Doubling a point on an elliptic curve












3












$begingroup$


I've a programming background and am just about to get into a project where Elliptic Curve Cryptography (ECC) is used. Although our libraries deal with the details I still like to do background reading so started with the ECC chapter of Understanding Cryptography. Everything was fine until I came upon this example of point doubling over $Z_{17}$:



enter image description here



I can't figure out how he gets $s$. For example the $(2cdot1)^{-1}(3cdot5^2+2)$ why does that evaluate to $2^{-1} cdot 9$? From looking at it I would have thought $(3cdot5^2+2)$ evaluates to $77$ so giving $2^{-1}cdot77$ or $77div2$ for the whole expression. Obviously the math doesn't work in the way I expect, is it something to do with the dot product not being normal multiplication? Or something else?



p.s. Sorry about formatting, I'm looking up how to do the Tex for the site now.






elliptic-curves







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are you familiar at all with modular arithmetic? $77=9$ in $mathbb{Z}_{17}$.
    $endgroup$
    – fretty
    Jun 27 '13 at 14:21








  • 1




    $begingroup$
    Yes, to be precise, the author should have written $(2cdot 1)^{-1}(3cdot 5^2+2)equiv 2^{-1}cdot 9 bmod 17$, and not just $=$.
    $endgroup$
    – Álvaro Lozano-Robledo
    Jun 27 '13 at 14:23










  • $begingroup$
    @ÁlvaroLozano-Robledo Thanks, can't believe it's that simple for that bit... So you have to apply the modulus operator the whole way through? So that bit is equivalent to $9cdot9$ because $9/2$ is 4.5 and $89mod17$ is 4 and you truncate in $Z$? Just not seeing how $9cdot9 = 13mod17$
    $endgroup$
    – Peanut
    Jun 27 '13 at 14:35








  • 1




    $begingroup$
    $9cdot 9 = 81 = 13 + 68 = 13 + 4cdot 17 equiv 13 bmod 17$.
    $endgroup$
    – Álvaro Lozano-Robledo
    Jun 27 '13 at 14:39






  • 1




    $begingroup$
    I asked nearly this identical question on Crypto crypto.stackexchange.com/questions/64456, and the answer there may also be worth looking at.
    $endgroup$
    – Brian M. Hunt
    Dec 1 '18 at 13:44
















3












$begingroup$


I've a programming background and am just about to get into a project where Elliptic Curve Cryptography (ECC) is used. Although our libraries deal with the details I still like to do background reading so started with the ECC chapter of Understanding Cryptography. Everything was fine until I came upon this example of point doubling over $Z_{17}$:



enter image description here



I can't figure out how he gets $s$. For example the $(2cdot1)^{-1}(3cdot5^2+2)$ why does that evaluate to $2^{-1} cdot 9$? From looking at it I would have thought $(3cdot5^2+2)$ evaluates to $77$ so giving $2^{-1}cdot77$ or $77div2$ for the whole expression. Obviously the math doesn't work in the way I expect, is it something to do with the dot product not being normal multiplication? Or something else?



p.s. Sorry about formatting, I'm looking up how to do the Tex for the site now.






elliptic-curves







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are you familiar at all with modular arithmetic? $77=9$ in $mathbb{Z}_{17}$.
    $endgroup$
    – fretty
    Jun 27 '13 at 14:21








  • 1




    $begingroup$
    Yes, to be precise, the author should have written $(2cdot 1)^{-1}(3cdot 5^2+2)equiv 2^{-1}cdot 9 bmod 17$, and not just $=$.
    $endgroup$
    – Álvaro Lozano-Robledo
    Jun 27 '13 at 14:23










  • $begingroup$
    @ÁlvaroLozano-Robledo Thanks, can't believe it's that simple for that bit... So you have to apply the modulus operator the whole way through? So that bit is equivalent to $9cdot9$ because $9/2$ is 4.5 and $89mod17$ is 4 and you truncate in $Z$? Just not seeing how $9cdot9 = 13mod17$
    $endgroup$
    – Peanut
    Jun 27 '13 at 14:35








