If $x,y$ are distinct real numbers

Question

If $x,y$ are distinct real numbers, prove that $x+y=-2$ if and only if $(x+1)^2=(y+1)^2$

How would I prove this? do I use contrapositive, contradiction or direct proof?

Answer 1

If $x+y=-2$ then $ x+1=-1-y $ and therefore $ (x+1)^2=(y+1)^2$.

For the other direction, if $ (x+1)^2=(y+1)^2$ then $(x+1)=(y+1) $ or $ (x+1)=-(y+1)$. Since x and y distinct the first option isn't valid and so $x+1=-y-1$ which implies that $x+y=-2$.

Answer 2

$(x+1)^2=(y+1)^2 \iff (x+1)^2-(y+1)^2=0\iff (x+1-(y+1))(x+1+y+1)=0$

$ \iff x+1=y+1$ or $x+y+2=0$

Since we are excluding $x+1=y+1$ we have that the first is equivalent to $x+y=-2$.