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wall e
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#1
solve

x^2+y^2-13(x-y)=0
x,y\in Z+

PostPosted: Wed Nov 04, 2009 7:16 am  Back to top 
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mavropnevma
Yang-Mills Theory
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#2
Very weak, since the discriminant of the equation in x is \Delta = 338 - (2y+13)^2.

Solutions (3,2), (10,2) and (0,0), (13,0).
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PostPosted: Wed Nov 04, 2009 7:41 am  Back to top 
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Arrange your tan
Riemann Hypothesis
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#3
Re: solve

wall e wrote:
x^2 + y^2 - 13(x - y) = 0
x,y\in Z +


mavropnevma,

you included too many many solutions. You note the restrictions
on x and y. So each of them must be positive.

The solutions are: (3,2) and (10,2).

PostPosted: Wed Nov 04, 2009 8:13 pm  Back to top 
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mavropnevma
Yang-Mills Theory
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#4
It is a matter of notation. For most mathematicians \mathbb{Z}_+ includes zero; if we want to exclude it, either use \mathbb{Z}_+^* or use plain English and say "positive integers".
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PostPosted: Wed Nov 04, 2009 10:41 pm  Back to top 
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aadil
Riemann Hypothesis
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#5
Quote:
It is a matter of notation. For most mathematicians includes zero;

how is this ?the definition of a positive no. is that it should be>0how can you include 0
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PostPosted: Thu Nov 05, 2009 5:02 am  Back to top 
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JBL
Birch & Swinnerton Dyer
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#6
Discussions about ambiguous notations are ridiculous: if you want people to know what you mean, use an unambiguous notation like \mathbb{Z}_{> 0} or \mathbb{Z}_{\geq 0} or just define what you mean in words. Of course, in this case it doesn't matter since if you personally don't think that 0 is in whatever the set in question is, you can simply throw out those solutions that include it.
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PostPosted: Thu Nov 05, 2009 6:11 am  Back to top 
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