22. Let $A$ and $B$ be two positive prime integers such that
$$
\frac{1}{A}-\frac{1}{B}=\frac{192}{2005^{2}-2004^{2}}
$$

Find the value of $B$.
22. Ans: 211
$$
\frac{1}{A}-\frac{1}{B}=\frac{192}{2005^{2}-2004^{2}} \Longleftrightarrow \frac{B-A}{A \times B}=\frac{192}{4009} \Longleftrightarrow 4009 \times(B-A)=192 \times A \times B .
$$

The prime factorization of 4009 is given by $4009=19 \times 211$. Since $192=2^{6} \times 3$ and $B>A$, it follows that $A=19$ and $B=211$.