# Diophantine equation and NP-Completness

I found the following question (and answer) on Mathoverflow: https://mathoverflow.net/questions/36420/is-the-solution-bounded-diophantine-problem-np-complete.

I cite:

Let a problem instance be given as $(\phi(x_1,x_2,\dots, x_J),M)$ where $\phi$ is a diophantine equation, $J\leq 9$, and $M$ is a natural number. The decision problem is whether or not a given instance has a solution in natural numbers such that $\sum_{j=1}^J x_j \leq M$. With no upper bound M, the problem is undecidable (if I have the literature correct). With the bound, what is the computational complexity? If the equation does have such a solution, then the solution itself serves as a polytime certificate, putting it in NP. What else can be said about the complexity of this problem?

According to the source it is NP-complete problem. But why? Is not that a mistake?

We let us take the following problem:

We have diophantine equation:

$\phi(A, B, C, x, y, z) = A^x + B^y = C^z$

We are looking for solutions according to the quotation, such that:

$A + B + C + x + y + z \le M$

M is determined.

Computational complexity measure after M.

Is it a problem NP-complete? I'm not sure whether I understand everything.

• The answer you link shows that deciding the following: "given $a,b,c$ in binary, does the diophantine equation $ax^2+by=c$ have solutions" is NP-complete, where the complexity is measured w.r.t. the size of the input (i.e., $\log_2 a+\log_2 b+ \log_c$). – Clement C. Feb 4 '17 at 21:07
• Yes, but how this response relates to the question? This is just one very special case. The question is general. Additionally, in the comment to the question is: " ... answered, the problem in the revised (v2) question is NP-complete." (v2, because you can see that the question was edited). – Aurelio Feb 4 '17 at 21:33
• v2 is the current version you link. As for the previous comment: if a problem P is known to be in NP, and a restricted version (special case) P' of P is NP-complete, then since P is at least as hard as P' it must be the case that P is NP-complete (being in NP and NP-hard). – Clement C. Feb 4 '17 at 21:35
• Ahhh. Ok. I understand. My diophantine equation is also a special case in the problem of the NPC. However, it not must be in NPC. Some idea of the proof is there in the NPC (or not?)? – Aurelio Feb 4 '17 at 22:00
• I see -- in your case, you have coefficients $a=b=1$. (The link above states one can still have NP-completeness with $a=1$, but it's not clear that enforcing $a=b=1$ preserves this). I'd suggest you have a look at the paper proving the NP-completeness in the first place. – Clement C. Feb 4 '17 at 22:04