- #1

Hyperreality

- 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...

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Hyperreality
- Start date

- #1

Hyperreality

- 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

Science Advisor

Homework Helper

- 43,021

- 970

- #3

Hyperreality

- 202

- 0

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

Hyperreality

- 202

- 0

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

- #5

MathematicalPhysicist

Gold Member

- 4,699

- 369

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

Share:

- Last Post

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 8

- Views
- 149

- Last Post

- Replies
- 7

- Views
- 208

- Last Post

- Replies
- 11

- Views
- 420

- Last Post

- Replies
- 11

- Views
- 623

- Last Post

- Replies
- 1

- Views
- 947

MHB
Number theory

- Last Post

- Replies
- 1

- Views
- 347

- Last Post

- Replies
- 11

- Views
- 572

- Last Post

- Replies
- 3

- Views
- 516

- Last Post

- Replies
- 19

- Views
- 708