14. Find all prime numbers $x$ and $y$ that satisfy $x^{3}-x=y^{7}-y^{3}$.

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$$
\begin{array}{l}
\text { Since } y>5 \text {, } y^{7}-y^{3}>5 y^{6}-y^{3}=y^{6}+4 y^{6}-y^{3}>y^{6}+20 y^{5}-y^{3} \\
=y^{6}+15 y^{5}+5 y^{5}-y^{3}>y^{6}+15 y^{5}+25 y^{4}-y^{3}=y^{6}+15 y^{5}+20 y^{4}+5 y^{4}-y^{3} \\
>y^{6}+15 y^{5}+20 y^{4}+25 y^{3}-y^{3}=y^{6}+15 y^{5}+20 y^{4}+20 y^{3}+4 y^{3} \\
>y^{6}+15 y^{5}+20 y^{4}+20 y^{3}+20 y^{2}=y^{6}+15 y^{5}+20 y^{4}+20 y^{3}+14 y^{2}+6 y^{2} \\
>y^{6}+15 y^{5}+20 y^{4}+20 y^{3}+14 y^{2}+30 y>y^{6}+15 y^{5}+20 y^{4}+20 y^{3}+14 y^{2}+4 y \\
=(y+1)^{6}-(y+1)^{2} \Rightarrow x^{3}-x>(y+1)^{6}-(y+1)^{2} \Rightarrow x>(y+1)^{2},
\end{array}
$$

and $y^{7}-y^{3}=y^{3}\left(y^{2}+1\right)(y+1)(y-1), y-1<y<y+1<y^{2}+1<(y+1)^{2}<x$ and $x$ is a prime number $\Rightarrow x \nmid\left(y^{7}-y^{3}\right)$, contradiction. Therefore, $y$ can only take $2,3,5$.
When $y=2$, $x(x+1)(x-1)=8 \cdot 5 \cdot 3 \Rightarrow x=5$;
When $y=3$, $x(x+1)(x-1)=27 \cdot 10 \cdot 4 \cdot 2 \Rightarrow x$ has no prime solution;
When $y=5$, $x(x+1)(x-1)=125 \cdot 26 \cdot 6 \cdot 4 \Rightarrow x$ has no prime solution.
In summary, the only prime solution to $x^{3}-x=y^{7}-y^{3}$ is $(x, y)=(5,2)$.
$$