  • 1




    $begingroup$
    $9cdot 9 = 81 = 13 + 68 = 13 + 4cdot 17 equiv 13 bmod 17$.
    $endgroup$
    – Álvaro Lozano-Robledo
    Jun 27 '13 at 14:39






  • 1




    $begingroup$
    I asked nearly this identical question on Crypto crypto.stackexchange.com/questions/64456, and the answer there may also be worth looking at.
    $endgroup$
    – Brian M. Hunt
    Dec 1 '18 at 13:44














3












3








3





$begingroup$


I've a programming background and am just about to get into a project where Elliptic Curve Cryptography (ECC) is used. Although our libraries deal with the details I still like to do background reading so started with the ECC chapter of Understanding Cryptography. Everything was fine until I came upon this example of point doubling over $Z_{17}$:



enter image description here



I can't figure out how he gets $s$. For example the $(2cdot1)^{-1}(3cdot5^2+2)$ why does that evaluate to $2^{-1} cdot 9$? From looking at it I would have thought $(3cdot5^2+2)$ evaluates to $77$ so giving $2^{-1}cdot77$ or $77div2$ for the whole expression. Obviously the math doesn't work in the way I expect, is it something to do with the dot product not being normal multiplication? Or something else?



p.s. Sorry about formatting, I'm looking up how to do the Tex for the site now.






elliptic-curves







share|cite|improve this question











$endgroup$




I've a programming background and am just about to get into a project where Elliptic Curve Cryptography (ECC) is used. Although our libraries deal with the details I still like to do background reading so started with the ECC chapter of Understanding Cryptography. Everything was fine until I came upon this example of point doubling over $Z_{17}$:



enter image description here



I can't figure out how he gets $s$. For example the $(2cdot1)^{-1}(3cdot5^2+2)$ why does that evaluate to $2^{-1} cdot 9$? From looking at it I would have thought $(3cdot5^2+2)$ evaluates to $77$ so giving $2^{-1}cdot77$ or $77div2$ for the whole expression. Obviously the math doesn't work in the way I expect, is it something to do with the dot product not being normal multiplication? Or something else?



p.s. Sorry about formatting, I'm looking up how to do the Tex for the site now.






elliptic-curves




elliptic-curves






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jun 27 '13 at 14:45







Peanut

















asked Jun 27 '13 at 14:14









PeanutPeanut

15029




15029












  • $begingroup$
    Are you familiar at all with modular arithmetic? $77=9$ in $mathbb{Z}_{17}$.
    $endgroup$
    – fretty
    Jun 27 '13 at 14:21








  • 1




    $begingroup$
    Yes, to be precise, the author should have written $(2cdot 1)^{-1}(3cdot 5^2+2)equiv 2^{-1}cdot 9 bmod 17$, and not just $=$.
    $endgroup$
    – Álvaro Lozano-Robledo
    Jun 27 '13 at 14:23










  • $begingroup$
    @ÁlvaroLozano-Robledo Thanks, can't believe it's that simple for that bit... So you have to apply the modulus operator the whole way through? So that bit is equivalent to $9cdot9$ because $9/2$ is 4.5 and $89mod17$ is 4 and you truncate in $Z$? Just not seeing how $9cdot9 = 13mod17$
    $endgroup$
    – Peanut
    Jun 27 '13 at 14:35








  • 1




    $begingroup$
    $9cdot 9 = 81 = 13 + 68 = 13 + 4cdot 17 equiv 13 bmod 17$.
    $endgroup$
    – Álvaro Lozano-Robledo
    Jun 27 '13 at 14:39






