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 2018-04-02, 11:32 #1 wildrabbitt   Jul 2014 19·23 Posts factors of surds Hello, 446 + 177$\sqrt{2}$ = (7 + 11$sqrt{2}$)*(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
2018-04-02, 11:42   #2
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts

Quote:
 Originally Posted by wildrabbitt Hello, 446 + 177$\sqrt{2}$ = (7 + 11$sqrt{2}$)*(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

 2018-04-02, 12:20 #3 Nick     Dec 2012 The Netherlands 26×3×7 Posts 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
 2018-04-02, 13:29 #4 wildrabbitt   Jul 2014 19·23 Posts Thanks to both posters.
 2018-09-30, 18:02 #5 wildrabbitt   Jul 2014 19×23 Posts 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 $a+b\sqrt{2}=(c+d\sqrt{2})(e+f\sqrt{2})$ & $a-b\sqrt{2}= (c-d\sqrt{2})(e-f\sqrt{2})$ ?
2018-10-01, 09:00   #6
Nick

Dec 2012
The Netherlands

101010000002 Posts

Quote:
 Originally Posted by wildrabbitt 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.

 2018-10-01, 09:17 #7 wildrabbitt   Jul 2014 19·23 Posts thanks
 2018-10-01, 13:42 #8 Dr Sardonicus     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]
 2018-10-01, 17:28 #9 wildrabbitt   Jul 2014 19·23 Posts Thanks.

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