June 26th, 2017, 04:46 AM  #1 
Newbie Joined: Jun 2017 From: Bucharest Posts: 5 Thanks: 0  Complex numbers
A = {a + ib√3  a, b ∈ Z}. If u, v ∈ A and u*v = 5+2i√3 then show that u or v needs to be 1 or 1 
June 26th, 2017, 09:14 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 18,962 Thanks: 1606 
Consider the complex conjugates of u and v.

June 26th, 2017, 12:14 PM  #3 
Newbie Joined: Jun 2017 From: Bucharest Posts: 5 Thanks: 0  I tried to take in considerations the formulas and substitute u with u^2/u(conjugate) and the same with v, but it gets me nowhere. I really can't figure it out. The only thing I accomplished is writing u=a+ib√3, v=c+id√3 and substitute in uv = 5+2i√3 and I get the system: ac3bd=5 ad+bc=2 and if I treat it like a system with a,b being the variables first then c,d I get the inequations: c^2+3d^2≠0 a^2+3b^2≠0 Last edited by skipjack; June 26th, 2017 at 04:54 PM. 
June 26th, 2017, 05:06 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 18,962 Thanks: 1606 
Hint: if u and v have conjugates m and n respectively, 37 = (5 + 2i√3)(5  2i√3) = (uv)(mn) = (um)(vn). 
June 26th, 2017, 11:57 PM  #5  
Newbie Joined: Jun 2017 From: Bucharest Posts: 5 Thanks: 0  Quote:
I cannot figure it out. If you may, please write the answer. It's not like I'm lazy, but I've spent way too much time on this problem and I have a lot more to cover. I appreciate the hints you gave me and maybe if you don't want to give me the answer maybe another hint? In any case, thank you! Last edited by skipjack; June 27th, 2017 at 01:06 AM.  
June 27th, 2017, 01:00 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 18,962 Thanks: 1606 
As um and vn are positive divisors of 37, one of them is 1 and the other is 37. If an integer of the form p² + 3q² (where p and q are integers) has value 1, p must be 1 or 1, and q must be 0. 

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