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June 26th, 2017, 05:46 AM   #1
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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
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June 26th, 2017, 10:14 AM   #2
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Consider the complex conjugates of u and v.
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June 26th, 2017, 01:14 PM   #3
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Quote:
Originally Posted by skipjack View Post
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
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 05:54 PM.
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June 26th, 2017, 06:06 PM   #4
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Hint: if u and v have conjugates m and n respectively,
37 = (5 + 2i√3)(5 - 2i√3) = (uv)(mn) = (um)(vn).
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June 27th, 2017, 12:57 AM   #5
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Quote:
Originally Posted by skipjack View Post
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 02:06 AM.
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June 27th, 2017, 02:00 AM   #6
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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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