<s>[INST] Is there a systematic way to solve in $\bf Z$: $x_1^2+x_2^3+...+x_{n}^{n+1}=z^{n+2}$ for all $n$? Is there a systematic way to solve in $\bf Z$ $$x_1^2+x_2^3+...+x_{n}^{n+1}=z^{n+2}$$ For all $n$?
It's evident that $\vec 0$ is a solution for all $n$.
But finding more solutions becomes harder even for small $n$:
When $n=2$,
$$
x^2+y^3=z^4
$$
I'm already pretty lost.
After this it gets even more complicated, has anyone encountered this problem before?
I want all the solutions, or at least infinitely many for every equation.
 [/INST] Here's a proof of infinitely many solutions $(x_1, x_2, x_3, \dots, x_n, z)$:
For $n = 1$ we have $(1, 1)$. For $n = 2$ we have $(3^3, 2 \cdot 3^2, 3^2)$.
For $n \ge 3$ we have $(2, 3, 1, 2, 2, 2, 2, \dots, 2)$.
(To verify: $2^2 + 3^3 + 1^4 + 2^5 + 2^6 + \dots + 2^n = 32 + 2^5 (1 + 2 + 4 + \dots + 2^{n-5}) = 32 + 2^5 (2^{n-4} - 1) = 2^{n+1}$, where we have used a geometric series to simplify the sum.)
Thus all $n \ge 1$ yields at least one solution in positive integers. To get infinitely many solutions, multiply term $x_k$ by $m^{\frac{(n+1)!}{k}}$ and $z$ by $m^{n!}$; then both sides of the Diophantine equation are multiplies by $m^{(n+1)!}$. Choosing arbitrary $m \ge 2$ gives infinitely many solutions, as desired.
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