0

I followed these steps
Assume $p$ is an arbitrary prime and $p > 3$,then $p$ can be written in form of,
$$p=6k+r,r ∈\left\{0,1,2,3,4,5\right\}$$
Then for $r∈\left\{0,2,3,4\right\}$ Asumption contradicts.
Therefore $r=1$ or $r=5$
My problem is when for $r∈\left\{0,2,3,4\right\}$ Assumption contradicts, How we say certainly any prime is either form $6k+1$ or $6k+5$?

Galaxylokka
  • 97
  • 2
    Well, any number of the form $6k+2$ is even, for example. Can you handle the other cases? – lulu Nov 16 '20 at 14:22
  • Of course, my problem is, If for $r∈\left{0,2,3,4\right}$ asumption contradicts.then how we certainly say r=1 or r=5. – Galaxylokka Nov 16 '20 at 14:28

0 Answers0