We have some attempt to numerically solve this math problem, which means that we like to count the number of independent solutions of this set of six of modular N algebraic equations:

$$ (1) x_1 y_2=x_2 y_1 \text{(mod $N$)}\\ (2) x_1 y_3=x_3 y_1 \text{(mod $N$)}\\ (3) x_4 y_1=x_1 y_4 \text{(mod $N$)}\\ (4) x_2 y_3=x_3 y_2 \text{(mod $N$)}\\ (5) x_2 y_4=x_4 y_2 \text{(mod $N$)}\\ (6) x_3 y_4=x_4 y_3 \text{(mod $N$)} $$

suppose that all variables are module $N$, i.e. $x_j=x_j$(mod $N$) and $y_j=y_j$(mod $N$). Here there are $x_1,x_2,x_3,x_4, y_1,y_2,y_3,y_4$ eight variables.

We may take $N=2$ or $N=3$ for simplicity. How should one do this to output a number $sol(N)$(the number of independent solutions)? I expect a number larger than $2N^4-1$ and larger than $N^5$, and smaller than $N^8$. So,

$$sol(N)\geq N^5,\;\;\; sol(N)\geq 2N^4-1,\;\;\; sol(N)\leq N^8$$

For $N=2$, I expect that the number of independent solutions of this set of six of modular $N$ algebraic equations = $48$ (is it correct?). For $N=3$, we expect the answer is 324 (is it correct?).

Thank you. :0)

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  • $\begingroup$ Did you intend this for Math.stackexchange.com? $\endgroup$ – rasher Apr 12 '14 at 22:54
  • $\begingroup$ I had tried, but numerical attempt is fine too. :o) $\endgroup$ – annie heart Apr 12 '14 at 22:56

Mathematica claims slightly different counts.

f[n_?PrimeQ] :=
 Module[{xvars = Array[x, 4], yvars = Array[y, 4]},
  Length[solns =
    Solve[Flatten[Minors[{xvars, yvars}, 2]] == 0, Join[xvars, yvars],
      Modulus -> n]]]

{f[2], f[3], f[5]}

(* Out[51]= {46, 321, 3745} *)

Given that you are solving over integer (prime only?) moduli, your second inequality is really not needed (it is implied by the first one).

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