Convert affine to projective coordinates and vice versa in ECC?

Question

I am working on a small project. An elliptic curve equation with y^2=x^3-3x+27 mod 43, a point $Q=(1,38)$, using point doubling method https://en.wikipedia.org/wiki/Elliptic_curve_point_multiplication#Point_doubling, i got $2Q(Q+Q)=(41,5)$. Now in projective coordinate $Q=(1,38,1)$ taking $z=1$.https://en.wikibooks.org/wiki/Cryptography/Prime_Curve/Standard_Projective_Coordinates

Calculating $2Q$, I get $2Q=(22,31,39)$ and converting to affine coordinate I can't get $2Q=(41,5)$.

So, where did I went wrong can anyone explain or am I following the wrong method??

Answer 1

Somewhere you seem to have made a small mistake. $39$ is already the inverse of $Z_{2Q}$. The correct value of $Z_{2Q}$ is $\mathbf{32}$.

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For the long story, let us first go through the doubling formula one step a time:

if (Y == 0)                     # nope
 return POINT_AT_INFINITY
W = a*Z^2 + 3*X^2               # W = 0
S = Y*Z                         # S = 38
B = X*Y*S                       # B = 25
H = W^2 - 8*B                   # H = 15
X' = 2*H*S                      # X'= 22
Y' = W*(4*B - H) - 8*Y^2*S^2    # Y'= 31
Z' = 8*S^3                      # Z'= 32
return (X', Y', Z')

This leaves us with $2Q = (22 : 31 : \mathbf{32})$.

To compute the affine representation, we compute $Z_{2Q}^{-1} = 39$. Now the affine representation of $2Q = (XZ^{-1}, YZ^{-1}) = (22 \cdot 39, 31 \cdot 39) = (41, 5)$.