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How many ordered triples $(x,y,z)$ of real numbers are there such that

$x+y^2=z^3$

$x^2+y^3=z^4$

$x^3+y^4=z^5$?

Adding those equations together and factorizing gives you $x(1+x+x^2)+y^2(1+y+y^2)=z^3(1+z+z^2)$

Which is of the same form as the 1st equation, but I don’t know how to use this to my advantage. Hints, suggestions and solutions would be appreciated.

Taken from the 2018 BIMC

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1 Answer 1

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Hint

From $z^3\cdot z^5=(z^4)^2$ we obtain $$(x+y^2)(x^3+y^4)=(x^2+y^3)^2$$which leads to $$xy^2(x-y)^2=0$$

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