Picking a Base Point in ECC

In ECC, we have an equation of y²=x³+ax+b (mod p). This gives us points on an elliptic curve. For a=2, b=3 and p=97, we get [here]:

(1,54) (1,43) (3,91) (3,6) (4,47) (4,50) (10,76) (10,21) (11,17) (11,80) (12,94) (12,3) (17,87) (17,10) (20,34) (20,63) (21,73) (21,24) (22,92) (22,5) (23,73) (23,24) (24,95) (24,2) (25,35) (25,62) (27,7) (27,90) (28,34) (28,63) (29,54) (29,43) (32,7) (32,90) (37,22) (37,75) (38,7) (38,90) (39,91) (39,6) (44,77) (44,20) (46,72) (46,25) (47,79) (47,18) (49,34) (49,63) (50,19) (50,78) (52,68) (52,29) (53,73) (53,24) (54,85) (54,12) (55,91) (55,6) (56,89) (56,8) (59,65) (59,32) (65,65) (65,32) (67,54) (67,43) (70,65) (70,32) (73,83) (73,14) (74,77) (74,20) (76,77) (76,20) (80,87) (80,10) (83,74) (83,23) (84,37) (84,60) (85,26) (85,71) (86,69) (86,28) (87,27) (87,70) (88,41) (88,56) (91,39) (91,58) (92,81) (92,16) (95,66) (95,31) (97,87) (97,10)

You can check with (1,54), as 54² (mod 97) is 6, and 1³+2 x 1 + 3 (mod 97) is 6. For (3,91), we have 91² (mod 97) and which is 36, and 3³ + 2 x 3 + 3 (mod 97) is 36. There is no solution for x=2, and x=5.

A basic plot is [here]: