* Find all prime numbers $p, q, r$ that satisfy the equation $p(p+1)+q(q+1)=r(r+1)$.
Analyzing the equation has only one solution, that is, $p=q=2, r=3$.
We consider a more general equation $p(p+1)+q(q+1)=n(n+1)$, where $p, q$ are prime numbers, and $n$ is a positive integer. Transform the equation to $p(p+1)=n(n+1)-q(q+1)=(n-q)(n+q+1)$. Clearly, $n>q$. Since $p$ is a prime, $p \mid n-q$ or $p \mid n+q+1$. If $p \mid n-q$, then $p \leqslant n-q$, $p(p+1) \leqslant(n-q)(n-q+1)1$. From (1), it is easy to get $2 q=(n+q)-(n-q)=k p-1-(n-q)=k[k(n-$ $q)-1]-1-(n-q)-(k+1)[(k-1)(n-q)-1]$, the positive integer factors on the left side are only 1, 2, $q$ and $2 q$. So $k+1=q$ or $k+1=2 q$. If $k+1=q$, then $(k-1)(n-q)=3$. So $(q-2)(n-q)=3$, thus we get (1) $q-2=1, n-q=3$ i.e., $q=3, n=6, k=q-1=$ 2, hence $p=5$; or (2) $q-2=3, n-q=1$ i.e., $q=5, n=6, k=4$, hence $p=3$.

If $k+1=2 q$, then $(k-1)(n-k)=2$. So $2(q-1)(n-q)=2$. Thus we get $q-1=1, n-q=1$, i.e., $q-2, n=3$, hence $p=2$. Therefore, we have three solutions (1) $p=q-$ $2, n=3$; (2) $p=5, q=3, n=6$; (3) $p=3, q=5, n=6$. Among them, only the first set of three numbers are all prime.