of $\zeta(s)$. Let $$N_s(T):=\displaystyle\sum_{\substack{0<\gamma\leq T\\ m_\rho=1}}1.$$ Using the fact that $${N_s(T)\geq \displaystyle\sum_{0<{\gamma}\leq T}(2-m_\rho)}=2N(T)-N^{*}(T).$$ we obtain the following corollary.

Assuming RH we have $${N_s(T)}\geq ({0.6792} + o(1)){N(T)}.$$ Assuming GRH we have $${N_s(T)}\geq ({0.6845} + o(1)){N(T)}. $$

Under the pair correlation approach the previous best result known is due by Cheer and Goldston [@CG] showing that $67.27\%$ of the zeros are simple. Assuming GRH, Goldston, Gonek, Özlük and Snyder [@GGOS] showed that $67.38\%$ are simple. In this way, we improved all these bounds. However, by a different technique, still assuming RH, Bui and Heath-Brown [@BHB] improved the result to $70.37\%$, which currently is the best.

Combining the above result of Bui and Heath-Brown with Theorem \[thmNstar\] and an argument of Ghosh, we can bound the proportion of distinct zeros of zeta. Let $$N_d(T) := \sum_{0<\gamma\leq T} \frac{1}{m_\rho},$$ be the number of distinct zeros of $\zeta(s)$ with $0<{\gamma}\leq T$. Using the inequality $$2N_s(T)\leq \displaystyle\sum_{0<\gamma\leq T}\dfrac{(m_\rho-2)(m_\rho-3)}{m_\rho}=N^{*}(T)-5N(T)+6N_d(T).$$ in conjunction with the estimate $$N_s(T) \geq (0.7037+o(1))N(T)$$ and Theorem \[thmNstar\], we deduce the following corollary.

Assuming RH we have $${N_d(T)}\geq ({0.8477} + o(1)){N(T)}.$$ Assuming GRH we have $${N_d(T)}\geq ({0.8486} + o(1)){N(T)}. $$

Using the pair correlation approach, the best previous result known is due to Farmer, Gonek 