Oh I see now.
so
5 mod 2 = 1
c mod 2 = 1
therefore c can be 5, 7, 9, 11, etc. since 7/2 has a remainder of 1 and 9/2 has a remainder of one.
Similarly,
3 mod 3 = 0 so;
d mod 3 = 0
therefore d can be 3, 6, 9, 12 etc.
So would 4 ordered pairs be;
(5,3), (7,6), (9,9)...