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?
 A: 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 $4n+1=2(2n-1)+3$
A: Hint
think about the terms mod 3.
2n +1 can only be congruent to 2 mod 3. why?
now if it's congruent to 2 mod 3, what are the other two terms congruent to mod 3?
A: When $2n+1$ is prime, $4n+2$ will only have $2$ and $2n+1$ as its prime factor.
From the question, $4n+3$ is also a prime.
As $4n+1,\,4n+2,\,4n+3$ are three consecutive numbers, one of them must be divisible by $3$ and we know that $4n+2$ and $4n+3$ are not, thus we have the conclusion that $4n+1$ is divisible by $3$.
$$\begin{align}
4n+1 \equiv 0 &\pmod 3\\
\implies 4n\equiv-1&\pmod 3\\
\implies n \equiv-1&\pmod 3 [\text{as}\, \gcd(3,4)=1].
\end{align}$$
And as a result, we can work out $2n-1$ and $4n+1$:
$$2(-1)-1 \equiv 0 \pmod 3\\
4(-1)+1 \equiv 0 \pmod 3$$
Hope it helps
