Solve system of congruences $k^2 + l^2 \equiv 0 \mod 17 \wedge k^3+l^3 \equiv 0 \mod 17$

Question

Solve system of congruences $$k^2 + l^2 \equiv 0 \mod 17 \wedge k^3+l^3 \equiv 0 \mod 17$$

Is there any faster way to solve that congurence than looking at table and finding such pairs of $i$ that in both cases it gives $0$? When we have this table it is very fast job but making it is... time-consuming \begin{array}{|c|c|c|c|} \hline i^2 \mod 17 & i^3 \mod 17 \\ \hline 0 & 0 \\ \hline 1 & 1 \\ \hline 4 & 8 \\ \hline 9 & 10 \\ \hline 16 & 13 \\ \hline 8 & 6 \\ \hline 2 & 12 \\ \hline 15 & 3 \\ \hline 13 & 2 \\ \hline 13 & 15 \\ \hline 15 & 14 \\ \hline 2 & 5 \\ \hline 8 & 11 \\ \hline 16 & 4 \\ \hline 9 & 7 \\ \hline 4 & 9 \\ \hline 1 & 16 \\ \hline 0 & 0 \\ \hline 1 & 1 \\ \hline 4 & 8 \\ \hline \end{array}

Answer 1

$\left.\begin{align}\rm Squaring &\ \ k^{\large 3}\equiv -l^{\large 3}\,\Rightarrow\, k^{\large 6}\,\equiv\, l^{\large 6}\\ \rm\&\ cubing &\ \ k^{\large 2}\equiv -l^{\large 2}\,\Rightarrow\, k^{\large 6}\!\equiv -l^{\large 6}\end{align}\right\} $ $\Rightarrow\, 2l^{\large 6}\equiv 0\,\Rightarrow\, l\equiv 0\ $ (in any domain with $2\not\equiv 0)$

Answer 2

Clearly if one of them is $0$ in modulo $17$ the other is. We show no other solutions exist.

Assume both $k, l$ are nonzero. Note that since $k^2 + l^2 \equiv 0$, we multiply both sides by $k+l$ and remove the $k^3+l^3$ to get $k^2l + l^2k =0$. Dividing both sides by $lk \not \equiv 0$ gives $k+l \equiv 0$. Squaring gives $k^2+l^2 + 2kl = 0$ which gives $2kl \equiv 0$; dividing by $2$ gives $lk \equiv 0$ which implies one of them is $0$ - a contradiction.

Answer 3

From $k^3 + l^3 \equiv 0 \pmod{17}$, we have $$k^3 \equiv (-l)^3 \pmod {17}.$$

But, as you can see from your table, cubes are distinct modulo $17$, which effectively lets us take cube roots: thus, $k \equiv -l \pmod {17}$. Therefore $$k^2 + (-k)^2 \equiv 0 \pmod {17},$$ so $2k^2 \equiv 0 \pmod {17}$. Divide by $2$ and conclude that $k \equiv l \equiv 0 \pmod{17}$.