Motivation: It has been shown that for $X = \left(X_1, X_2, X_3\right)$, where $X$ is distributed uniformly over the surface of a sphere, we have marginally $X_i \sim U[-1, 1]$ and $X_1^2 + X_2^2 + X_3^2 = 1$.
Furthermore, taking $X_1$ and $X_2$ from above, if we let $Y_1 = \frac{X_1 + 1}{3}$, $Y_2 = \frac{2 - X_1 - X_2\sqrt{3}}{6}$, and $Y_3 = \frac{2 - X_1 - X_2\sqrt{3}}{6}$ it can be shown that $Y_i \sim U[0, \frac{2}{3}]$ and $Y_1 + Y_2 + Y_3 = 1$. This is done by projecting $X$ onto the $X_1, X_2$-plane and circumscribing the unit circle with an equilateral triangle. Then, letting the point $(X_1, X_2)$ divide the equilateral triangle into three smaller triangles, $Y_1$, $Y_2$, and $Y_3$ are the ratios of areas of those three triangles. This can be done with any normal $d$-gon to generate $Y_i \sim U[0, \frac{2}{d}]$ with $\sum_{i=1}^d Y_i = 1$.
Problem: Given $d$ and $n$, is there a general method to generate a distribution of $X = \left(X_1, \dots, X_d\right)$ such that $\sum_{i=1}^d X_i^n = 1$ and $X_i \sim U[a, b]$ for some $a, b$? For what values of $n$ and $d$ is this not possible?
Example of not possible: For $n=2$, $X$ will be on the surface of a $d$-sphere. It has been shown that the $X_i$ are uniform only if $d=3$.
Example 2 of not possible: For $d=2$, we must have $X_2 = 1 - X_1^n$. If $X_1$ is uniform, then $X_2$ cannot be uniform for $n \geq 2$.