       • To: mathgroup at smc.vnet.net
• Subject: [mg117693] Re: About C[i]
• From: Peter <petsie at dordos.net>
• Date: Wed, 30 Mar 2011 04:08:22 -0500 (EST)
• References: <imshf8\$5r8\$1@smc.vnet.net>

```Am 29.03.2011 13:57, schrieb olfa:
> Hi Mathematica community,
>
> I have this solution set:
> ((C | C) \[Element] Integers&&  i == C&&  k == 7 C&&
>     kP == 0&&  iP == C + C) ||((C | C) \[Element] Integers
> &&
>   i == C&&  k == 2 + 7 C&&    kP == 2&&  iP == C + C)
>
> it means that it is an infinite solution set.
> I need to expess it without C and C since C==k/7||(k-2)/7
> and C==i . consequently the solution set would be expressed as
> (kP==0&&  iP==k/7+i)||(kP==2&&iP==(k-2)/7+i)
>   is it possible? how?can we say that is now transformed into finite
> set?
>
> thank you very much for your help.
>

Hi,

this is far from being finite. You've put the property "being integer"
from C,C to (i,k/7) and (i,(k-2)/7) respectively. It is just a bit
less effort to calculate e.g.:

Table[{{c2, 7 c1, 0, c1 + c2}, {c2, 2 + 7 c1, 2, c1 + c2}}, {c1, -5,
5}, {c2, -5, 5}]

than a table where you have to test for either k/7 or (k-2)/7 to be Integer.

hth,
Peter

```

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