15. Let $p \geqslant 3$, try to calculate the value of the following expression:
$$\left(\frac{1 \cdot 2}{p}\right)+\left(\frac{2 \cdot 3}{p}\right)+\cdots+\left(\frac{(p-2)(p-1)}{p}\right)$$
15. Solution: To solve this problem, we need to study the properties of the Legendre symbol with the general term $\left(\frac{n(n+1)}{p}\right)$, where $(n, p)=1$.

By $(n, p)$ being coprime, we know there must exist an integer $r_{n}$ such that $p \nmid r_{n}$ and $n r_{n} \equiv 1(\bmod p)$. This $r_{n}$ is called the modular inverse of $n$ modulo $p$. From the properties of the Legendre symbol, it is easy to see that
$$\begin{aligned}
\left(\frac{n(n+1)}{p}\right) & =\left(\frac{n\left(n+n r_{n}\right)}{p}\right)=\left(\frac{n^{2}\left(1+r_{n}\right)}{p}\right) \\
& =\left(\frac{n}{p}\right)^{2}\left(\frac{1+r_{n}}{p}\right)=\left(\frac{1+r_{n}}{p}\right)
\end{aligned}$$

We will prove that for $n \neq m(\bmod p), p \nmid n m$, it must also be true that
$$r_{n} \not\equiv r_{m}(\bmod p)$$

That is, different numbers in the reduced residue system modulo $p$ must correspond to different inverses. We use proof by contradiction. If $r_{n} \equiv r_{m}(\bmod p)$, multiplying both sides by $n m$ gives
$$n\left(m r_{m}\right) \equiv m\left(n r_{n}\right)(\bmod p)$$

Then, by the definition of the inverse, we get
$$n \equiv m(\bmod p)$$

This leads to a contradiction. This proves that when $n$ runs through $1,2, \ldots, p-1$, the corresponding inverse $r_{n}$ also runs through $1,2, \ldots, p-1(\bmod p)$, just in a different order. Also, note that from $p-1 \equiv-1(\bmod p)$, we immediately get
$$(p-1)^{2} \equiv(-1)^{2} \equiv 1(\bmod p)$$

So $r_{p-1}=p-1$. Therefore, when $n$ takes $1,2, \ldots, p-2$, the inverse $r_{n}$ of $n$ also takes $1,2, \ldots, p-2(\bmod p)$, just in a different order. Thus, we get
$$\begin{array}{l}
\left(\frac{1.2}{p}\right)+\left(\frac{2.3}{p}\right)+\ldots+\left(\frac{(p-2)(p-1)}{p}\right) \\
= \sum_{n=1}^{p-2}\left(\frac{1+r_{n}}{p}\right)=\sum_{r=1}^{p-2}\left(\frac{1+r}{p}\right) \\
=\sum_{r=1}^{p-1}\left(\frac{r}{p}\right)-\left(\frac{1}{p}\right)
\end{array}$$

Since in a reduced residue system modulo $p$, there are exactly $\frac{p-1}{2}$ quadratic residues and $\frac{p-1}{2}$ quadratic non-residues, we have $\sum_{r=1}^{p-1}\left(\frac{r}{p}\right)=0$. Therefore, the required sum is $-\left(\frac{1}{p}\right)=-1$.