Three variable equation using only natural numbers I am a high school student that has been working on amateur mathematics for about a year now. Lately, I have been encountering a particular type of equation that I have not yet solved. If I could figure out this equation then I could move on to the next part to solving the conjecture.
$$z = Axy \pm x \pm y$$
You can fill in any value for $A$. $x$ and $y$ are integers greater than $0$.
So for example $A=6$
$$z = 6xy \pm x \pm y$$
Does anyone know of a method I could use to prove that there is an infinite amount of numbers that do not equal $z$. I have been working on this problem for about two months with not much gain.
Thank You
 A: Maria Suzuki found (and subsequently published in the January 2000 edition of American Mathematical Monthly) that the Twin Primes Conjecture is equivalent to the statement
"There exist an infinite number of values $k\in\Bbb Z^+$ such that for all $a,b\in\Bbb Z\setminus\{0\}$,
$$k\ne 6|ab|+a+b$$
" (end quote, variable names possibly changed).
I've worked on this particular direction for a long time myself.  Let me know if you'd like more information on what I've found...
The direct creation of this inequality is as follows:
If we assume that $p,p+2$ are each prime numbers (thus in $\Bbb Z^+$), then the number $p+1$ must in general be divisible by $6$, since otherwise either $p$ or $p+2$ will be divisible by either $2$ or $3$.  So then we have $p=(6k-1),p+2=(6k+1)$, and we also have $p=(6a\pm1)(6b\pm1),p+2=(6a\pm1)(6b\pm1)$ if either $p,p+2$ are composite.  The expansion of these equations results in four equations, but if we take $a,b\in\Bbb Z\setminus\{0\}$, then they can be rewritten as
$$6k\pm1\ne (6a\pm1)(6b\pm1)=36|ab|+6a+6b\pm1$$
For the equation to be consistent modulo $6$, the $\pm1$ terms must cancel, and then we can divide by $6$ everywhere to get
$$k\ne 6|ab|+a+b$$
which demonstrates the required result.
Note that the prime numbers fall under the equations
$$k\ne ab+a+b\\
k\ne 2ab+a+b$$
for any prime $k+1$ in the first equation and for any odd prime $2k+1$ in the second equation, $\forall a,b\in\Bbb Z^+$.
In particular, if $k+1$ is prime, then there are no divisors of the form $a+1,b+1$ for $a,b\in\Bbb Z^+$, therefore $k+1\ne(a+1)(b+1)=ab+a+b+1\implies k\ne ab+a+b$.  The same logic holds for odd primes $2k+1$ where the divisors must be odd in the form $2a+1,2b+1$.
