TITLE: For every prime $p$, does there exist a knot $K$ such that $K$ is $p$-colourable?
QUESTION [1 upvotes]: Once there is a $p$-colourable knot $K$, $K\#K$ is also $p$-colourable and, iteratively, there are infinite many $p$-colourable knots. But I am not sure if there really is a $p$-colourable knot for any prime $p$.

REPLY [1 votes]: For each $p$ a prime, there is a knot which is $p$-colorable.  From A.-L. Breiland, L. Oesper, and L. Taalman, “p-coloring classes of torus knots”,

Theorem 1:
Suppose $T_{m,n}$ is a torus knot and $p$ is prime.

If $m$ and $n$ are both odd, then $T_{m,n}$ is not $p$-colorable.
If $m$ is odd and $n$ is even, then $T_{m,n}$ is $p$-colorable if and only if $p|m$.


In particular, for your chosen $p$, the $T_{2,p}$ torus knot is $p$-colorable.
And one more note which might be of use: their proof uses the crucial fact that a knot is $p$-colorable iff $p$ divides $\det(K)$.
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