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I was looking at an example of the ElGamal encryption operation here (page 24), but I can't seem to understand why: $$\beta = 3(10, 3) = (10, 8)$$

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we have $a=5, b=4, (x_P,y_P) = (10,3) and p = 11$ Back to the page 22 calculate $s\equiv (3x_P^2+a)(2y_P)^{-1} \mod p \equiv 5\mod 11$ calculate in $\mod p$: $2(10,3) = (5^2- 2\cdot 10,-3+5(10-5)) = (5,0) $ now add (5,0) to (10,3) (use the rule in the page 21) now we have $s\equiv (3-0)(10-5)^{-1}\mod p \equiv 5\mod p $ then $x_\beta \equiv (5^2 - 10 - 5) \equiv 10\mod p$ and $y_\beta \equiv -3 + 5 (10 -10) \equiv 11-3 \equiv 8 \mod p$ Attention to use $(5,0)$ instead of $(10,3)$ while calculating $y_\beta$.

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