On the difference between an integer and its inverse modulo \(n\)
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.
difference between an integer and its inverse
scientific article; zbMATH DE number 765156