How to solve for the numerical value of $a,b,c,$ or $d$ where $p * a = q * b = r * c = s * d $, derived from examples If you have the following, where $a,b,c,d,$ and $z$ are unknown values while all other values are known, how do you solve for the actual numerical value of at least one of $a,b,c,d,$ or $z$.  $m$ is a prime number.
$(p * a) \bmod m = z$
$(q * b) \bmod m = z$
$(r * c) \bmod m = z$
$(s * d) \bmod m = z$
Obviously,
$(p * a) \bmod m = (q * b) \bmod m = (r * c) \bmod m = (s * d) \bmod m$
All values, known and unknown, $\ne 0$ and $\ne 1.$
If additional examples are found then each other example would be identical except that the second variable (such as 'd') would be a new unknown value and each first variable (such as 'p') be a new known value.  If it cannot be determined with these 4 examples then how many examples would I need to figure this out mathematically as opposed to testing virtually every possible value.  How would one go about this?  If you can, please answer in relatively layman's terms.
 A: No matter the values of $p,q,r,s$, in the above system, $a\equiv b\equiv c\equiv d\equiv z\equiv 0 \bmod m$ is a solution; so we can assume that $z\not\equiv 0 \bmod m$ to look for more solutions, which would also require that none of $p,q,r,s$ are divisible by $m$.
You specify that $m$ is a prime number. This means that, for $p\not\equiv 0 \bmod m$, there will be an inverse value $p^{-1}$ such that $p^{-1}p\equiv 1 \bmod m$.
Similarly $q^{-1},r^{-1},s^{-1}$ will exist. Then we can choose $z$ coprime to $m$ arbitrarily and calculate $a\equiv p^{-1}z,$ $b\equiv q^{-1}z,$ $c\equiv r^{-1}z$  and $d\equiv s^{-1}z \bmod m$ as a solution set corresponding to that $z$.

With your added condition of none of the variables being allowed to have values of $0$ or $1$, and taking as implied that all values are in the inclusive range $[2\ldots m{-}1]$, the choices may be a little more restricted but the above argument essentially holds. The modular inverses of $p,q,r,s$ will allow a suitable solution set for any $z$ not equal to any of the $p,q,r,s$.
For example, taking $m=7$ and $(p,q,r,s)=(2,3,4,5)$:
$2a \equiv 3b \equiv 4c \equiv 5c \bmod 7$
To avoid any of $a,b,c,d\equiv 1$, we need $z\equiv 6 \bmod 7$ (as the only permitted value left in range).
This then solves as $(a,b,c,d)\equiv (3,2,5,4) \bmod 7$
Another example: taking $m=17$ and $(p,q,r,s)=(3,8,11,14)$ (all calcs $\bmod 17$)
Then $z$ can be any of $(2,4,5,6,7,9,10,12,13,15,16)$. We can assess each quickly by first calculating $(p^{-1},q^{-1},r^{-1},s^{-1})\equiv(6,15,14,11)$ and then for example taking $z=10$ we can find $(a,b,c,d) \equiv (6\cdot 10,15\cdot 10,14\cdot 10,11\cdot 10)\equiv (9,14,4,8)$
