We can suppose that we have four naturals not equal to zero: $a, b, c, d$. Further, we're working modulo a prime $p$.
Now if we find $a, b, c, d$ that satisfy:
$$f \equiv a \cdot c \equiv b \cdot d \ne 0, \mod p \tag{1}$$ $$g \equiv a \cdot d \equiv b \cdot c \ne 0, \mod p \tag{2}$$ $$\text{and } a \cdot c \ne b \cdot c, \tag{3}$$
It seems to me (from experiments) that $f$ always must equal $p-g$. Further, we can find numbers that can make equations (1), (2) and (3) true for any $f$ not equal to $0$.
Can someone prove this? In other words, can someone prove that we can satisfy the above equations for any given $f$ modulo $p$? Also, can we prove that $f \equiv p-g$?
MY IDEAS
If we solve $a \cdot c = b \cdot d$, we find $a = \frac{bd}{c}$. Then, plugging in $a$ into $a \cdot d = b \cdot c$, we find $d^2 \equiv c^2 \mod p$. Similarly, solving (1) for $c$, we find $c = \frac{bd}{a}$. Then, plugging into $a \cdot d = b \cdot c$, we find $a^2 \equiv b^2 \mod p$. So I believe we can collect the terms as:
$$c^2 \equiv d^2 \mod p \tag{4}$$ $$a^2 \equiv b^2 \mod p \tag{5}$$
I'm not sure where to proceed from here. Perhaps there is another approach...
Hint $\,\ ac\not\equiv bc\,\Rightarrow\, a\not\equiv b,\,$ so $\, 0\equiv a^2-b^2 = (a-b)(a+b)\,\Rightarrow\, \color{#c00}{a+b\equiv 0},\,$ therefore $\,f+g \equiv ac+bc = (\color{#c00}{a+b})c\equiv 0,\,$ so $\,f\equiv -g.$