1
$\begingroup$

I am reading the following part:

Diophantine sets

A subset of a power $\mathbb{Z}^n$ of the set $\mathbb{Z}$ of integers is diophantine if it can be written as $$\{\overline{x} \in \mathbb{Z}^n : (\exists \overline{y} \in \mathbb{Z}^m ) P(\overline{x}, \overline{y})=0\}$$ where $P$ is a diophantine polynomial (that is, a polynomial in several variables-here the variables $\overline{x}$ and $\overline{y}$-with integer coefficients).

If $P$ is a diophantine polynomial, then the equation $P=0$ is called a diophantine equation.

It should be noted that the intersection and the union of diophantine sets are diophantine; this is because the intersection of the zero sets of two diophantine polynomials $P$ and $Q$ coincides with the zero set of the single diophantine polynomial $P^2+Q^2$ while the union of these sets coincides with the zero set of $P \cdot Q$.

Therefore, the question of existence of integr solutions of any disjunction of systems of diophantine equations is equivalent to the question foe one equation.

Moreover, the set of nonnegative integers is diophantine (by Lagrange's Theorem, which asserts that an intere is nonnegative if and only if i is the sum of four integer squares), and therefore, the set of nonzero integes is diophantine ($X$ is nonzero if and only if there is a nonnegative integer $y$ so that $x=y+1$ or $x=-y-1$).

Therefore, the solvability of any disjunction of systems of diophantine equations and inequations is equivalent to teh solvability of one diophantine equation.

$$$$


At the following part:

the set of nonnegative integers is diophantine (by Lagrange's Theorem, which asserts that an integer is nonnegative if and only if it is the sum of four integer squares), and therefore, the set of nonzero integers is diophantine ($X$ is nonzero if and only if there is a nonnegative integer $y$ so that $x=y+1$ or $x=-y-1$)

does it mean that the set of nonnegative integers is diophantine because we can write each integer $x$ in the form $x=a^2+b^2+c^2+y^2$, so it is the root of the diophantine polynomial $P(x,y)=x-a^2-b^2-c^2-y^2$ ?

And the set of nonzero integers is diophantine because each nonzero $x$ can be written as $x=y+1$, so it is the root of the diophantine polynomial $P(x, y)=x-y-1$ ?

$\endgroup$
1
$\begingroup$

For the first, you had it. The set of non-negative integers is the set of all (integers) $x$ such that $$\exists t_1\exists t_2\exists t_3\exists t_4 (P(x,t_1,t_2,t_3,t_4)=0),$$ where $P$ is the polynomial $x-t_1^2-t_2^2-t_3^2-t_4^2$.

The set of non-zero integers is diophantine because $x$ is a non-zero integer if and only if there is a non-negative $y$ such that $(x-y-1)(x+y+1)=0$. This is the case precisely if $$\exists y\exists t_1\exists t_2\exists t_3\exists t_4((y-t_1^2-t_2^2-t_3^2-t_4^2)^2 + ((x-y-1)(x+y+1))^2=0).$$

Note that we have used the two tricks you quoted for encoding "and" and "or."

$\endgroup$
  • $\begingroup$ So, at the first case we use the intersection of the polynomials $x$, $t_1^2$, $t_2^2$, $t_3^2$, $t_4^2$ and at the second case we use the union of the polynomials $x-y-1$ and $x+y+1$. Is this correct? $$$$ Could you explain to me further why at the second case it is $$\exists y\exists t_1\exists t_2\exists t_3\exists t_4((y-t_1^2-t_2^2-t_3^2-t_4^2)^2 + ((x-y-1)(x+y+1))^2=0)$$ ? $\endgroup$ – Mary Star Jul 24 '15 at 22:50
  • 1
    $\begingroup$ The first paragraph above is not really clear. For the second question, note that the above sum $P^2+Q^2$ of squares is equal to $0$ if and only if $P=0$ and $Q=0$, that is, if and only if $y$ is the sum of the squares of $t_1$ up to $t_4$, and one of $x-y-1=0$ or $x+y+1=0$ holds. (Soon this will snap into place. And soon you will be upset that we are summing squares, and/or multiplying, meaning that for some basic logical operations we are multiplying, or summing squares, non-linear operations.) $\endgroup$ – André Nicolas Jul 24 '15 at 23:25
  • $\begingroup$ I got it now... Thanks for the clarification!! :-) $\endgroup$ – Mary Star Jul 25 '15 at 0:32

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.