Забыл предупредить, действительно первый запрос довольно суровый на таблице с большим количеством записей.
Более оптимизированный вариант этого запроса
SELECT i.member_id, i.ip_address FROM ( SELECT member_id, ip_address FROM core_members_known_ip_addresses GROUP BY ip_address, member_id ) as i INNER JOIN ( SELECT member_id, ip_address FROM core_members_known_ip_addresses GROUP BY ip_address, member_id ) as z ON i.ip_address=z.ip_address AND i.member_id!=z.member_id GROUP BY i.ip_address, i.member_id
Исключить пользователей с ip 127.0.0.1
SELECT i.member_id, i.ip_address FROM ( SELECT member_id, ip_address FROM core_members_known_ip_addresses GROUP BY ip_address, member_id ) as i INNER JOIN ( SELECT member_id, ip_address FROM core_members_known_ip_addresses GROUP BY ip_address, member_id ) as z ON i.ip_address=z.ip_address AND i.member_id!=z.member_id WHERE i.ip_address!='127.0.0.1' GROUP BY i.ip_address, i.member_id
и вариант с выведением имени пользователя
SELECT m.name, i.member_id, i.ip_address FROM ( SELECT member_id, ip_address FROM core_members_known_ip_addresses GROUP BY ip_address, member_id ) as i INNER JOIN ( SELECT member_id, ip_address FROM core_members_known_ip_addresses GROUP BY ip_address, member_id ) as z ON i.ip_address=z.ip_address AND i.member_id!=z.member_id LEFT JOIN core_members m ON i.member_id=m.member_id GROUP BY i.ip_address, i.member_id