Are there two non-isotopic knots $K,K'$ in $S^3$ with the same $\mathrm{SL}\_2(\mathbb C)$ $A$-polynomials? If it's an open problem, has anyone suggested a method for finding them, or a reason why no such pair should exist (i.e., why the $A$-polynomial should be a complete knot invariant)?


My first guess for how to find such a pair $K, K'$ would be to try mutation, but I wasn't able to find much about (or figure out on my own) how mutation affects the $A$-polynomial.



Distinct knots with same $A$-polynomial
The torus knots $T\_{7,15}$ and $T\_{3,35}$ have the same A-polynomials. In general, if $p,q>1$ are coprime and odd then $T\_{p,q}$ has A-polynomial $(L-1)(LM^{pq}+1)(LM^{pq}-1)$, which only depends on the product $pq$. This follows from a few observations:


1. The [original paper](https://link.springer.com/article/10.1007%2FBF01231526) on the A-polynomial by Cooper-Culler-Gillet-Long-Shalen asserts (Proposition 2.7) that the A-polynomial of $T\_{p,q}$ is a multiple of $LM^{pq}+1$.
2. That factor comes from a family of representations $\pi\_1(S^3\setminus T\_{p,q})\to SL\_2(\mathbb{C})$ sending $\mu^{pq}\lambda$ to $-I$, and if we multiply all of these by the nontrivial character $\pi\_1(S^3\setminus T\_{p,q}) \to \{\pm1\}$ then (since $pq$ is odd) we get a different family for which $\mu^{pq}\lambda \mapsto +I$, contributing another factor of $LM^{pq}-1$.
3. Any irreducible representation $\rho: \pi\_1(S^3\setminus T\_{p,q}) \to SL\_2(\mathbb{C})$ with $\rho(\mu),\rho(\lambda)$ diagonal has to send $\mu^{pq}\lambda$ to $\pm I$, so every irreducible is accounted for by these two factors. This is because the complement $S^3\setminus T\_{p,q}$ is Seifert fibered, with $\mu^{pq}\lambda$ the class of a generic fiber, and so $\mu^{pq}\lambda$ is central in the knot group.


So in fact you can get arbitrarily many odd torus knots with the same A-polynomial. Namely, if $p\_1,\dots,p\_n$ are distinct odd primes, then you can get $2^{n-1}-1$ different such knots of the form $T\_{r,s}$ by letting $r$ and $s$ be the products of any two complementary, nonempty subsets of $\{p\_1,\dots,p\_n\}$. (This gives $2^n-2$ pairs $(r,s)$, but only half as many torus knots because $T\_{r,s}$ is isotopic to $T\_{s,r}$.) These all have the same A-polynomial, determined completely by the product $p\_1\dots p\_n$.


