Walk through a concrete point DOUBLING on $y^2 \equiv x^3 + 8x + 5 \mod 11$ with $P(0, 4)$. Compute $2P$.

Use the doubling formula with slope $\lambda = \frac{3x_1^2 + a}{2y_1} \mod p = \frac{3 \cdot 0 + 8}{8} = 8 \cdot 8^{-1} = 8 \cdot 7 = 56 \equiv 1 \mod 11$. Then $x_3 = 1 - 0 = 1$ and $y_3 = 1(0-1) - 4 = -5 \equiv 6 \mod 11$. Result: $2P = (1, 6)$.

Given: Curve $y^2 = x^3 + 8x + 5 \mod 11$, point $P(0, 4)$, curve parameter $a = 8$

Step 1 — Doubling slope (different formula from addition!): $$\lambda = \frac{3x_1^2 + a}{2y_1} \mod 11 = \frac{3 \cdot 0^2 + 8}{2 \cdot 4} \mod 11 = \frac{8}{8} \mod 11$$

Need $8^{-1} \mod 11$: since $8 \cdot 7 = 56 = 5 \cdot 11 + 1 \equiv 1 \mod 11$, so $8^{-1} = 7$.

$$\lambda = 8 \cdot 7 \mod 11 = 56 \mod 11 = 1$$

Step 2 — x-coordinate (note: $2x_1$ not $x_1 + x_2$): $$x_3 = \lambda^2 - 2x_1 \mod 11 = 1^2 - 2 \cdot 0 = 1$$

Step 3 — y-coordinate: $$y_3 = \lambda(x_1 - x_3) - y_1 \mod 11 = 1 \cdot (0 - 1) - 4 = -5 \mod 11 = 6$$

Result: $2P = (1, 6)$

Verification: $6^2 = 36 \equiv 3 \mod 11$. $1^3 + 8 \cdot 1 + 5 = 14 \equiv 3 \mod 11$. ✓

Key difference from addition: The slope formula uses $\frac{3x_1^2 + a}{2y_1}$ (tangent line) instead of $\frac{y_2-y_1}{x_2-x_1}$ (secant line). Everything else is the same. Watch out for the $-2x_1$ in the x-formula instead of $-x_1-x_2$.

Go deeper:

• Corbellini — ECC part 2 (point doubling over $\mathbb{F}_p$) — the tangent-slope formula in the same worked style.