I found a mistake in my work for 5.1c
but I'm not sure how to deal with it:
4x^{3}  3x^{2}  = 0  

4y^{3} + 2y  = 0   
Which factor into
x^{2}(4x  3)  = 0  

2y(y^{2} + 1)  = 0   
[yielding]
Yeah... that second equation is supposed to be
So the actual solutions are
Let's test out this new pair of ys then. If we plug it into the curve equation
y^{2} + x^{4} + y^{4}  x^{3}, we get

+ x^{4} +
 x^{3}  (1)

 =
+
+ x^{4}  x^{3}  (2)

 =
+ x^{4}  x^{3}  (3)

  (4) 
Clearly, setting
x = 0 doesn't work. So the only other possibility is
x =
:

+ (
)^{4}  (
)^{3}  

 =
+

 

 =
+

 

 =
 

 =
 

  
Which can't be
0 unless
91 = 0. The prime factorization of
91 is
7 ⋅ 13...........
which means....... if the field has characteristic
7 or
13........ this is actually.......
possible...........?
FUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUCK.
Reader: I do not know how to rule these cases out. Maybe I should just accept
that
(
, +∕ 
) actually works if char
k = 7 or char
k = 13?
Arrrggghh, fuck that. the solution I saw online (which I'm actually not putting
too much trust in atm) said that
(0, 0) is the only singular point, so I'm just
going to assume that moving forward for now (i.e. 5.3, COMING SOON)