Find the chance that $a^3 + b^3 \equiv 0 (\mod 3)$ We are given set of integer numbers $\{1,2, \dots N\}$. $N \ge 3$
Then perform a drawing with replacement of two elements $a$ and $b$.
Problem is to find the probability of following statement holding true:
$a^3 + b^3 \equiv 0\space (\mod 3)$
Solution draft
I believe Little Fermat theorem will be of help here, $ a^3 \equiv a \space(\mod 3)$.
From there we conclude that either $a$ and $b$ both must be divisible by 3, or one must have residue 1 and other must have 2.
Total amount of possible $a$ and $b$ pairs is $n^2$.
Case when $3|a$ and $3|b$ is achievable for ${\lfloor \frac n 3 \rfloor}^2$ pairs.
Last case holds for the same amount of pairs.
Hence, answer would be $\frac {2{\lfloor \frac n 3 \rfloor}^2} {n^2} = \frac 2 9$
Is that a correct answer and is that legal to drop lowerbound around $\frac n 3$ here?
Thanks!
Edit
Solution above is wrong.
Suppose, $N=4$
than (3,3) (2,1) (2,4) (1,2) (4,2) fits, hence $P= \frac 5 {16}$.
I would appreciate some suggestions.
Edit 2
$$
    P = \begin{cases}
    \frac 1 3 & if N \equiv 0 (\mod 3 )\\
    \frac {{\lfloor \frac n 3 \rfloor}^2 + 2 \cdot \lfloor \frac n 3 \rfloor \lceil \frac n 3 \rceil}{n^2} & N \equiv 1 (\mod 3 )\\
\frac {{\lfloor \frac n 3 \rfloor}^2 + 2 \cdot \lceil \frac n 3 \rceil \lceil \frac n 3 \rceil}{n^2} & N \equiv 2 (\mod 3 )
\end{cases}$$

you can write all possible remainders by divisibility of $3$ these are $0,1,2$ and you will get
$$a^3+b^3 \equiv \mod 0+0;1+0;0+1;1+1;2+0;0+2;2+1;1+2;2+2, 3$$
now we will find $$0+0;2+1;1+2\equiv 0 \mod 3$$
