solve for every k and l: $k^{2} + l^{2} \equiv_{17} 0$ and $k^{3} + l^{3} \equiv_{17} 0$

Question

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?

$$k^{3} \equiv_{17} - l^{3}$$ $$k^{2} \equiv_{17} - l^{2}$$ $$k^{2}l \equiv_{17} - l^{3}$$ $$k^{2}l \equiv_{17} k^{3}$$ $$k \equiv_{17} l$$ $$k - l\equiv_{17} 0$$

Answer 1

If $l \equiv 0$ then $k^2 \equiv 0$ and, since $\Bbb Z _{17}$ is a field, it follows that $k \equiv 0$.

Suppose, then, that $l \not\equiv 0$ and divide the first equation by $l^2$ and the second by $l^3$. Letting $x \equiv k l^{-1}$ gives you

$$x^2 \equiv -1 \\ x^3 \equiv -1 .$$

Subtracting the first equation from the second gives $x(x-1) \equiv 0$ and, since $k,l \not\equiv 0$, we shall also have $x \not\equiv 0$, so $x^2 \not\equiv 0$, so we may divide by it and get $x-1 \equiv 0$, so $x \equiv 1$ - but this is easily seen not to be a solution.

Therefore, the only solutions remain the one we found first, namely $k \equiv l \equiv 0$.

Answer 2

$k^3 + l^3 \equiv 0 \pmod {17}\\ (k+l)(k^2 - lk+ l^2) \equiv 0 \pmod {17}$

either$(k+l)\equiv 0 \pmod {17}$ or$(k^2 + l^2 + lk) \equiv 0 \pmod {17}$

Case: $(k+l)\equiv 0 \pmod {17}$

$k^2 + l^2 \equiv 0 \pmod {17}\\ (k+l)^2\equiv 0 \pmod {17}\\ 2kl\equiv 0 \pmod {17}$

since 17 is prime, $k$ or $l$ are a multiple of 17 and since (k+l) \equiv 0 \pmod {17}$

both $k$ and $l\equiv 0 \pmod {17}$

Case: $(k^2 - lk+ l^2) \equiv 0 \pmod {17}$ has identical implications.

Answer 3

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.