14. Let $a$ be a prime number, $b$ be a positive integer, and $9(2 a+b)^{2}=509(4 a+511 b)$, find the values of $a, b$.
14. Since $9(2 a+b)^{2}=3^{2}(2 a+b)^{2}$ is a perfect square, therefore, $509(4 a+511 b)$ is a perfect square. Since 509 is a prime number, we can let
$4 a+511 b=509 \times 3^{2} k^{2}$.
Thus, the original equation becomes
$9(2 a+b)^{2}=509^{2} \times 3^{2} k^{2}$, which simplifies to $2 a+b=509 k$.
Hence, $b=509 k-2 a$, substituting into equation (1) gives
$4 a+511(509 k-2 a)=509 \times 3^{2} k^{2}$,
solving for $a$ yields $a=\frac{k(511-9 k)}{2}$.
Since $a$ is a prime number, i.e., $\frac{k(511-9 k)}{2}$ is a prime number, we have the following cases:
(1) When $k=1$, $a=\frac{k(511-9 k)}{2}=\frac{511-9}{2}=251$ is a prime number, which meets the condition. In this case,
$b=509 k-2 a=509-502=7$.
(2) When $k=2$, $a=\frac{k(511-9 k)}{2}=511-18=493=17 \times 29$ is not a prime number, so it is discarded.
(3) When $k>2$ and $k$ is odd, since $a=\frac{k(511-9 k)}{2}=k \cdot \frac{511-9 k}{2}$ is a prime number, and $k>1$, so $\frac{511-9 k}{2}=1$. But $\frac{511-9 k}{2}=1$ has no integer solution, so it is discarded.
(4) When $k>2$ and $k$ is even, since $a=\frac{k(511-9 k)}{2}=\frac{k}{2}(511-9 k)$ is a prime number, and $\frac{k}{2}>1$, so $511-9 k=1$. But $511-9 k=1$ has no integer solution, so it is discarded.
In summary, $a=251, b=7$.