- #1

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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...

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- Thread starter Hyperreality
- Start date

- #1

- 202

- 0

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...

- #2

HallsofIvy

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- #3

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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?

- #4

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

- #5

MathematicalPhysicist

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p^2=P^2+2pq+q^2Originally posted by Hyperreality

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?

-2pq=q^2

-2p=q

p=-q/2

not p=-q

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