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

$\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 $2\not\equiv 0)$
