An integer sequence with integer $k$ norms

Question

Find the maximum value of $n$(if exists) such that there exists a sequence $a_1,a_2,\ldots,a_n$ of positive integers such that for every $2\leq k \leq n$

$$\sqrt[k]{a_1^k+a_2^k+\cdots+a_k^k}$$

is itself an integer.

I know the sequence 3,4,5 for $n=3$ works, but I do not know anything for larger values.

If there is no maximum value, is there exist an infinite sequence having such a property?

Answer 1

This is not a complete answer, but is too long for a comment. The case $n=4$ entails solving the simultaneous system,

$$\begin{align} &x_1^2+x_2^2 = y_1^2\tag1\\ &x_1^3+x_2^3+x_3^3 = y_2^3\tag2\\ &x_1^4+x_4^4+x_3^4+x_4^4 = y_3^4\tag3 \end{align}$$

Solving $(3)$ in particular is no trivial matter. Fortunately, Jarek Wroblewski has a complete table of the 1009 primitive solutions with $y_3<220000$. After checking them, it turns out there is no subset of the terms $x_i$ such that $(1)$ or $(2)$ has solutions.

If indeed $n=4$ has solutions, then it should involve larger terms.

P.S. Incidentally, there is the curiosity,

$$15935^2 + 27022^2 + 57910^2 + 59260^2 = 88597^2$$

$$15935^4 + 27022^4 + 57910^4 + 59260^4 = 70121^4$$

though this is only one of its kind in the table.

Answer 2

Partial answer: There does not exist such an infinite sequence.

Proof. Let $p\geq3$ be prime. We have, for some $x\in\mathbb Z$, $$x^{p-1}=a_1^{p-1}+\cdots+a_{p-1}^{p-1}.$$ By Little Fermat, each term is $0$ or $1$ mod $p$. Clearly, for the LHS to be $0$ or $1$, at most one term in the RHS can be not divisible by $p$. In particular, $p\mid a_1a_2$. So $a_1a_2$ would be divisible by every prime $p\geq3$, contradiction. $\square$

Maybe, by making explicit the obtained lower bounds on the $a_k$, this observation allows to prove that there exists a maximal length for such sequences...