Prove that every prime larger than $3$ gives a remainder of $1$ or $5$ if divided by $6$

Question

Can we prove that every prime larger than $3$ gives a remainder of $1$ or $5$ if divided by $6$ and if so, which formulas can be used while proving?

Answer 1

Yes:

• $N \equiv 0 \pmod 6 \implies N$ is divisible by $6 \implies$ $N$ is not a prime
• $N \equiv 2 \pmod 6 \implies N$ is divisible by $2 \implies$ $N$ is not a prime (or $N=2$)
• $N \equiv 3 \pmod 6 \implies N$ is divisible by $3 \implies$ $N$ is not a prime (or $N=3$)
• $N \equiv 4 \pmod 6 \implies N$ is divisible by $2$ and $N\geq4 \implies$ $N$ is not a prime
• $N$ is a prime larger than $3 \implies N \not\equiv 0,2,3,4 \pmod 6 \implies N \equiv 1,5 \pmod 6$

Answer 2

HINT:

Every positive integer can be written as $$6n,6n\pm1,6n\pm2=2(3n\pm1),6n+3=3(2n+1)$$

Answer 3

Hint $ $ By Euclid $\ (n,\,6) = (n\ {\rm mod}\ 6,\,6).\,$ Yours is special case $\,(n,6)=1,\,$ e.g. prime $\,n>3.$

So integers coprime to $\,6\,$ have residues (mod $6)\,$ that are coprime to $\,6,\,$ i.e. $\,1$ or $\,5\pmod{\! 6}.$

Generally if $\,n\,$ is coprime to $\,m\,$ then so too is its remainder $\,n\bmod m,\,$ so the remainder must be one of the $\phi(m)$ residues coprime to $m$, because the gcd remains unchanged upon applying a single descent step of the Euclidean algorithm for the gcd (proved in the prior linked post). In the OP we have $\,\phi(2\cdot 3) = 1\cdot 2$.

Answer 4

A proof without word : $\begin{bmatrix} 0 & \textbf1 & 2 & 3 & 4 & \textbf5 \\ 6 & 7 & 8 & 9 & 10 & 11 \\ 12 & 13 & 14 & 15 & 16 & 17 \\ 18& 19 & 20 & 21 & 22 & 23\\ 24 & \color{red}{25} & 26 & 27 & 28 & 29\\ 30 & 31 & 32 & 33 & 34 & \color{red}{35} \\ 36& 37 & 38 & 39 & 40& 41 \\ 42& 43 & 44& 45 & 46 & 47 \\ 48& \color{red}{49} & 50 & 51& 52 & 53 \\ 54& \color{red}{55}&56 &57 & 58 & 59\\..&..&..&..&..&.. \end{bmatrix}$