If $p \ge 5$ is prime prove that $$\sum_{i=1}^{p-2}\sum_{j=i+1}^{p-1}ij$$
Attempt. We have $$\sum_{i=1}^{p-2}\sum_{j=i+1}^{p-1}ij = \sum_{i=1}^{p-2}i \left[ \frac{(p-1)p - i(i+1)}{2}\right] = \frac{(p-1)(p)}{2}\sum_{i=1}^{p-2}i - \sum_{i=1}^{p-2}\frac{i^2(i+1)}{2}$$
We see that the first sum is divisible by $p$ since $p-1$ is divisible by $2$. Now we only need to check that $\sum_{i=1}^{p-2}\frac{i^2(i+1)}{2}$ is divisible by $p$.We have: $$\sum_{i=1}^{p-2}\frac{i^2(i+1)}{2} = \frac{1}{2}\sum_{i=1}^{p-2}i^3 + \frac{1}{2}\sum_{i=1}^{p-2}i^2$$
My thought is to use formulas for sums of consecutive cubes and squares, so we get $$\frac{(p-1)^2(p-2)^2}{8} + \frac{(p-2)(p-1)(2p-3)}{12}$$
My only next step is to notice that $p$ is odd, so we get $$\left[\frac{(p-1)^2}{4} \right]\frac{(p-2)^2}{2} + \left[\frac{(p-1)(2p-3)}{4}\right]\frac{(p-2)}{3}$$ Where terms in brackets are integers, however i am not sure how to continue. I also wonder if there is a smarter way to approach this.
Here is another way to tackle the problem.
Another way to state the problem would be to say $$\sum\limits_{1 \le i < j \le {p-1}}ij \equiv 0 \pmod{p}$$ And notice that this is equivalent to $$\dfrac{1}{2}\left(\left(\sum_{n=1}^{p-1} n \right)^2-\sum_{n=1}^{p-1} n^2\right)$$ By algebraic manipulation.
Since $\gcd(p,2)=\gcd(p,3)=1$ so we have that $\sum\limits_{n=1}^{p-1} n=\frac{p(p-1)}{2}$ and $\sum\limits_{n=1}^{p-1} n^2=\frac{p(p-1)(2p-1)}{6}$ are both divisible by $p$.
Thus, the statement is true.