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

Solution 1:

I have numerically computed, there is no better intuition for me, but your guess $N^4+N^5$ is close. Let us say the number of independent solutions $\equiv Sol(N)$, then

$$ Sol(2)=46=2^4+2^5-2 $$ $$ Sol(3)=321=3^4+3^5-3 $$ $$ Sol(5)=3745=5^4+5^5-5 $$ $$ Sol(7)=19201=7^4+7^5-7 $$

thus my conjecture is that for the prime $N$ $$ Sol(N)=N^4+N^5-N $$ I hope it helps.

Solution 2:

Modulo a prime $N$, the integers form a field. For simplicity, use conventional algebraic notation (equals, forget about modulus). Take first the case that $x_1$ and $y_1$ aren't zero, then your first three equations reduce to: \begin{align} \frac{x_2}{x_1} &= \frac{y_2}{y_1} \\ \frac{x_3}{x_1} &= \frac{y_3}{y_1} \\ \frac{x_4}{x_1} &= \frac{y_4}{y_1} \end{align} So you can define the three ratios independently ($N^3$ options), and then fix the values of $x_1$ and $y_1$ at will ($(N - 1)^2$ options), so the total is $N^3 (N - 1)^2$. If $x_1 = 0$, then either $y_1 = 0$ and $x_2$, $x_3$ are arbitrary ($N^2$ options) or $y_1 \ne 0$ ($N - 1$ possible values) and $x_2$, $x_3$, $x_4$ arbitrary ($N^3$ options).

Adding all up gives $N^3 (N - 1)^2 + N^2 + (N - 1) N^3 = N^5 - N^4 + N^2$.