Re: modulo solving lacking domain?

• To: mathgroup at smc.vnet.net
• Subject: [mg126860] Re: modulo solving lacking domain?
• From: Andrzej Kozlowski <akozlowski at gmail.com>
• Date: Thu, 14 Jun 2012 05:29:52 -0400 (EDT)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• References: <201206120659.CAA25892@smc.vnet.net> <C49675CE-DFFB-4759-A1FC-F853D06D2456@mimuw.edu.pl> <4FD7DD77.6040101@eecs.berkeley.edu> <88093364-D782-410D-8920-E66714C94086@gmail.com> <4FD8A1D4.2030201@eecs.berkeley.edu>

```On 13 Jun 2012, at 15:21, Richard Fateman wrote:

> The argument you are presenting is that  Modulus-> ...  in Solve changes
> not the method of computation but the domain for the answer.  Since the answer
> exists outside the Solve and is not marked as a "modular" solution only,
> it is a potential troublespot, though I understand the rationale.
>
> It is more like
> (context: Change semantics to assume all integers are to be treated mod N;
>    consider a*x=b,  solve for x.  restore context returning the integer x).
>
> and what I want is
>
> Solve (Mod(a*x-b,N)==0 ,x)    which Mathematica 8 refuses to do.

But Mathematica does not "refuse" to do it - you are just using the wrong approach.

Reduce[12*n - 8 == 20*k, {k, n}, Integers]

Element[C[1], Integers] && k == 3*C[1] + 2 && n == 5*C[1] + 4

The condition on k can be ignored: it only tells you additional information. The answer is the same as the one you wanted but now you do get the Element[=85.] thing.

Andrzej Kozlowski

```

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