On the difference between an integer and its inverse modulo \(n\) (Q1892566)

If \(n>2\), \(x>0\) are integers satisfying \(x<n\) and \((n,x)=1\), then there exists a unique integer \(\overline {x}\) such that \(0< \overline {x}<n\) and \(x\overline {x}\equiv 1\bmod n\). Let us write \[ M(n,k)= \sum^n _{\substack{ a=1\\ (a,n)=1}} (a- \overline {a})^{2k} \qquad \text{for} \quad k\in \mathbb{N}. \] The author derives the asymptotic formula \[ M(n,k)= {{\varphi (n) n^{2k}} \over {(2k+1) (k+1)}}+ O(4^k n^{(4k+ 1)/2} \tau^2 (n)\log^2 n), \] where \(\varphi\) denotes Euler's function, and \(\tau\) is the divisor function. The proof uses estimates for Kloosterman sums and properties of trigonometric sums.