# Thread: primes and unique factorization

1. ok..i understand where you're coming from now. so since p / b^2 it also divides b..
indeed. I hope you understood the steps leading upto $\displaystyle p|b^2$.

The definition of prime-numbers say: if $\displaystyle p|mn \Rightarrow p|m$ or $\displaystyle p|n$

Apply that idea to $\displaystyle p|b\cdot b$ then follows that $\displaystyle p|b$.

2. yes..i do..understand how we got where we did..

for b, your saying counter-example, as in prove it is tru by saying that if p does not divide a^2 + b^2 then it does not divide a^2 + b^2 nor c^2 + d^2.

3. No. It means you're giving an example that shows the given statement is false.

A statement like: For all $\displaystyle n$ we have $\displaystyle n+1 = 5$ is clearly false. Then $\displaystyle n= 1$ is an counterexample against that statement.

4. ok.. i think i see where you're coming from

5. Are you sure that a is false..if p divides thoroughly isnt the statement tru?

6. so that would make a and b false then..correct..

7. (a) and (b) is false. For (a) I gave you a counterexample in the first post. For (b) try to come up with a counter-example yourself.

(c) correct.

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