## Odd permutations $\tau\in S_n$ with $\sum_{k=1}^nk\tau(k)$ a square

For any positive integer $$n$$, as usual we let $$S_n$$ be the symmetric group of all the permutations of $$\{1,\ldots,n\}$$.

QUESTION: Is it true that for each integer $$n>3$$ there is an odd permutation $$\tau\in S_n$$ such that $$\sum_{k=1}^n k\tau(k)$$ is a square?

Let $$a(n)$$ denote the number of odd permutations $$\tau\in S_n$$ with $$\sum_{k=1}^nk\tau(k)$$ a square. Via a computer I find that $$(a(1),\ldots,a(11))=(0,1,0,2,8,22,82,718,4583,53385,513897).$$ For example, $$(2,4,1,3)$$ and $$(3,1,4,2)$$ are the only two odd permutations in $$S_4$$ meeting our requirement; in fact, $$1\cdot2+2\cdot4+3\cdot1+4\cdot3=5^2\ \ \text{and}\ \ 1\cdot3+2\cdot1+3\cdot4+4\cdot2=5^2.$$ For $$n=2,3,4,5$$, there is no even permutation $$\tau\in S_n$$ such that $$\sum_{k=1}^nk\tau(k)$$ is a square.

I conjecture that the question has a positive answer. Any comments are welcome!