Does an expression exist such that... Can you prove or disprove the existence of an expression P, such that 
$Z=6ab+a+b-P$ 
Makes Z expressible in the form;
$Z=6xy\pm x \pm y$ 
for all a and b
where $a,b,x,y∈N $
Finding an example of P would act as sufficient proof.
$P$ cannot involve the variables $a$ and $b$
Clarification:
Assume instead we are using the expression $2K+1$
Hence;
$N=2k+1-P$
In this scenario, letting P= 2r would suffice, since then ; 
$N=2(k-r)+1$
Hence N would be of the form $2Q+1$
 A: I have rewritten this, as the previous version was a little rushed.
If zero may appear then, for any expression $P$,
$6ab + a + b - P = 6(6ab + a + b - P)(0) + (6ab + a + b - P) + (0)$
Otherwise suppose we have non-zero expression $P$ such that for any non-zero $a$, $b$
$(6ab + a + b) - P = (6xy + x + y)$ for some non-zero $x$, $y$
then 
$(36ab + 6a + 6b) - 6P = (36xy + 6x + 6y)$
$(6a+1)(6b+1) - 6P = (6x+1)(6y+1)$
So $6P$, when subtracted from any number expressible as the product of two numbers $\equiv 1 \mod 6$, neither equal to 1, (i.e. a composite number $\equiv 1 \mod 6$*) gives another such number. Hence we can show this is impossible if for any $P$ there exists a prime number $n$, and a composite number $m \equiv 1 \mod 6$, such that $n + 6P = m$.
* Note that since negative numbers are allowed every composite $\equiv 1 \mod 6$ can be expressed with factors $\equiv 1 \mod 6$ e.g. $55 = -5 * -11 = (6(-1)+1)*(6(-2)+1)$ 
Take any prime number $p \equiv 1 \mod 6$ and consider the sequence $p$, $p+6P$, $p+12P$, ..., $p+6(p-1)P$, $p+6pP$. Now $p$ is prime, and $p+6pP$ is composite ($P$ non-zero) , so somewhere the sequence switches from prime to composite, and we have a composite $\equiv 1 \mod 6$ that is $6P$ greater than a prime.
Hence there is no such $P$.
