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What's the most straightforward way to ask Mathematica to find all solutions of an equation like

$$3x + 2y + 4z = 0 \pmod {11}$$

(for instance), where either $x$, $y$, $z$ can be considered to be integers in the range $-5\dots 5$, or equivalently they belong to the ring $Z/(11Z)$ of integers modulo $11$?

(I've tried a number of obvious things with no success.)

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2 Answers 2

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Solve[Mod[3 x + 2 y + 4 z, 11] == 0 &&
      -5 <= x <= 5 && -5 <= y <= 5 && -5 <= z <= 5, Integers]

(*    {{x -> -5, y -> -5, z -> -2},
       {x -> -5, y -> -4, z -> 3},
       {x -> -5, y -> -3, z -> -3},
       ...
       {x -> 5, y -> 5, z -> 2}}    *)

(121 solutions)

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Solve[3 x + 2 y + 4 z == 0, Modulus -> 11] produces {{z -> 2 x + 5 y}} (as residue classes modulo 11).

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  • $\begingroup$ Yes, multiplying 3x + 2y + 4z = 0 by 3 (in the field Z/11Z) gives -2x + 6y + z = 0, or in other words z = 2x + 5y. $\endgroup$
    – Daniel Asimov
    Commented Feb 22, 2021 at 6:25
  • $\begingroup$ Which is the set of solutions! Just like linear algebra over the field of real numbers, we have two free variables x and y, and the value of z is uniquely determined by them. In my opinion, this is a much better form than listing the 121 solutions one by one. $\endgroup$
    – Greg Martin
    Commented Feb 22, 2021 at 6:48
  • $\begingroup$ I agree, this is much better than what I proposed. $\endgroup$
    – Roman
    Commented Feb 22, 2021 at 12:38
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    $\begingroup$ I personally prefer Reduce[] for this: Reduce[3 x + 2 y + 4 z == 0, {x, y, z}, Modulus -> 11]. This highlights the fact that there is a parametric solution. $\endgroup$
    – J. M.'s missing motivation ♦
    Commented Feb 22, 2021 at 16:14

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