Diophantine Sets

 Diophantine set

is simply set of positive integers for which a particular Diophantine equation has solution.

For example, if we have a polynomial equation of single variable,

$x^2=4$

Then the Diophantine set for this equation contains $4$ as above equation can be factored into integers.

Now, instead of $4$, we use a $z \in \mathcal{Z}$, then we get lots of elements for Diophantine set.

In two variables, say our equation of interest is,

$(x+1)(y+1)=z$.

Here, we are looking for $z$. Based on above equation, we are seeking composite numbers.

The following Mathematica code generates Diophantine set for the first 20 integers - that is $z=1,2,...$.

f[z_] := FindInstance[z == (1 + y) (1 + x), {x, y}, PositiveIntegers]
S = Map[f, Range[20]]
Complement[Range[20], Position[S, {}] // Flatten]

Yields
$\{4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20 \}$

Which are elements of Diophantine set, that is set of integers for which above polynomial has solution. Notice we restricted the range to first 20 integers. Same Mathematica expression can be extended for higher ranges.

From Wik, lets take the sum of squares example.

$z = x^2+y^2$

We can reuse above Mathematica expresions to find the Diophantine set elements when we restrict to first 20 integers.

ss[z_] := FindInstance[z == x^2 + y^2, {x, y}, PositiveIntegers]
SS = Map[ss, Range[20]]
Complement[Range[20], Position[SS, {}] // Flatten]

Yields
{2, 5, 8, 10, 13, 17, 18, 20}
Another example is beautiful Pell equation which manages to avoid all the perfect squares.

Mathematica code given below.

pel[z_] := FindInstance[z == (y^2 - 1)/x^2, {x, y}, PositiveIntegers]
PEL = Map[pel, Range[20]]
Complement[Range[20], Position[PEL, {}] // Flatten]

Yields
{2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20}

Mathematica code written above looks for solutions in positive integer space. However, in general, both positive and negative integer solutions are permitted.

Based on these examples, one can understand the definition of Diophantine set.

A set $A \subset Z$ is diophantine if there exists a ploynomial $p(t,\overrightarrow{x})=\{a \in Z|(\exists \overrightarrow{x} \in \mathcal{Z}^n) p(a,\overrightarrow{x})=0$.

Here we think of $p$ as a family of polynomials depending on parameter $t$. Then $A$ captures values of $t$ for which $p$ has solution.