The Conversation

Let x and y be the two distinct real POSITIVE values

(Eqn 1) $x+y=xy$

(Eqn 2) $x+y=x^2-y^2$

$x+y=x^2-y^2$ $x+y=\left(x+y\right)\left(x-y\right)$ $1=x-y$ (Eqn 3) $y=x-1$

Substitute y from Eqn3 in Eqn1 $x+\left(x-1\right)=x\left(x-1\right)$ $2x-1=x^2-x$ $x^2-3x+1=0$

By quadratic formula $x=\frac{-(-3)\pm \sqrt{{(-3)}^2-4\left(1\right)(1)}}{2\left(1\right)}$ $x=\frac{3\pm \sqrt{5}}{2}$

$x\approx 2.618$

$x\approx 0.3819$ (Extraneous solution because it is paired to a negative value of y.)

Substitute x in Eqn3 $y\approx 2.618-1$ $y\approx 1.618$

$y\approx 0.3819-1$ $y\approx -0.6181$ (Extraneous solution because we need only positive values)

Taking the floor function of each value.

$\left\lfloor x\right\rfloor =\left\lfloor 2.618\right\rfloor =2$ and $\left\lfloor y\right\rfloor =\left\lfloor 1.618\right\rfloor =1$

Adding the result

$\left\lfloor x\right\rfloor +\left\lfloor y\right\rfloor =2+1=3$