The system is the following:
$$x^{10}+y^{21}=1$$ $$x^8+y^{15}=1$$
I know it' s non empty, since $(1,0)$ and $(0,1)$ are two solutions, but how can I show that it has finitely many solutions?
Solutions of a system of $2$ non-linear equations.
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algebraic-geometry
computational-algebra
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1By Bezout's theorem, this can have at most $15 \times 21$ solutions. In fact, $15 \times 21$ is the number of complex solutions, counted with multiplicity in case the graphs are tangent, etc., and with solutions at "points at infinity" included. See https://en.wikipedia.org/wiki/B%C3%A9zout's_theorem – 2017-02-28
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1The solutions are $(x,y)=(0,1),(\pm 1,0),(0,\pm(\sqrt{-3} - 1)/2)$ and the complex solutions given by a polynomial equation $x^2=2y^{51} + 5y^{48}+\cdots + 11$ with $y^{57} + 4y^{54}+\cdots +5=0$, coming from a resultant equation. – 2017-02-28
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0Is there a more fundamental solution, for example without using Bezout's theorem? @KennyWong – 2017-02-28
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0How did you find the solution? Is there an algorithm or something? @DietrichBurde – 2017-02-28
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1Yes, of course. We may apply [Buchberger's algorithm](https://en.wikipedia.org/wiki/Buchberger's_algorithm), which gives all soltions (up to finding the roots of a polynomial $f(y)=0$). – 2017-02-28
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0@DietrichBurde: The poly $y^{57}+4y^{54}+\ldots + 5$ is odd and has a real root $\approx - 1.076$, giving two more real points $(\pm 1.18932793584109, - 1.076067350438112)$. – 2017-03-30