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Let $n>1$ be an integer, and let $S_n$ be the symmetric group of all the permutatins of $\{1,\ldots,n\}$. I'm curious whether there is a permutation $\tau\in S_n$ such that $$\tau(1)^{\tau(2)}+\cdots+\tau(n-1)^{\tau(n)}+\tau(n)^{\tau(1)}$$ is a square. (Without loss of generality we may assume that $\tau(n)=n$.) For $n=2,3$ there is no such a permutation $\tau$. But my computations for $n=4,5,\ldots,11$ lead me to formulate the following conjecture.

Conjecture. For any integer $n>3$, there is a permutation $\tau\in S_n$ such that $$\tau(1)^{\tau(2)}+\cdots+\tau(n-1)^{\tau(n)}+\tau(n)^{\tau(1)}$$ is a square.

For example, $$2^1 + 1^3 + 3^4 + 4^2 = 10^2, \ \ 1^5 + 5^2 + 2^4 + 4^3 + 3^6 + 6^1 = 29^2,$$ and $$1^3 + 3^2 + 2^{10} + 10^5 + 5^7 + 7^8 + 8^6 + 6^9 + 9^4 + 4^{11} + 11^1 = 4526^2.$$ For more examples and related data, one may consult http://oeis.org/A342965.

QUESTION. Is the above conjecture true?

You are welcome to check the conjecture further.

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    $\begingroup$ Why is this interesting? You have $(n-1)!$ shots in the range $[0,n^n]$ (or so). The chances of hitting a square grow fast towards $1$ as $n$ grows. $\endgroup$
    – Yaakov Baruch
    Commented Apr 1, 2021 at 15:23
  • $\begingroup$ Maybe there is a secret hope to find a pattern which yields a solution for each n... $\endgroup$
    – Wolfgang
    Commented Apr 1, 2021 at 15:34
  • $\begingroup$ (But such a pattern would surely be VERY surprising) $\endgroup$
    – Wolfgang
    Commented Apr 1, 2021 at 16:08
  • $\begingroup$ Let $a(n)$ denote the number of permutations $\tau\in S_n$ with $\tau(n)=n$ satisfying the requirement in the conjecture. I have computed $a(n)$ for $n=4,\ldots,11$, namely, $$a(4)=1,\ a(5)=2,\ a(6)=1,\ a(7)=a(8)=6,\ a(9)=10,\ a(10)=27.$$ $\endgroup$
    – Zhi-Wei Sun
    Commented Apr 1, 2021 at 16:25
  • $\begingroup$ I have verified the conjecture for $n=12$ with the example $$1^2 + 2^5 + 5^6 + 6^8 + 8^4 + 4^{11} + 11^9 + 9^7 + 7^{10} + 10^3 + 3^{12}+ 12^1 = 51494^2$$. Jinyuan Wang has computed $a(n)$ for $n=11,12,13$ (cf. oeis.org/A342965), their values are $$a(11)=105,\ a(12)=245,\ a(13)=525.$$ In view of the values of $a(4),\ldots,a(13)$, it seems that $a(n)$ does not grow quickly. By Stirling's formula, $n!\sim (n/e)^n\sqrt{2\pi n}$. $\endgroup$
    – Zhi-Wei Sun
    Commented Apr 3, 2021 at 23:59

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