal of signs of its variables.

The only values that satisfy for these conditions are x = y = 0