Suppose that you had a machine that could find all four solutions for some given $a$. How could you use this machine to factor $n$? Question: Suppose $n = pq$ with $p$ and $q$ distinct odd primes.
Suppose that you had a machine that could find all four solutions for some given $a$. How could you use this machine to factor $n$? 
Proof: Suppose that $n = pq$ with $p$ and $q$ distinct odd primes and that a machine that could find all four solutions for some given $a$. Let the four solutions be denoted as $a, b, c, d$ such that $a \not = b \not = c \not = d$. Then, the sum $s_{k}$ of any two given solutions is not zero in $\mathbb{Z}_{n}$, for all $k \in \mathbb{Z}$. The sums $s_{k}$ are divisible by either $p$ or $q$ such that gcd$(n,s_{k}) = p$ or $q$, which are the factors of $n$. Therefore, we can find the factors of $n$ by giving the machine gcd$(n,s_{k})$. \blacklozenge
This is a follow-up proof in addition to the other proof I've written on another post. Thoughts on this proof I've written?

Hint $\ $ Suppose $\,f(x)\in\Bbb Z_n[x]\,$ has more roots than its degree. Iterating the Factor Theorem we can write $\,f(x) = c(x-r_1)\cdots (x-r_k)\,$ By hypothesis it has at least one more root $\,r\not\equiv r_i\,$ so $\,c(r-r_1)\cdots (r-r_k)\equiv 0\pmod n,\,$ so $\,n\,$ divides that product, but does not divide any factor,  hence the gcd of $\,n\,$ with some factor must yield a proper factor of $\,n.\,$  See here for more.
Remark $\ $ The inductive proof using the Factor Theorem  involves cancellation of $\,r_i-r_j,\,$ assuming it is coprime to $\,n\,$ (if not then taking a gcd already yields a proper factor of $\,n).\,$ The idea is that the proof breaks down due to the discovery of a zero-divisor, yielding a proper factor of $\,n.\,$
For example, $\,x^2\equiv 1\,$ has roots $\,x\equiv \pm1,\pm 4 \pmod{15}\,$ Let's see  what happens it we try to use the Factor Theorem to attempt to deduce that $\,f(1)\equiv 0\equiv f(4)\,\Rightarrow\, f(x) \equiv (x-1)(x-4).\,$ First $\, f(1)\equiv 0\,\Rightarrow\,f(x) = (x-1)g(x).\,$  Next $\, 0\equiv f(4)\equiv 3g(4).\,$ To deduce that $\,g(4)\equiv 0\,$ (so $\,x\!-\!4\mid g)\,$ requires cancelling $\,3,\,$ so we use the Euclidean Algorithm to test if $\,3\,$ is coprime to $\,n\,$ (so invertible, so cancellable mod $\,n).\,$ It is not, since $\,\gcd(3,15) = 3 \ne 1.\,$ But that's ok, since we have achieved our goal: we found a proper factor $\,3\,$ of $\,n=15.$
Alternatively, if we chose the roots $\,\pm1$ then iterating the Factor Theorem yields the factorization $\,f(x) = x^2-1 = (x-1)(x+1).\,$ Let's see what we can deduce from the quadratic's $\rm\color{#c00}{third}$ root $\,x\equiv \color{#c00}{4\not\equiv \pm1}:\ $  $\,0\equiv f(4)\equiv (4\!-\!1)(4\!+\!1)\,$ so $\,n\mid(4\!-\!1)(4\!+\!1),\,$ but $\,n\nmid 4\!-\!1,4\!+\!1\,$ (for otherwise $\,\color{#c00}{4\equiv 1}\,$ or $\,\color{#c00}{4\equiv -1}).\,$ It follows that $\,\gcd(n,4\!-\!1)\,$ and  $\,\gcd(n,4\!+\!1)\,$ are proper factors of $\,n\,$ (neither gcd can be $1$ else $\,n\,$ would divide the other factor). Again, we have found a proper factor of $\,n.$
Generally, in the same way, we can show one of the $\, r_i-r_j\,$ will have a nontrivial gcd with $\,n.$
