10. Let real numbers $x, y$ be such that $x-y, x^{2}-y^{2}, x^{3}-y^{3}$ are all prime numbers, then the value of $x-y$ is
Let $x-y=p, x^{2}-y^{2}=q, x^{3}-y^{3}=r$ be prime numbers,
then $\left\{\begin{array}{l}x+y=\frac{q}{p}, \\ x-y=p\end{array} \Rightarrow\left\{\begin{array}{l}x=\frac{1}{2}\left(\frac{q}{p}+p\right), \\ y=\frac{1}{2}\left(\frac{q}{p}-p\right) .\end{array}\right.\right.$ Thus, $x^{3}-y^{3}=\frac{1}{8}\left(2 p^{3}+6 \cdot \frac{q^{2}}{p^{2}} \cdot p\right)=\frac{1}{4}\left(p^{3}+\frac{3 q^{2}}{p}\right)=r$ $\Rightarrow 4 p r=p^{4}+3 q^{2} \Rightarrow p \mid 3 q^{2}$, hence $p=3$ or $p=q$.

If $p=3$, then $27+q^{2}=4 r$, taking $q=5, r=13$ satisfies the condition;
If $p=q$, then $q\left(q^{2}+3\right)=4 r \Rightarrow q=r=1$, which is a contradiction. In summary, the value of $x-y$ is 3.