  • 1




    $begingroup$
    I asked nearly this identical question on Crypto crypto.stackexchange.com/questions/64456, and the answer there may also be worth looking at.
    $endgroup$
    – Brian M. Hunt
    Dec 1 '18 at 13:44


















  • $begingroup$
    Are you familiar at all with modular arithmetic? $77=9$ in $mathbb{Z}_{17}$.
    $endgroup$
    – fretty
    Jun 27 '13 at 14:21








  • 1




    $begingroup$
    Yes, to be precise, the author should have written $(2cdot 1)^{-1}(3cdot 5^2+2)equiv 2^{-1}cdot 9 bmod 17$, and not just $=$.
    $endgroup$
    – Álvaro Lozano-Robledo
    Jun 27 '13 at 14:23










  • $begingroup$
    @ÁlvaroLozano-Robledo Thanks, can't believe it's that simple for that bit... So you have to apply the modulus operator the whole way through? So that bit is equivalent to $9cdot9$ because $9/2$ is 4.5 and $89mod17$ is 4 and you truncate in $Z$? Just not seeing how $9cdot9 = 13mod17$
    $endgroup$
    – Peanut
    Jun 27 '13 at 14:35








  • 1




    $begingroup$
    $9cdot 9 = 81 = 13 + 68 = 13 + 4cdot 17 equiv 13 bmod 17$.
    $endgroup$
    – Álvaro Lozano-Robledo
    Jun 27 '13 at 14:39






  • 1




    $begingroup$
    I asked nearly this identical question on Crypto crypto.stackexchange.com/questions/64456, and the answer there may also be worth looking at.
    $endgroup$
    – Brian M. Hunt
    Dec 1 '18 at 13:44
















$begingroup$
Are you familiar at all with modular arithmetic? $77=9$ in $mathbb{Z}_{17}$.
$endgroup$
– fretty
Jun 27 '13 at 14:21






$begingroup$
Are you familiar at all with modular arithmetic? $77=9$ in $mathbb{Z}_{17}$.
$endgroup$
– fretty
Jun 27 '13 at 14:21






1




1




$begingroup$
Yes, to be precise, the author should have written $(2cdot 1)^{-1}(3cdot 5^2+2)equiv 2^{-1}cdot 9 bmod 17$, and not just $=$.
$endgroup$
– Álvaro Lozano-Robledo
Jun 27 '13 at 14:23




$begingroup$
Yes, to be precise, the author should have written $(2cdot 1)^{-1}(3cdot 5^2+2)equiv 2^{-1}cdot 9 bmod 17$, and not just $=$.
$endgroup$
– Álvaro Lozano-Robledo
Jun 27 '13 at 14:23












$begingroup$
@ÁlvaroLozano-Robledo Thanks, can't believe it's that simple for that bit... So you have to apply the modulus operator the whole way through? So that bit is equivalent to $9cdot9$ because $9/2$ is 4.5 and $89mod17$ is 4 and you truncate in $Z$? Just not seeing how $9cdot9 = 13mod17$
$endgroup$
– Peanut
Jun 27 '13 at 14:35






$begingroup$
@ÁlvaroLozano-Robledo Thanks, can't believe it's that simple for that bit... So you have to apply the modulus operator the whole way through? So that bit is equivalent to $9cdot9$ because $9/2$ is 4.5 and $89mod17$ is 4 and you truncate in $Z$? Just not seeing how $9cdot9 = 13mod17$
$endgroup$
– Peanut
Jun 27 '13 at 14:35






1




1




$begingroup$
$9cdot 9 = 81 = 13 + 68 = 13 + 4cdot 17 equiv 13 bmod 17$.
$endgroup$
– Álvaro Lozano-Robledo
Jun 27 '13 at 14:39




$begingroup$
$9cdot 9 = 81 = 13 + 68 = 13 + 4cdot 17 equiv 13 bmod 17$.
$endgroup$
– Álvaro Lozano-Robledo
Jun 27 '13 at 14:39




