 Solution

The usual quadratic variation convention includes $⟨M{⟩}_0=0$. If $A,C$ are two continuous finite-variation processes with $A_0=C_0=0$ such that $M^2-A$ and $M^2-C$ are continuous local martingales, then $A-C$ is a continuous local martingale of finite variation. It must be constant, and its initial value is zero, so $A=C$ up to indistinguishability of stochastic processes.

Here is a proof of the fact used. Subtract the initial value of a continuous local martingale $N$ of finite variation, and stop so that both $|N|$ and its total-variation process are bounded by deterministic constants. The stopped local martingale is a bounded true martingale, so orthogonality of its increments gives, on any deterministic partition of an interval of $[0,t]$,Continuity and the dominated convergence theorem justify the last limit. Remove stopping and use continuity on a countable dense set of times to conclude $N\equiv 0$.

As literally printed, uniqueness is false without normalization. If $A$ works then $A+c$ also works for any deterministic constant $c$, since subtracting a constant preserves the local martingale property; this is a genuine omitted convention.

With normalization, existence always holds for a continuous local martingale. Its quadratic variation $[M]$, obtained as the limit in probability of squared increments, is continuous, increasing, starts at zero and satisfies $M^2-[M]$ is a continuous local martingale, so taking $⟨M⟩=[M]$ supplies the required finite-variation process.