system of congruence - my approach We have:
$$k^3 + l^3 \equiv 0 \pmod{17}\\
k^2 + l^2 \equiv 0 \pmod{17} $$
And I get:
$$k = 17n+r_k\\
l = 17m+r_l$$
And I analyzed possible rests respect to system of congruences.
My result is: 
$$ k,l \equiv 0\pmod{17}$$
But is it properly ? Is my way is ok ? Is exists other, better solution ?
 A: When we are working with a prime modulus, we can use the fact that $p|ab$ implies that either $p|a$ or $p|b$. In terms of modular arithmetic this is the same as $ab\equiv 0 \mod p$ implies $a\equiv 0$ or $b\equiv 0$. Here we can use this observation to avoid having to deal with lots of cases.
Note that the first equation is equivalent to: $$(k+l)(k^2-kl+l^2)\equiv 0\mod 17$$
Because $17$ is prime we have either $k+l\equiv 0$ or $k^2-kl+l^2\equiv 0$
In the first case we use the second (original) congruence as well and find $(k+l)^2-k^2-l^2=2kl \equiv 0$ so that $kl\equiv 0$
In the second case $k^2-kl+l^2-k^2-l^2=-kl \equiv 0$
In either case $kl\equiv 0$ so one of $k,l$ must be divisible by $17$ - and it is easy to see that this implies both must be multiples of $17$.
A: For any integer $a,a\equiv0,\pm1,\pm2,\pm3,\pm4,\pm5,\pm6,\pm7,\pm8\pmod{17}$
$$\implies a^2\equiv0,1,4,9,16\equiv-1,8,2,15,13$$
Find which combinations of $(k,l)$ gives $k^2+l^2\equiv1\pmod{17}$ namely, obvious sets being $(0,\pm1);(\pm3,\pm3),(\pm4,\pm6)$
Find which of them, satisfy  $k^3+l^3\equiv1\pmod{17}$
Answer to the EDITED Question
Observe that  $17|k\iff 17|l$
Else we have $$(k\cdot l^{-1})^2\equiv-1\ \ \ \ (1),(k\cdot l^{-1})^3\equiv-1\pmod{17}\ \ \ \ (2)$$
$k\cdot l^{-1}\equiv\pm4,$ satisfy $(1)$ but not $(2)$
A: Let $m\equiv k+l \pmod{17}$
Now $(k+l)^3 = 3(k+l)(k^2 + l^2) - 2(k^3 + l^3)$ therefore:
$$m^3 \equiv 3m - 2$$
Try what values in $\{0\ldots 16\}$ satisfy that equation and then see what combinations of $m= k+l$ satisfy the original equations.
A: I try present my approach:
$$a\equiv0,\pm1,\pm2,\pm3,\pm4,\pm5,\pm6,\pm7,\pm8\pmod{17}\\
\implies a^2\equiv0,1,4,9,16\equiv-1,8,2,15,13$$
Possible pairs:
$$(0,0),(\pm4,\pm1),(\pm16,\pm8),(\pm3,\pm5),(\pm6,\pm7) $$
And now for $k^3+l^3$:
$$(0,0),(2,-2),(3,-3),(4,-4),(5,-5),(6,-6),(7,-7),(8,-8) $$
And the joint is $(0,0)$
Ok ?
