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 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: 19,299 Thanks: 1688 Consider the complex conjugates of u and v.
June 26th, 2017, 12:14 PM   #3
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Quote:
 Originally Posted by skipjack Consider the complex conjugates of u and v.
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:
ac-3bd=5
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: 19,299 Thanks: 1688 Hint: if u and v have conjugates m and n respectively, 37 = (5 + 2i√3)(5 - 2i√3) = (uv)(mn) = (um)(vn). Thanks from topsquark and Alexxandru
June 26th, 2017, 11:57 PM   #5
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Quote:
 Originally Posted by skipjack Hint: if u and v have conjugates m and n respectively, 37 = (5 + 2i√3)(5 - 2i√3) = (uv)(mn) = (um)(vn).
I tried like this and it gets me to (a^2+3b^2)(c^2+3d^2)=37
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: 19,299 Thanks: 1688 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. Thanks from Alexxandru

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