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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...
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...