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I have a family of Diophantine equations that I am trying to solve, and I am trying to figure out what methods could be used to prove existence of solutions. Unfortunately, the equations are inhomogeneous and the techniques that I am used to using don't seem to apply. For one example, I am curious for which integers $n$ there exist integers $x,y,z$ all strictly greater than one so that $xyz-xy-xz-yz+y=n$. I can come up with some ad hoc methods for particular examples in my family, but I was curious if someone could point me to more general references on approaches to this type of question.

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  • $\begingroup$ To those people who are voting to close: could you please explain why? Perhaps I am missing something, but this question seems non-trivial to me. $\endgroup$ Commented May 19, 2015 at 20:40
  • $\begingroup$ Thanks, but what if I cannot do this? Are there other approaches for various families? Alternately, what if I wanted to limit solutions to integers greater than 2? $\endgroup$
    – user61388
    Commented May 19, 2015 at 21:52
  • $\begingroup$ I am not sure what kind of answer you are looking for. In general one cannot expect a simple description for the set of numbers represented by a specific function. Are you looking for things like whether a given polynomial represents all numbers? $\endgroup$ Commented May 20, 2015 at 1:45
  • $\begingroup$ I know that in general this is a hard question, but in the cases of both Homogeneous quadratic forms and Frobenius-number types of problems there are techniques that can be used for large families of examples. I was curious if anyone knew of any such techniques here... $\endgroup$
    – user61388
    Commented May 20, 2015 at 3:03

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$(x,y,z)=(n+3,3,2)$ solves your example. Your form is of the form $x\cdot a(y,z) + b(y,z)=n$. If you can choose $y,z$ such that $a(y,z)=1$ you have a solution for all $n$ large enough.

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  • $\begingroup$ $a(y,z)=yz-y-z=(y-1)(z-1)-1$, so the only way to get $a(y,z)=1$ is $\{\,y,z\,\}=\{\,2,3\,\}$. $\endgroup$ Commented May 19, 2015 at 23:20

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