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Sep 29, 2011 at 6:46 history closed user6976
Felipe Voloch
Andreas Blass
David Loeffler
Andrés E. Caicedo
too localized
Sep 28, 2011 at 15:59 comment added David Loeffler @Carl: Watch your logical connectives! The question you just asked is totally vacuous. I think you meant "... only if there exists one to the whole system". Here the answer is (almost as obviously) no: consider the equations $x+y=1, x=0, y=0$ -- any two of these are simultaneously solvable but the three are obviously not simultaneously solvable. But to reiterate what others have already said, with coefficients in $\mathbb{R}$ these would be totally trivial undergraduate linear algebra exercises, and linear algebra over $\mathbb{F}_p$ works just the same.
Sep 28, 2011 at 13:44 comment added Carl For example, is it true that there exists a solution to each pair of congruences if there exists one to the whole system?
Sep 28, 2011 at 13:40 comment added Igor Rivin You only have one prime -- you can't get much more local than that.
Sep 28, 2011 at 13:27 comment added Carl That's not what I'm looking for. I'm looking for local conditions as in the case of proofwiki.org/wiki/Solution_to_Simultaneous_Linear_Congruences
Sep 28, 2011 at 12:50 comment added Martin Bright Carl - this question will probably get closed soon since it is more appropriate for math.stackexchange.com. However, two comments: (1) What are the conditions for such a system of equations to be soluble over the real numbers? If you can't answer this, you should read up on linear algebra first. (2) The integers modulo p form a field, so pretty much everything you know about linear algebra over the reals works mod p without any changes.
Sep 28, 2011 at 12:48 answer added Igor Rivin timeline score: 2
Sep 28, 2011 at 12:47 history edited Carl CC BY-SA 3.0
added 129 characters in body
Sep 28, 2011 at 12:42 history asked Carl CC BY-SA 3.0