Consider the function $x^m \pm y^n \pm z^p$, where $x, y, z, m, n, p$ are integers such that $m, n, p \geq 2$. The question is, are all numbers expressable using this function? Are there any exceptions? (I know about numbers conjectured to be not expressable as $x^m  y^n$, so, I am asking an extension question). Is this formulation something new, or, is it known before?

2$\begingroup$ Every integer is of the form $x^2y^2$ or $x^2y^2+1$, and thus of the form $x^2y^2+z^p$ for any $p$ (where we can take $z=0$ or $z=1$), so you need to rule out this situation. $\endgroup$ – Richard Stanley Dec 22 '12 at 4:01

$\begingroup$ You are right, we will exclude that solution. $\endgroup$ – Salahuddin559 Dec 22 '12 at 4:19

$\begingroup$ Particularly, $z$ can be zero, but none of $x, y$ and $z$ can be $1$. $\endgroup$ – Salahuddin559 Dec 22 '12 at 5:05
Your equation has infinitely many efficiently computable solutions $(m,n,p)=(2,2,p)$ and maybe infinitely many $(m,n,p)=(3,3,2)$ for all integers $a$ s.t. $x^m  y^n \pm z^p =a$.
Every odd integer $b$ is the difference of two squares $(1+b)/2,(b1)/2$ so take any $p$, any $z$ the opposite parity of $a$ and express $a \mp z^p$ as difference of two squares.
For the $(3,3,2)$ case, pick random $c$ and set $x=u+c,y=u$ and pick the minus sign. Your equation is $x^3y^3z^2=3u^2c + 3uc^2 + c^3 z^2=a$ leading to $ 3 u^{2} c + 3 u c^{2} + c^{3}  z^{2}  a=0$. The last equation is genus $0$ curve which might have infinitely many integral points unless I am missing some obstruction and you can try another $c$. For $c=3 c'^2$ the curve has explicit rational parametrization.