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Old 2018-04-02, 11:32   #1
wildrabbitt
 
Jul 2014

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Default factors of surds

Hello,


446 + 177\sqrt{2} = (7 + 11sqrt{2})*(4 + 19\sqrt{2})

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\sqrt{2}, how can I work out factors of the same form?

Last fiddled with by science_man_88 on 2018-04-02 at 12:22
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Old 2018-04-02, 11:42   #2
science_man_88
 
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Quote:
Originally Posted by wildrabbitt View Post
Hello,


446 + 177\sqrt{2} = (7 + 11sqrt{2})*(4 + 19\sqrt{2})

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\sqrt{2}, how can I work out factors of the same form?
Knowing (a+b)(c+d)= ac+bc+ad+bd may help. Edit: ex. In your own example 146= 7*4+11*19*2, the two came out divide 146 by 2 you get 73, that means you have one product even one product odd in what's left, namely 7*2 and 11*19. 177√2 is 7*19√2+4*11√2 which again points to an even odd split.

Last fiddled with by science_man_88 on 2018-04-02 at 12:22
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Old 2018-04-02, 12:20   #3
Nick
 
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If \(x=a+b\sqrt{2}\), define \(N(x)=a^2-2b^2\).
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/Algebraic-N.../9781498738392
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Old 2018-04-02, 13:29   #4
wildrabbitt
 
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Thanks to both posters.
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Old 2018-09-30, 18:02   #5
wildrabbitt
 
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Hi again,

does anyknow if this is true

(a+b\sqrt{2})(a-b\sqrt{2})=(c+d\sqrt{2})(e+f\sqrt{2})(c-d\sqrt{2})(e-f\sqrt{2})

implies

a+b\sqrt{2}=(c+d\sqrt{2})(e+f\sqrt{2})
&
a-b\sqrt{2}= (c-d\sqrt{2})(e-f\sqrt{2}) ?
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Old 2018-10-01, 09:00   #6
Nick
 
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Quote:
Originally Posted by wildrabbitt View Post
Hi again,

does anyknow if this is true

(a+b\sqrt{2})(a-b\sqrt{2})=(c+d\sqrt{2})(e+f\sqrt{2})(c-d\sqrt{2})(e-f\sqrt{2})

implies

a+b\sqrt{2}=(c+d\sqrt{2})(e+f\sqrt{2})
&
a-b\sqrt{2}= (c-d\sqrt{2})(e-f\sqrt{2}) ?
No - for example, put a=7, b=0, c=e=3, d=f=1.
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Old 2018-10-01, 09:17   #7
wildrabbitt
 
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thanks
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Old 2018-10-01, 13:42   #8
Dr Sardonicus
 
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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]
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Old 2018-10-01, 17:28   #9
wildrabbitt
 
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Thanks.
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