Four consecutive odd numbers

Let there be an odd prime $x$

Case 1 $x\equiv0(\mod3)$

Either $x=3$ or it is composite, as it must be prime $\therefore x=3$

Other odd numbers are $5,7,\red{9}$

And $9$ is not a prime

Case 2 $x \equiv 1 (\mod 3)$

$x+2\equiv3 (\mod 3)\Rightarrow x+2\equiv0(\mod 3)\Rightarrow 3|(x+2)$

Therefore either $x+2=3$ or itis not a prime. If $x+2=3\Rightarrow x=1$ which is not a prime

Case 3 $x \equiv 2 (\mod3)$

$x+2\equiv4(\mod3)$

$x+4\equiv6(\mod3)\Rightarrow x+4\equiv 0(\mod3)\Rightarrow 3|(x+4)$

Therefore either $x+4=3$ or it is not a prime. If $x+4=3\Rightarrow x=-1$ which is not a prime

$\therefore$ Four consecutive odd numbers can never be primes

Four consecutive odd numbers are such that, odd , even, odd , even, odd , even, odd

They are a total of seven consecutive numbers, since 3 is not divisible by two, it'll not repeatedly fall on even numbers, but on both odd and even alternatively, in seven, 3 must appear twice (3*2=6<7), and both can't be even, so one of four consecutive odd numbers is definitely a multiple of 3