Problem 1. Find the natural number $n$ that is the product of the primes $p, q$, and $r$, knowing that

$$
r-q=2 p \quad \text { and } \quad r q+p^{2}=676
$$
Solution:

We take $x=r-p=q+p$. Then,

$$
x^{2}=(r-p)(q+p)=r q+(r-q) p-p^{2}=r q+2 p^{2}-p^{2}=r q+p^{2}=276
$$

so $x=26$. And $p$ is a prime such that $26-p$ and $26+p$ are primes. We try the possible primes $p$ less than 26 and see that this only holds for $p=3$.

Thus, $p=3, q=23, r=29$ and $n=p . q . r=2001$.