1




1




$begingroup$
I asked nearly this identical question on Crypto crypto.stackexchange.com/questions/64456, and the answer there may also be worth looking at.
$endgroup$
– Brian M. Hunt
Dec 1 '18 at 13:44




$begingroup$
I asked nearly this identical question on Crypto crypto.stackexchange.com/questions/64456, and the answer there may also be worth looking at.
$endgroup$
– Brian M. Hunt
Dec 1 '18 at 13:44










1 Answer
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$begingroup$

All arithmetic is done modulo $17$, so
$$(2cdot 1)^{-1}(3cdot 5^2+2)equiv 2^{-1}cdot 9bmod 17$$
is a congruence, not an equality in $mathbb{Q}$.



Moreover, $9cdot 9equiv 13 bmod 17$ because
$$9cdot 9 = 81 = 13 + 68 = 13 + 4cdot 17 equiv 13 bmod 17,$$
or, equivalently,
$$9cdot 9equiv (-8)cdot (-8)equiv 64equiv 16cdot 4equiv -4equiv 13bmod 17.$$






share|cite|improve this answer









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    $begingroup$

    All arithmetic is done modulo $17$, so
    $$(2cdot 1)^{-1}(3cdot 5^2+2)equiv 2^{-1}cdot 9bmod 17$$
    is a congruence, not an equality in $mathbb{Q}$.



    Moreover, $9cdot 9equiv 13 bmod 17$ because
    $$9cdot 9 = 81 = 13 + 68 = 13 + 4cdot 17 equiv 13 bmod 17,$$
    or, equivalently,
    $$9cdot 9equiv (-8)cdot (-8)equiv 64equiv 16cdot 4equiv -4equiv 13bmod 17.$$






    share|cite|improve this answer









    $endgroup$


















      6












      $begingroup$

      All arithmetic is done modulo $17$, so
      $$(2cdot 1)^{-1}(3cdot 5^2+2)equiv 2^{-1}cdot 9bmod 17$$
      is a congruence, not an equality in $mathbb{Q}$.



      Moreover, $9cdot 9equiv 13 bmod 17$ because
      $$9cdot 9 = 81 = 13 + 68 = 13 + 4cdot 17 equiv 13 bmod 17,$$
      or, equivalently,
      $$9cdot 9equiv (-8)cdot (-8)equiv 64equiv 16cdot 4equiv -4equiv 13bmod 17.$$






      share|cite|improve this answer









      $endgroup$
















        6












        6








        6





        $begingroup$

        All arithmetic is done modulo $17$, so
        $$(2cdot 1)^{-1}(3cdot 5^2+2)equiv 2^{-1}cdot 9bmod 17$$
        is a congruence, not an equality in $mathbb{Q}$.



        Moreover, $9cdot 9equiv 13 bmod 17$ because
        $$9cdot 9 = 81 = 13 + 68 = 13 + 4cdot 17 equiv 13 bmod 17,$$
        or, equivalently,
        $$9cdot 9equiv (-8)cdot (-8)equiv 64equiv 16cdot 4equiv -4equiv 13bmod 17.$$






        share|cite|improve this answer









        $endgroup$



        All arithmetic is done modulo $17$, so
        $$(2cdot 1)^{-1}(3cdot 5^2+2)equiv 2^{-1}cdot 9bmod 17$$
        is a congruence, not an equality in $mathbb{Q}$.



        Moreover, $9cdot 9equiv 13 bmod 17$ because
        $$9cdot 9 = 81 = 13 + 68 = 13 + 4cdot 17 equiv 13 bmod 17,$$
        or, equivalently,
        $$9cdot 9equiv (-8)cdot (-8)equiv 64equiv 16cdot 4equiv -4equiv 13bmod 17.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jun 27 '13 at 14:46









        Álvaro Lozano-RobledoÁlvaro Lozano-Robledo

        12.4k21839




        12.4k21839






























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