Recent content by omega16

Can anyone show me the proof of FLT for n=3? Thanks

A GCD question- urgent help! Find the GCD of 24 and 49 in the integers of Q[sqrt(3)], assuming that the GCD is defined. (Note: you need not decompose 24 or 49 into primes in Q[sqrt(3)]. Please teach me . Thank you very much.

Thanks . I have already solved this question.

Suppose 32 = alpha*beta for alpha, beta reatively prime quadratic integers in Q[i] . Show that alpha = epsilon*gamma^2 for some unit epsilon and some quadratic integers gamma in Q[i].

If alpha is a quadratic integers in Q[sqrt(-d)] , then define a notion of congruence (mod alpha). Furthermore, define +, -, and X for congruence classes , and show that this notion is well-defined. Could you please guide me to begin this question? Thank you very much.

because 2 and 3 are prime.

Find the prime factorization of 6 in Q[sqrt(-1)]. Ans : Since 6 = 2*3 so 6 = (1+sqrt(-1)) (1-sqrt(-1)) *3 Q.E.D. Do I need to add anything to it? Am I done with this question? Please kindly advise me. Thank you very much.

Thank you very much for your opinion. I have solved this question.

Consider Q[sqrt(2)]. Does every element of Q[sqrt(2)] have a square root in Q[sqrt(2)] ? Prove if true, and give a counterexample if false.My solution: sqrt(sqrt(2)) = a + bsqrt(2) if I square both sides then I will have : sqrt(2) = (a + b*sqrt(2))^2 = a^2 + 2ab*sqrt(2) + 2b^2...

Please guide me to start - a simple clue please , if possible. Thank you. I really don't know what to start with for this question.

Can anyone guide me how to prove this question?? Question: We want to describe via a picture a set of subsets of a square which are something like diagonals, but are not quite the same. We'll call them steep diagonals. One of them, labelled e, is illustrated in the square below; the other 6...

thank you very much. I have solved this question.

No , don't know what to go on next? Does the below theorem useful for go on my prove? if p is prime, the equation x^2 ≡ - 1 mod p has solution iff p ≡ 1 mod 4

That is what I can think of: Since p|(a^2+b^2), so a^2+b^2≡ 0 (mod p) and a^2 ≡ -b^2 mod p

No, still can't figure out how to start this proof. Could you please teach me how to do it? Thank you very much.