What is the simplest Diophantine equation equivalent to N is not zero? Given an integer N. What is the simplest Diophantine equation equivalent to the statement that integer $N\neq0$ ?
I can do one in 5 variables. Using the fact that any integer can be written as the sum of four squares.
$$N^2-(1+A^2+B^2+C^2+D^2)^2=0$$
So this can only be solved for integers $A,B,C,D,N$ when $N\neq 0$ and can be solved for all other positive and negative integers $N$.
Can it be done with fewer variables?
 A: Consider the equation:
$$
NX = (2Y - 1)(3Y - 1)
$$
Clearly, as $2$ and $3$ are not invertible in the ring of integers, if there is a solution $(X, Y)$ to this equation, then $N \neq 0$.
Conversely, suppose $N \neq 0$; we have to show that $(2Y - 1)(3Y - 1)$ has a root modulo $N$. This follows from the Chinese Remainder Theorem, since clearly $(2Y - 1)(3Y - 1)$ has a root modulo every prime power.
A: For every prime power $p^n$, the congruence $A^2+AB+B^2 \equiv -7 \pmod{p^n}$ has a solution by Hensel's lemma.(*)
By the Chinese remainder theorem, we can take
$$A^2+AB+B^2+7= NC$$
because the LHS is always $>0$.

(*)


*

*For $p \neq 7$ odd this is because a nondegenerate binary quadratic form represents the whole of $\mathbb F_p^\times$. The (nonsingular) solution lifts to $p^n$ by Hensel. 

*For $p=2$ it suffices to lift $(1, 0)$ from $2^3$ to $2^n$.

*For $p=7$ use that the discriminant $-3$ is a quadratic residue, because $7 \equiv1\pmod3$



You can replace $7$ by any prime congruent to $7$ or $13$ mod $24$.
