Proving all primes are 1 or -1 modulo 6 [duplicate]

Yes. Generally if $\rm\:p\:$ is prime, then modulo $\rm\,2p,\,$ any prime $\rm\:q\ne 2,p\:$ must lie in one of the $\rm\:\phi(2p) = \phi(p) = p\!-\!1\:$ residue classes that are coprime to $2$ and $\rm p,$ i.e. all odd residue classes excluding $\rm p,\:$ viz. $\rm\:1,3,5,\ldots,\hat p,\ldots,2p\!-\!1.\:$ Indeed, integers in other classes are divisible by $2$ or $\rm p\:$ hence, if prime, must be $2$ or $\rm p,\:$ resp. More succinctly, exploiting negation reflection symmetry: $\rm\: q\equiv \pm\{1,3,5,\cdots,p\!-\!2\}\ \ (mod\ 2\:\!p),\ $ e.g. $\rm\,\ q\equiv \pm 1\ \ (mod\ 6),\:$ $\,\rm q\equiv \pm\{1,3\}\ \ (mod\ 10),\:$ $\,\rm q\equiv\pm \{1,3,5\}\ \ (mod\ 14),\:$ etc.

Generally, if $\rm\,q\,$ is any integer coprime to $\rm\:m\:$ then its remainder mod $\rm\:m\:$ lies in one of the $\rm\:\phi(m)\:$ residue classes coprime to $\rm\:m,\:$ where $\phi$ is the Euler totient function.

We have $6| 6n$, $2| 6n+ 2$, $3|6n+3$, and $2|6n + 4$ for all integers $n$. That leaves the $6n+1$s and $6n+5$s as possible primes, save for 3 and 2.