Denote by \(\delta _{k}(n)\) the greatest divisor of a positive integer n which is coprime to a given \(k\ge 2\). In the case of \(k=p\) (a prime) Joshi and Vaidya studied \(E_{p}(x):=\sum _{n\le x}\delta _{p}(n)-\frac{p}{2(p+1)}x^{2}\) (as \(x\rightarrow \infty \)) and obtained \(E_{p}(x)=\Omega _{\pm }(x)\) by an elementary and beautiful approach. Here we study \(R_{p}^{(2)}(x):=\sum _{n\le x}\delta _{p}^{2}(n)-\frac{p^{2}}{3(p^{2}+p+1)}x^{3}+\frac{p}{6}x\) and show \(R_{p}^{(2)}(x)=\Omega _{\pm }(x^{2})\). Moreover, using a method of Adhikari and Balasubramanian we consider a bound of \(|R_{k}^{(2)}(x)|/x^{2}\) for any square-free integer k.

用\(\delta _{k}(n)\) 表示与给定\(k\ge 2\)互质的正整数n的最大除数。在\(k=p\)（素数）的情况下，Joshi 和 Vaidya 研究了\(E_{p}(x):=\sum _{n\le x}\delta _{p}(n)-\ frac{p}{2(p+1)}x^{2}\)（如\(x\rightarrow \infty \)）并获得\(E_{p}(x)=\Omega _{\pm } (x)\)通过一种基本而美丽的方法。这里我们研究\(R_{p}^{(2)}(x):=\sum _{n\le x}\delta _{p}^{2}(n)-\frac{p^{2 }}{3(p^{2}+p+1)}x^{3}+\frac{p}{6}x\)并显示\(R_{p}^{(2)}(x) =\Omega _{\pm }(x^{2})\)。此外，使用 Adhikari 和 Balasubramanian 的方法，我们考虑了一个界限\(|R_{k}^{(2)}(x)|/x^{2}\)对于任何无平方整数k。