As we know that primes other than 2 and 3 can be expressible as: $p \equiv 1\pmod{6}$ or $p \equiv -1\pmod{6}$. In other words, 6|(p-1) or 6|(p+1). Or, p = 6h+1 or 6h-1.
Now, for any integer h, 6h-1 or 6h+1 may be decomposable. Say 6h -1 = $p_1$p_2$ with $p_1$ = $p_2$ or $p_1$ < $p_2$. In the same way, 6h +1 = $p_3$p_4$ with $p_3$ = $p_4$ or $p_3$ < $p_4$.
By observation, one can realize that, 5 is the only smallest factor.
So, $p_2$= [(6h -1)/$p_1$] $\le $ [(6h -1)/5] Or $p_4$= [(6h +1)/$p_3$] $\le $ [(6h +1)/5].
If we consider m = h+6-($p_1$+$p_2$) and n = h+6-($p_3$+$p_4$) the following are true. I am not sure how far I am true first of all.
(1) $p_1$ = ½ (h-m+6) - square root of {$(n-m+6)^2$-4(6h-1)}
(2) $p_2$ = ½ (h-m+6) + square root of {$(n-m+6)^2$-4(6h-1)}
If (1) and (2) are correct, please explain with proof. If any one or both are wrong, please let me know, where, I am wrong.
Thanking you one and all.