5

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?

zooby
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2 Answers2

2

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.

Bib-lost
  • 3,890
1

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 $24$.

Bart Michels
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