I'm currently reading Discrete Mathematics: Elementary and Beyond of Springer Mathematical Series; and come across a fact in the review questions that seems quite obvious, but difficult to prove.
It goes like:
All prime numbers greater than 3 give remainder 1 or -1 when divided by 6.
How about proving this?
Won't it require us to know the general term, i.e. the formula for nth prime number?
The only possibilities are $$x\equiv 0,1,2,3,4,5\bmod 6.$$
If $x \equiv 0 \bmod 6$ then $x$ is divisible by $6$ and so not prime.
If $x \equiv 2 \bmod 6$ or $x \equiv 4 \bmod 6$ then $x$ is divisible by $2$ and so not prime (unless $x=2$).
If $x \equiv 3 \bmod 6$ then $x$ is divisible by $3$ and so not prime (unless $x=3$).
So we are only left with $x \equiv 1 \bmod 6$ or $x \equiv 5 \equiv -1 \bmod 6$ if $x>3$ is prime.
