Remainder theory proof

[Hyperreality]

If p^2 is exactly divisible by p+q, then proof q^2 is exactly divisible by p+q.

How do I proof this, and how do I apply the remainder theorem?

I know if f(x) = x^2 + 2x + 1, since f(-1) = 0 there fore (x + 1) is a factor of f(x).

So in this case
p^2 = p x p or p^2 x 1...

[HallsofIvy]

Since you specifically mention x2+ 2x+ 1, haven't you looked at (p+q)²= p²+ 2pq+ q²?

[Hyperreality]

This is what I've done afterwards, but I'm not sure if it is right.

Let p^2 = x^2 + 2xy + y^2.

where y^2 = y.y or -y.-y or x^2 = x.x or -x.-x.

When y = -x, or x = -y, p^2 = 0

Therefore (x + y) is a factor of p^2, also (x + y) = (p + q)

Therefore p^2 = p^2 + 2pq + p^2. p^2 = 0 when p = -q,

but (p)^2 = (-q)^2
p^2 = q^2
therefore (p+q)mod q^2 = 0

Are there any flaw in my argument or have I made any calcuation error?

[Hyperreality]

Sorry, it's q^2mod(p+q) = 0

[MathematicalPhysicist]

Originally posted by Hyperreality
This is what I've done afterwards, but I'm not sure if it is right.

Let p^2 = x^2 + 2xy + y^2.

where y^2 = y.y or -y.-y or x^2 = x.x or -x.-x.

When y = -x, or x = -y, p^2 = 0

Therefore (x + y) is a factor of p^2, also (x + y) = (p + q)

Therefore p^2 = p^2 + 2pq + p^2. p^2 = 0 when p = -q,

but (p)^2 = (-q)^2
p^2 = q^2
therefore (p+q)mod q^2 = 0

Are there any flaw in my argument or have I made any calcuation error?
p^2=P^2+2pq+q^2
-2pq=q^2
-2p=q
p=-q/2
not p=-q