Code:
SELECT P1.some, P1.fields
FROM patterns AS P1,
( SELECT lnktxt, category, COUNT(*) AS howmany
FROM patterns
GROUP BY linktxt, category
HAVING howmany > 1 ) AS P2
WHERE P1.lnktxt = P2.lnktxt AND P1.category = P2.category
???
If I understood what you were saying.