# nt.number theory – Is there a permutation \$tauin S_n\$ with \$tau(1)^{tau(2)}+cdots+tau(n-1)^{tau(n)}+tau(n)^{tau(1)}\$ a square?

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 $$tauin S_n$$ 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 $$tauin S_n$$ 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?

Your are welcome to check the conjecture further.