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Suppose A is Diophantine Equations which have a variable x,such that every other variable has function of x the variable as it's solution.For example,a Diophantine equation in three variables has two variables y,z, $$ y = x^2+1$$ $$ z = 4x^6+2$$ as their solutions.

All Diophantine Equations like A form B, a special class of Diophantine Equations.

EDIT:suppose $$P_1(x,y_1,\cdots,y_n)=P_2(x,y_1,\cdots,y_n)=\cdots=P_k(x,y_1,\cdots,y_n)=0$$ is Diopantine equations ,and there are functions$$y_1=f_1(x),\cdots,y_n=f_n(x)$$,which satisify the Diopantine equations.

Question: is any function of x for solution of other variables of such Diophantine Equations algebraic? or are $$y_1=f_1(x),\cdots,y_n=f_n(x)$$ all algebraic?

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    $\begingroup$ The question really isn't well posed. The theory of diophantine equations for all curves of genus 0 (i.e. all "space curves" that are rational curves from a geometric point of view) was written down by Hilbert and Hurwitz. It is in Mordell's books. If that is not essentially the question you are asking, I'm not really understanding the question. $\endgroup$
    – Charles Matthews
    Commented May 30, 2013 at 12:42
  • $\begingroup$ @François,thank you very much for your edit $\endgroup$
    – XL _At_Here_There
    Commented May 30, 2013 at 13:05
  • $\begingroup$ @Charles,I think there must be some misunderstanding,I think I am not asking about question of curves of genus 0.maybe some of them are about curves of genus 0.The example is just a very special case. $\endgroup$
    – XL _At_Here_There
    Commented May 30, 2013 at 13:27
  • $\begingroup$ If each variable can be written as a rational function of x, then the curve has genus zero. If not, then what does "every other variable has function of x the variable as it's solution" mean? $\endgroup$
    – quim
    Commented May 30, 2013 at 14:29
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    $\begingroup$ So we have x, f(x), g(x), h(x) ... where f, g, h are certain functions. What now? Which equations are concerned? I have a feeling you want something about implicit algebraic functions. $\endgroup$
    – Charles Matthews
    Commented May 30, 2013 at 15:00

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