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Kurve : EC : $y^2=x^3 + x + 1$
Generator:$(1,7)$
$p=23$

Result in Affine use Excel: $P=(1,7)$, $Q=(7,11) \implies P+Q=(18,20)$

Result in Projective use Excel: $P=(1:7:1), Q=(7:11:1) \implies P+Q=(15:2:9)$

$U = U_1 - U_2\ $ where $\ U_1 = Y_2 ⇤ Z_1, U_2 = Y_1 ⇤ Z_2$
$V = V_1 - V_2\ $ where $\ V_1 = X_2 ⇤ Z_1, V_2 = X_1 ⇤ Z_2$
$W = Z_1 ⇤ Z_2$
$A = U^2 ⇤ W-V^3-2V^2 ⇤ V_2$
Then
$X_3 = V ⇤ A$
$Y_3 = U ⇤ (V^2 ⇤ V2 - A) - V^3 ⇤ U2$
$Z_3= V^3 ⇤ W$

Is the result correct for projective coordinates (X/Z,Y/Z), because it should be same with affine (18:20:1)?

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The projection coordinates you got are wrong. We can tell that because those coordinates do not correspond to a point on the curve.

We have $15 \cdot 9^{-1} \equiv 17$ and $2 \cdot 9^{-1} \equiv 13$; when we plug $x = 17$ and $y = 13$ into the original curve equation, we do not get equality, and so we know something went wrong.

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  • $\begingroup$ Sorry it's my bad. I sloppy with the calculation, the right result is (1:19:9). And the affine coordinates of the presentation is (18,20) and it lays in the curve. Thanks Poncho. $\endgroup$ – alfred Feb 6 '16 at 7:41

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