If $2n+1$ and $4n+3$ are prime, then $2n-1$ and $4n+1$ are not when $n>2$ How do you prove that, for $n>2$, if $2n+1$ and $4n+3$ are prime numbers, then $2n-1$ and $4n+1$ are composite numbers?

HINT:  $3$ divides $2n(2n+1)(2n-1)$ and $3$ divides $(4n+1)(4n+2)(4n+3)$
As $n>2, 4n+1>2n+1>3$
Hence, $4n+1, 2n+1$ are not divisible by $3$
If $4n+3$ is prime, $3$ does not divide $n\implies 3$ divide  $2n-1$
If $2n+1$ is prime, $3$ does not divide $2(2n+1)\implies 3$ divide $4n+1$ 
which can also be derived from $4n+1=2(2n-1)+3$
