Let $S_n=\sum_{i=1}^n X_i$, where the $X_i$ are identically and independently distributed with $P(X_i=1)=p$ and $P(X_i=-1)=1-p$.
Show that there does not exist such $c(n)$ that makes $M_n=S_n^2+c(n)$ a martingale.
So, if $M_n$ were to be a martingale, then
$$E[M_{n+1}|{\cal{F_n}}]=M_n \quad\!\! \\\Rightarrow \quad \!\! E[S^2_{n+1}+c(n+1)|{\cal{F_n}}]\\=E\left[(S_n+X_{n+1})^2+c(n+1)|{\cal{F_n}}\right] \\=E\left[(S_n^2+2S_n\cdot X_{n+1}+X_{n+1}^2+c(n+1)|{\cal{F_n}}\right]\\=S^2_n+2S_nE[X_{n+1}]+E[X^2_{n+1}]+c(n+1)$$
Since $E(X_i)=2p-1\quad \text{and} \quad E(X_i^2)=1$:
$$E[M_{n+1}|{\cal{F_n}}]=S_n^2+2S_n(2p-1)+1+c(n+1)=M_n=S^2_n+c(n)$$
This requires that $c(n+1)=c(n)-2S_n(2p-1)-1$ if all the steps were shown correctly.
However, what I don't understand is why $c(n+1)$ has to be a random quantity which prevents it from taking on the value we require it to be and thus prevents $M_n$ from being a martingale.
Isn't $S_n$ already defined by the conditional expectation?
That's the point; $c(n)$ can't be a random quantity. Therefore you have shown that there is no $c(n)$ such that $M_n$ is a martingale.
$$E[S_{n+1}^2|F_n]=E[(S_n+X_{n+1})^2|F_n]$$
$$S_n^2 + 2(2p-1)S_n+ 1 = S_n^2 +(c(n)-c(n+1)) \implies c(n)-c(n+1)=2(2p-1)S_n+1$$
No deterministic $c(n)$ can exist unless $X_i$ were constant, but this contradicts the given mean and variance. Therefore $S_n^2$ cannot be made into a martingale via the addition of a function.