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?

Your are welcome to check the conjecture further.

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.

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?

Your are welcome to check the conjecture further.

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?

Your are welcome to check the conjecture further.