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Topic: A complex problem (Read 1297 times) |
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Mickey1
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A complex problem
« on: Dec 11th, 2012, 2:12pm » |
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I believe this is an adequate problem in its own right (although I have a special interest in it).
I cannot find anything about it on the internet, yet it is an interesting problem (and probably well known to some).
It is about complex numbers (a+ib) and (c+id) as solutions to Pell’s equation. a to d are natural non-zero numbers. (2i)^2-5*i^2=-4+5=1 is otherwise an obvious solution for a=c=0.
(a+ib)^2 –n (c+id)^2 =1 implies that 2iab=n2icd and therefore ab=ncd
The real part is (aa-bb) –n(cc-dd)=1
I don’t believe aa-bb-n(cc-dd)=1 for n=1
We can get the squares to be 1, 25-14 - (9-1)=9-8=1 but ab is not equal to cd.
If n, c and d are primes then I cannot a solution. a=n implies that b=cd, and the LHS becomes
nn-(cd)(cd) – n(cc-dd)= nn-ncc + ndd-(cd)(cd) = n(n-cc) +dd(n-cc)= (n+dd)(n-cc)
(n+dd)(n-cc)=1 which is impossible. b=n leads to a similar contradiction.
Can non-primes factors or other ideas leave room for a solution?
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