TITLE: solve for every k and l: $k^{2} + l^{2} \equiv_{17} 0$ and $k^{3} + l^{3} \equiv_{17} 0$
QUESTION [0 upvotes]: solve set of equations for every k and l: $$k^{2} + l^{2} \equiv_{17} 0$$ and  $$k^{3} + l^{3} \equiv_{17} 0$$
Any hints?
$$k^{3} \equiv_{17} - l^{3}$$
$$k^{2} \equiv_{17} - l^{2}$$
$$k^{2}l \equiv_{17} - l^{3}$$
$$k^{2}l \equiv_{17} k^{3}$$
$$k \equiv_{17} l$$
$$k - l\equiv_{17} 0$$

REPLY [1 votes]: Obviously in the field $\Bbb F_{17}$ the equation $x^2+y^2=0$  and $x^3+y^3=0$ have solution $x=y=0$
Supposing $x\ne 0$ one has $x^2+y^2=0\iff x^2=-y^2\Rightarrow y\ne 0$ But then
$$x^2=-y^2\\x^3=-y^3$$ implies $$x(-y^2)=-y^3\Rightarrow x=y$$ It follows
$$2x^2=0$$ which it is absurde in the field $\Bbb F_{17}$ for $x\ne 0$.
Thus the only solution is the obvious one.
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