Elliptic Curve #2: Elliptic Curves over Finite Fields

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  • เผยแพร่เมื่อ 24 ม.ค. 2025

ความคิดเห็น • 18

  • @accucc2990
    @accucc2990 2 ปีที่แล้ว +1

    I can't believe it took me only a few minutes to understand these concepts, thanks

  • @kareemshehata
    @kareemshehata 3 ปีที่แล้ว +5

    Is it just me, or is there something wrong with the points given? For example (6, 3) is not a point on that curve since 6^3 +1 mod 11 = 8, and 3^2 mod 11 = 9. On the other hand, (2, 3) and (2, 8) should be points on the curve.

    • @AdvancedMath
      @AdvancedMath  3 ปีที่แล้ว +3

      I have not properly calculated the points on elliptic curve for the example. You are right (6, 3) does not lie on the curve.

  • @eyob_g
    @eyob_g 10 หลายเดือนก่อน

    In point addition (using chord method), is it always the case that the line passing through the points P and Q crosses the elliptic curve at some point?
    How about in the case of point doubling (using tangent method), what do we say if the tangent at point P doesn't cross elliptic curve at another point?

  • @anusha5788
    @anusha5788 4 ปีที่แล้ว +1

    Sir, great explanation! How did you obtain the points of the elliptic curve at 2:20 ?

    • @AdvancedMath
      @AdvancedMath  4 ปีที่แล้ว +3

      Sorry, I have not made any video to explain that. I will make it in future. You may simply choose an x value from the set {0, 1, 2, .... m-1} where m is the mod on which elliptic curve is defined. Then find corresponding y value(s) for that specific x using simple modular arithmetic.

    • @anusha5788
      @anusha5788 4 ปีที่แล้ว

      @@AdvancedMath So I simply choose values of x from {0,1, 2, ..., m-1} one by one and substitute in the curve equation to get the corresponding y value. At the end I will get all points on the curve. Got it. Your effort, sincerity and prompt response is much appreciated and very inspiring. Thank you very much and lots of respect.

    • @AdvancedMath
      @AdvancedMath  4 ปีที่แล้ว

      @@anusha5788 True.

    • @AdvancedMath
      @AdvancedMath  4 ปีที่แล้ว

      @@anusha5788 True. But do not forget to use modular arithmetic while simplifying equation.

    • @anusha5788
      @anusha5788 4 ปีที่แล้ว

      @@AdvancedMath Thank you I have learnt calculation of points without given initial point.

  • @minhanhle2863
    @minhanhle2863 8 หลายเดือนก่อน

    Can I ask about the case of singularity? Why does the curve have no well-defined tangents when 4a^3+27b^2=0?

  • @paraghaldankar4988
    @paraghaldankar4988 3 ปีที่แล้ว

    What is simplified equation to find value of y when x is known? Or without using modular square foot?

  • @marcomoldenhauer7903
    @marcomoldenhauer7903 2 ปีที่แล้ว

    Thanks a lot sir for the video. Only one question at [02:38] in the video. Why is (6,3) an element in your set? I added x=6 and y=3 in your equation y^2≡(x^3+1)mod(11) to check if (6,3) is an element. I did the following: 3^2≡(6^3+1)mod(11) 9≡(216+1)mod(11) 9≡(216+1)mod(11) 9≡(217)mod(11) 9≡(217)mod(11) 9≡8. As you can see 9 and and 217 are not in the same class of reminders. So, why is then (6,3) an element in your set? THX and BR

    • @AdvancedMath
      @AdvancedMath  2 ปีที่แล้ว +2

      That is a mistake. Sorry about that.

  • @alejandroamoros5339
    @alejandroamoros5339 2 ปีที่แล้ว

    I have tried to find numbers on the curve but for (3,6) and (6,8) I find that when using x=3: y^2=6 and x=6:y^2=8. What am I missing? Why is it y^2 and not y?

    • @shwetangacharya
      @shwetangacharya ปีที่แล้ว

      y^2=6 then y= sqrt(6), this means u have to find a number mod 11 whose square is 6, but u have only {1,3,4,5,9} as QR and NQR= {2,6,7,8,10)

  • @ANOWARALI-t9m
    @ANOWARALI-t9m ปีที่แล้ว

    Thanks for the video but the given points were big disaster.