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**Prove that if x and y are ....**

## Homework Statement

*Prove that if x and y are distinct real numbers, then (x+1)*

^{2}=(y+1)^{2}if and only if x+y=-2. How does the conclusion change if we allow x=y?## Homework Equations

...

## The Attempt at a Solution

Suppose x and y are real numbers. If x≠y then

x+2≠y+2

----> (x+2)/(y+2)≠1

----> (x+2)/(y+2)≠ x/y or y/x (since x/y=y/x=1)

----> (x+2)x=(y+2)y

or

(x+2)/x=(y+2)/y

----> (x+2)x=(y+2)y (since (x+2)/x=(y+2)/y ---> x=y)

----> x

^{2}+2x=y

^{2}+2y

---->x

^{2}+2x+1=y

^{2}+2y+1

---->(x+1)

^{2}=(y+1)

^{2}

----> x+1 = (y+1) or (-y-1)

----> x+1=-y-1 (since x+1=y+1 ---> x=y)

----> x+y=-2

?????

But I'm not sure if I've done everything it asked. I know for "P if and only Q," it needs to be proven that P--->Q and Q--->P, but it seems here that if P is (x+1)

^{2}=(y+1)

^{2}and Q is x+y=-2, I'm just kind of going in circles by proving the "if and only if" part. See what I'm saying?