The truth of Nyonyon Theorem Let $s= ab$ be a semiprime number. Then the Nyonyon Theorem states that $s+a$, $s+b$, $s+a+b$ are not all coprime to three.
(In other words: there exist no $s= ab$ semiprimes such that $s+a$, $s+b$ and $s+a+b$ are coprime to three.)
I've checked plenty of numbers, and I think Nyonyon Theorem is absolutely true, but I don't understand why this phenomena happens. Can someone explain/proof to me why this thing happens ?
 A: If $s$ is coprime to $3$, then both $a,b$ are coprime to $3$. We have three possibilities. 


*

*If $a \equiv b \equiv 1 \pmod 3$, then $s \equiv 1 \pmod 3$ and so $s + a + b$ is divisible by $3$.

*If $a \equiv 1 \pmod 3$ and $b \equiv 2 \pmod 3$, then $s \equiv 2 \pmod 3$ and so $s + a$ is divisible by $3$. (Similarly if the roles of $a,b$ are reversed).

*If $a \equiv b \equiv 2 \pmod 3$, then $s \equiv 1 \pmod 3$ and $s + a$ and $s + b$ are divisible by $3$.


This proves the theorem. $\diamondsuit$
A: If none of $$ab+a,\quad ab+b,\quad ab+a+b $$
are divisible by three, then none of 
$$ a(b+1),\quad b(a+1),\quad (a+1)(b+1)-1 $$
are divisible by three. So $a,b\not\in\{0,2\}\pmod{3}$, that implies $a\equiv b\equiv 1\pmod{3}$. 
But in such a case:
$$ (a+1)(b+1)-1 \equiv 2\cdot 2-1\equiv 0\pmod{3}.$$
A: Assume the existence of such a semiprime $s=ab$,we have $s+b=b(a+1)$ and $s+a=a(b+1)$ are all coprime to $3$ hence $a\equiv b\equiv 1 \mod 3$ but in this case $s+a+b\equiv 1+1+1 \equiv 0\mod 3 $ contradiction
