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 $x^2-y^2$ or $x^2-y^2+1$, and thus of the form $x^2-y^2+z^p$ 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 $(1+b)/2,(b-1)/2$ 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 $x^3-y^3-z^2=3u^2c + 3uc^2 + c^3 -z^2=a$ 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.