Recent content by Baba-k

Hi guys, Thanks for the responses, I think I see now. So the angle formed at z_{1} is the same as the argument of \frac{z_{1} - z_{2}}{z_{1} - z_{3}} ? thanks babak

Hi, Homework Statement Interpret the angle of the complex number (z_{1} - z_{2}) / (z_{1} - z_{3}) in the triangle formed by the points z_{1}, z_{2}, z_{3}. Homework Equations The Attempt at a Solution I'm not entirely sure what to do in this question, I've done a couple of...

Riteo don't worry, was just confusing my self hehe. thanks for all your help!

Hi, Okay thanks, think I see what you're saying. Is it enough to say.. since (a+b)(a-b)\equiv 0 \pmod p then a+b=np or a-b=kp where a, b \in \left \{ 1, ..., p-1 \right \} if a-b=kp then a-b will always be < p and hence k is 0. While 0 < a+b < 2p . Hence n is 1 as we've already said...

Hi Dick, Thanks for your comments. i) The product (a-b)(a+b)\equiv 0 \pmod p is divisible by the prime p only if one of the factors is, so either a-b\equiv 0 or a+b\equiv 0 \pmod p must be divisible by p. ii) I chose a+b=p as a \in \left \{1,..,p-1 \right \} so will always be < p but...

Hi Petek, Heres what I've come up with, what do you think ? a^{2} \equiv b^{2} \pmod p a^{2} - b^{2}\equiv 0 \pmod p (a-b)(a+b)\equiv 0 \pmod p (a-b)\equiv 0 \pmod p \texttt{ or } (a+b)\equiv 0 \pmod p a+b=p \texttt{ or } a-b=p a=p-b \texttt{ as } a\in \left \{ 1, 2,...,p-1...

Hi Petek, Thanks the response. I'm not sure how atm but will give it ago hehe cheers babak

Hi, I'm slowly reading through the book What is Mathematics which asks the following question at the end of its quadratic residues section. I'm not sure how to begin it really, so any hints/suggestions would be greatly appreciated. Homework Statement We have seen that x^{2} \equiv (p -...