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.