# (Counting problem) very interesting Modular N algebraic eqs - for combinatorics experts

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)

• Did you intend this for Math.stackexchange.com? – rasher Apr 12 '14 at 22:54
• I had tried, but numerical attempt is fine too. :o) – 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, f, f}

(* Out= {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).