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You are correct. Their answer is instead the sum of the first k terms of 1, 2, 6, 10, 14, 18, ..., for positive k.

You're misreading the solution, the first part reads n=1, a trivial special case, not n congruent to 1 mod 4.

The statement doesn't hold for e.g. n=5. Taking m=2 gives the permutation (1 2 4 3), which is odd, and thus cannot have a square root.

Very astute. Did you communicate this to the authors?