$3\tau(K_1$#$K_2)$=$\tau(K_1)\tau(K_2)$ Suppose we have two knots $K_1$ and $K_2$. Then look to the connected sum of $K_1$ and $K_2$ denoted by $K_1\#K_2$ (defined for knots). Suppose $\tau$ is the number of $3$-colourings (definition for knots). I want to prove the formula above, namely: $3\tau(K_1\#K_2)=\tau(K_1)\tau(K_2)$. I read about some proofs with help of $n$-tangles. But i ask myself if there are proofs without such tangles?? Is it possible to come to the solution just with counting??
 A: Method 1: Wirtinger presentation.  The thing about Fox $3$-colorings is that they actually correspond to homomorphisms $\pi_1(S^3-K)\to D_{3}$, where $D_{3}$ is the symmetries of the triangle, and where meridians are sent to flips.  Take a diagram of a connect sum where it is obviously a connect sum (so one can draw a circle bounding the $K_1$ part that intersects $K_1\# K_2$ in exactly two points), and the meridian generators for the two intersected arcs are equal.  Thus, they are sent to the same thing in $D_3$ under any homomorphism, and hence they have the same coloring in any Fox $3$-coloring.  In the following picture, the blue circle demonstrates the connect sum, and the green loops are the meridian generators for the two intersected arcs.

The result follows since in the connect sum the two arcs have to be the same color, but individually the knots have no constraint on the color of the arcs.
Method 2: Tangles without the tangles.  Jozef Przytycki has a nice argument in 3-coloring and other elementary invariants of knots.  I'll give a variation on the argument which never uses the word "tangle."  Consider taking $K_1\#K_2$ and and adding an extra unknotted component.  If $K_1\#K_2$ can be $3$-colored, then the new diagram can be $3$-colored in three times the ways since there are no constraints on the new unknot.

The unknot can be isotoped to be under and around $K_1$ in the diagram.  This implies the colors $a$ and $b$ in the diagram are equal, since otherwise $c$ passed under $a$ and $b$ would be a different color from itself.  Therefore, the result follows.
Method 3: the Jones polynomial.  Przytycki and Lickorish-Millet proved that the Jones polynomial at the sixth root of unity gives the number of 3-colorings (see https://mathoverflow.net/a/176866).  If you know that the Jones polynomial is multiplicative, then the result follows.

I don't know how or if the next one can be finished by a direct argument.
Method 4: representation theory of $S_3$. Since $3$-colorings are invariant under permutations of the colors, we could develop a planar algebra (sorry... involves tangles and tensor diagrams).  Let $V$ be the $3$-dimensional permutation representation of $V$, with $e_1,e_2,e_3$ the basis representing the three colors.  We can decompose knots into cups, caps, crossings, and identity strands and replace them with tensor operators on $V$.  With the following replacements

the resulting element of $\mathbb{C}$ is an integer -- the number of Fox $3$-colorings.  A tangle gives a map $V^{\otimes n}\to V^{\otimes m}$ that is an intertwiner since each of the pieces of the decomposition are $S_3$ intertwiners.
In a connect sum, each component corresponds to a map $V\to V$.  The trace of the composition of the maps is the number of $3$-colorings of the connect sum.  The claim is that $V\to V$ is a scale multiple of the identity map, which means that the "input strand" and the "output strand" always have to be the same color.
The representation decomposes as $V=T\oplus W$ into irreducible representations, where $T$ is the $1$-dimensional trivial representation spanned by $e_1+e_2+e_3$ and $W$ is the orthogonal complement.  An intertwiner $f:V\to V$ is determined by how it scales $T$ and $W$.
If one could show it has to scale $T$ and $W$ by the same amount, the result would follow.
