Re: modulo solving lacking domain?

*To*: mathgroup at smc.vnet.net*Subject*: [mg126844] Re: modulo solving lacking domain?*From*: Dana DeLouis <dana01 at me.com>*Date*: Wed, 13 Jun 2012 04:54:11 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com

> Solve[12*n==8,n,Modulus->20] > {{n->4+5*C[1]}} > > It omits C[1] element of Integers. Hi. Don't know where one would find this, so try: Solve[12*n==8,n,Integers,Modulus->20] Solve::mdom: Warning: Mathematica is ignoring domain specification Integers; option setting Modulus -> 20 implies domain integers modulo 20. There you go! .. Modulus -> 20 implies domain integers :>) // And the solution appears to check ok. Table[Mod[12(4+5 c), 20], {c, -3, 10}] {8,8,8,8,8,8,8,8,8,8,8,8,8,8} = = = = = = = = = = HTH :>) Dana DeLouis Mac & Math 8 = = = = = = = = = = On Jun 12, 3:02 am, Richard Fateman <fate... at cs.berkeley.edu> wrote: > Solve[12*n==8,n,Modulus->20] > > returns > {{n->4+5*C[1]}} > > It omits C[1] element of Integers. > I doubt that this is a feature; is it a bug? > > C[1] is not necessarily a member of the finite field of > integers modulo 20. It is obvious not an arbitrary Real.