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

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Could you give me an example of such an polynomial?? How can it be that $x \in \mathbb{Z}^k$ and $y \in \mathbb{Z}^m$ ?

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EDIT:

Could you explain to me the following sentence??

It is evident that a diophantine set is the projection, in the above notation onto the $x$ hyperplane, of the solution set of a diophantine polynomial.

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  • $\begingroup$ I have not encountered the phrase "diophantine polynomial" before. Can you define it or give a reference? $\endgroup$ – lulu Jul 14 '15 at 17:16
  • $\begingroup$ A diophantine polynomial is a polynomial in several variables with integer coefficients. @lulu $\endgroup$ – Mary Star Jul 14 '15 at 17:18
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    $\begingroup$ Do you need example of polynom with integer coefficients? Really? $\endgroup$ – Michael Galuza Jul 14 '15 at 17:21
  • $\begingroup$ But, then what's the problem? $P(x_1,x_2,y_1) = x_1^2+ x_2 ^2+ y_1^2 +10$ or anything else you could write down, really. $\endgroup$ – lulu Jul 14 '15 at 17:22
  • $\begingroup$ No... I want to understand how it can be that $x \in \mathbb{Z}^k$ and $y \in \mathbb{Z}^m$... How can i be that they have different number of coordinates? @MichaelGaluza $\endgroup$ – Mary Star Jul 14 '15 at 17:23
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Suppose k = 2, m = 3. So, $x = (x1, x2), y = (y1, y2, y3)$, now consider this equation, $P(x,y) : (3,4).x + (-9,1,0).y = 0$. Here, dot stands for inner vector product. If $x1, x2, y1, y2, y3$ are all integers for some $x$, we can say $x$ is an element of the Diophantine Set defined for $P(x,y)$

the projection, in the above notation onto the x hyperplane, of the solution set of a diophantine polynomial.

Now, projecting the x would require a 2D plane, as k = 2. We can find infinitely many x vectors for the given P(x, y) , which can be plotted at this 2D plane. Please note that projecting on x hyperplane doesn't require reduction of dimensionality, and we are only plotting x tuples. Why? Because, your definition means, we will only add those x tuples, for which there is some y tuples so that P(x,y) = 0.

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  • $\begingroup$ I see... Thank you very much!! :-) $\endgroup$ – Mary Star Jul 15 '15 at 1:23

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