20180402, 11:32  #1 
Jul 2014
19·23 Posts 
factors of surds
Hello,
446 + 177 = (7 + 11)*(4 + 19) I picked some numbers and calculated the LHS. That's why I can write it the other way around. Can someone tell what kind of maths I need to use to work from the left to the right. i.e Given a + b, how can I work out factors of the same form? Last fiddled with by science_man_88 on 20180402 at 12:22 
20180402, 11:42  #2  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
Last fiddled with by science_man_88 on 20180402 at 12:22 

20180402, 12:20  #3 
Dec 2012
The Netherlands
2^{6}×3×7 Posts 
If \(x=a+b\sqrt{2}\), define \(N(x)=a^22b^2\).
Then for all x and y we have \(N(xy)=N(x)N(y)\) so looking at the absolute value of \(N(x)\) enables you to determine a finite list of possible factors of \(x\) You can find out more about this by looking up quadratic fields in a good book on Algebraic Number Theory, for example: https://www.crcpress.com/AlgebraicN.../9781498738392 
20180402, 13:29  #4 
Jul 2014
19·23 Posts 
Thanks to both posters.

20180930, 18:02  #5 
Jul 2014
19×23 Posts 
Hi again,
does anyknow if this is true implies & ? 
20181001, 09:00  #6 
Dec 2012
The Netherlands
10101000000_{2} Posts 

20181001, 09:17  #7 
Jul 2014
19·23 Posts 
thanks

20181001, 13:42  #8 
Feb 2017
Nowhere
3×1,009 Posts 
In addition to Nick's example, based on primes of the form a^2  2*b^2 [namely p == 1 or 7 (mod 8)], a "trivial" class of counterexamples uses "units of norm 1"; that is, solutions to
e^2  2*f^2 = 1, e.g. e = 3, f, = 2. We can take c = a, d = b, and e = 3, f = 2 [or c = a, d = b, and e = 17, f = 12; or c = a, d = b, and e = 99, f = 70, etc] 
20181001, 17:28  #9 
Jul 2014
19·23 Posts 
Thanks.

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