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.
On
For $n=2$ and $n=3,4$ we can give infinite families using a Pell equation and elliptic curves, respectively,
$n=2$:
$$\big(4q^2(p^2-2)\big)^2+(4dq^2)^3 = (2pq)^4$$
where $p,q$ solve $p^2-d^3q^2=1\tag1$.
$n=3$:
$$(a y)^2 + (ma)^3 + a^4 = a^5$$
and the elliptic curve solvable for an appropriate constant $m$,
$$a^3 - a^2 - m^3 a = y^2\tag2$$
Example: For $m=2$,
$$\begin{aligned} (4\cdot4)^2 + (2\cdot4)^3 + 4^4 &=4^5\\ (9\cdot24)^2 + (2\cdot9)^3 + 9^4 &=9^5\\ \big(\tfrac{10252a}{125}\big)^2+(2a)^3 + a^4 &= a^5 \end{aligned}$$
and $a=\big(\tfrac{22}{5}\big)^2$. This elliptic curve has an infinite number of rational points and by multiplying by an appropriate factor (such as $5^{60}$), then one can easily clear denominators and get integer solutions. (Note: The initial solution of $(2)$ may have a large height. For example, for $m=13$ we have $a=(\frac{45539}{6612})^2$.)
$n=4$:
Similarly, one can use,
$$(a y)^2 + (2a)^3 + a^4 + a^5 = a^6$$
$$(a y)^2 + (2a)^3 + (3a)^4 + a^5 = a^6$$
both of which entail solving an elliptic curve.
$n>4$:
A nice identity with almost consecutive exponents is one by Enrico Jabara,
$$(7^3m^{11} n^4)^3 + (7\cdot14^4m^{11}n)^4 + (7m^4n^4)^5 + (m n^4)^7 + (14^4 m^6-n^4)^8 = (14^4 m^6+n^4)^8$$
where $m = 28^4$ and arbitrary $n$, and which he probably found by expanding,
$$(an^4)^3 + (bn)^4 + (cn^4)^5 + (dn^4)^7 + (e-n^4)^8 = (e+n^4)^8$$
collecting powers of $n$, and solving for $a,b,c,d,e$.
If $n+2$ is composite, say $n+2=mk$ with $m,k>1$, take any $z$, $x_m = z^k$, all other $x_j = 0$.
EDIT: Suppose we have a solution $(a_1, \ldots, a_{n+1})$ in positive integers with $a_1^{d_1} + \ldots + a_n^{d_n} = a_{n+1}^{d_{n+1}}$, and positive integer $k$ so that $d_i+k$ are pairwise coprime. By the Chinese Remainder Theorem there are positive integers $m_1, \ldots, m_{n+1}$ such that $m_i \equiv 0 \mod d_j + k$ for $j \ne i$ while $m_i \equiv k \mod d_i + k$. Let $b = a_1^{m_1}\ldots a_{n+1}^{m_{n+1}}$ and $t_i = (b/a_i^k)^{1/(d_i+k)}$, which is a positive integer. Then $$(t_1 a_1)^{d_1+k} + (t_2 a_2)^{d_2+k} + \ldots + (t_n a_n)^{d_n+k} = b (a_1^{d_1} + \ldots + a_n^{d_n}) = b a_{n+1}^{d_{n+1}} = (t_{n+1} a_{n+1})^{d_{n+1}+k}$$
For example, starting from $$ 28^2 + 8^3 = 6^4$$ ($a_1 = 28$, $a_2 = 8$, $a_3 = 6$), for $k=1$ where $3,4,5$ are pairwise coprime (it will work for any odd $k$) we take $m_1,m_2,m_3 = 40, 45, 36$. Then $$\eqalign{b &= 28^{2} 8^3 6^4 = 2^{251} 3^{36} 7^{40} \cr t_1 &= (b/a_1)^{1/3} = 2^{83} 3^{12} 7^{13}\cr t_2 &= (b/a_2)^{1/4} = 2^{62} 3^9 7^{10}\cr t_3 &= (b/a_3)^{1/5} = 2^{50} 3^7 7^8\cr (t_1 a_1)^{3} &+ (t_2 a_2)^4 = (t_3 a_3)^5 \cr 0^2 + (2^{85} 3^{12} 7^{14})^3 &+ (2^{65} 3^9 7^{10})^4 = (2^{51} 3^8 7^8)^5 \cr}$$