4. Given $3^{a}+13^{a}=17^{a}, 5^{b}+7^{b}=11^{b}$, then the four numbers $a, b, 0,1$ from smallest to largest are
4. Solution: The function $f(x)=\left(\frac{3}{17}\right)^{x}+\left(\frac{13}{17}\right)^{x}$ is a decreasing function for $x \in \mathbf{R}$, and $f(1)=\frac{16}{17}$. Therefore, $0<a<1<b$.