Difference on montgomery curve equation between EFD and RFC7748












3












$begingroup$


There is a subtle difference between the 2 implementations for a Montgomery curve defined from the 2 following links



https://hyperelliptic.org/EFD/g1p/auto-montgom-xz.html





      A = X2+Z2

      AA = A^2

      B = X2-Z2

      BB = B^2

      E = AA-BB

      C = X3+Z3

      D = X3-Z3

      DA = D*A

      CB = C*B

      X5 = (DA+CB)^2

      Z5 = X1*(DA-CB)^2

      X4 = AA*BB

      Z4 = E*(BB+a24*E)


https://tools.ietf.org/html/rfc7748





       A = x_2 + z_2

       AA = A^2

       B = x_2 - z_2

       BB = B^2

       E = AA - BB

       C = x_3 + z_3

       D = x_3 - z_3

       DA = D * A

       CB = C * B

       x_3 = (DA + CB)^2

       z_3 = x_1 * (DA - CB)^2

       x_2 = AA * BB

       z_2 = E * (AA + a24 * E)


This AA / BB change on the last line does affect the result of a point multiplication with same input parameters.



Is there a reason for that difference ?






elliptic-curves x25519 rfc7748 x448







share|improve this question











$endgroup$












  • $begingroup$
    It looks to be a typo in RFC. When BB is used (as in EFD and original P.L. Montgomery paper), the test vectors can be reproduced. Submitted a review comment to RFC. Errare humanum est. How many existing implementations will fail to inter-operate ?
    $endgroup$
    – Pierre
    2 hours ago


















3












$begingroup$


There is a subtle difference between the 2 implementations for a Montgomery curve defined from the 2 following links



https://hyperelliptic.org/EFD/g1p/auto-montgom-xz.html





      A = X2+Z2

      AA = A^2

      B = X2-Z2

      BB = B^2

      E = AA-BB

      C = X3+Z3

      D = X3-Z3

      DA = D*A

      CB = C*B

      X5 = (DA+CB)^2

      Z5 = X1*(DA-CB)^2

      X4 = AA*BB

      Z4 = E*(BB+a24*E)


https://tools.ietf.org/html/rfc7748





       A = x_2 + z_2

       AA = A^2

       B = x_2 - z_2

       BB = B^2

       E = AA - BB

       C = x_3 + z_3

       D = x_3 - z_3

       DA = D * A

       CB = C * B

       x_3 = (DA + CB)^2

       z_3 = x_1 * (DA - CB)^2

       x_2 = AA * BB

       z_2 = E * (AA + a24 * E)


This AA / BB change on the last line does affect the result of a point multiplication with same input parameters.



Is there a reason for that difference ?






elliptic-curves x25519 rfc7748 x448







share|improve this question











$endgroup$












  • $begingroup$
    It looks to be a typo in RFC. When BB is used (as in EFD and original P.L. Montgomery paper), the test vectors can be reproduced. Submitted a review comment to RFC. Errare humanum est. How many existing implementations will fail to inter-operate ?
    $endgroup$
    – Pierre
    2 hours ago
















3












3








3





$begingroup$


There is a subtle difference between the 2 implementations for a Montgomery curve defined from the 2 following links



https://hyperelliptic.org/EFD/g1p/auto-montgom-xz.html





      A = X2+Z2

      AA = A^2

      B = X2-Z2

      BB = B^2

      E = AA-BB

      C = X3+Z3

      D = X3-Z3

      DA = D*A

      CB = C*B

      X5 = (DA+CB)^2

      Z5 = X1*(DA-CB)^2

      X4 = AA*BB

      Z4 = E*(BB+a24*E)


https://tools.ietf.org/html/rfc7748





       A = x_2 + z_2

       AA = A^2

       B = x_2 - z_2

       BB = B^2

       E = AA - BB

       C = x_3 + z_3

       D = x_3 - z_3

       DA = D * A

       CB = C * B

       x_3 = (DA + CB)^2

       z_3 = x_1 * (DA - CB)^2

       x_2 = AA * BB

       z_2 = E * (AA + a24 * E)


This AA / BB change on the last line does affect the result of a point multiplication with same input parameters.



Is there a reason for that difference ?






elliptic-curves x25519 rfc7748 x448







share|improve this question











$endgroup$




There is a subtle difference between the 2 implementations for a Montgomery curve defined from the 2 following links



https://hyperelliptic.org/EFD/g1p/auto-montgom-xz.html





      A = X2+Z2

      AA = A^2

      B = X2-Z2

      BB = B^2

      E = AA-BB

      C = X3+Z3

      D = X3-Z3

      DA = D*A

      CB = C*B

      X5 = (DA+CB)^2

      Z5 = X1*(DA-CB)^2

      X4 = AA*BB

      Z4 = E*(BB+a24*E)


https://tools.ietf.org/html/rfc7748





       A = x_2 + z_2

       AA = A^2

       B = x_2 - z_2

       BB = B^2

       E = AA - BB

       C = x_3 + z_3

       D = x_3 - z_3

       DA = D * A

       CB = C * B

       x_3 = (DA + CB)^2

       z_3 = x_1 * (DA - CB)^2

       x_2 = AA * BB

       z_2 = E * (AA + a24 * E)


This AA / BB change on the last line does affect the result of a point multiplication with same input parameters.



Is there a reason for that difference ?






elliptic-curves x25519 rfc7748 x448




elliptic-curves x25519 rfc7748 x448






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 3 hours ago









puzzlepalace

2,8601133




2,8601133










asked 4 hours ago









PierrePierre

36718




36718












  • $begingroup$
    It looks to be a typo in RFC. When BB is used (as in EFD and original P.L. Montgomery paper), the test vectors can be reproduced. Submitted a review comment to RFC. Errare humanum est. How many existing implementations will fail to inter-operate ?
    $endgroup$
    – Pierre
    2 hours ago




















  • $begingroup$
    It looks to be a typo in RFC. When BB is used (as in EFD and original P.L. Montgomery paper), the test vectors can be reproduced. Submitted a review comment to RFC. Errare humanum est. How many existing implementations will fail to inter-operate ?
    $endgroup$
    – Pierre
    2 hours ago


















$begingroup$
It looks to be a typo in RFC. When BB is used (as in EFD and original P.L. Montgomery paper), the test vectors can be reproduced. Submitted a review comment to RFC. Errare humanum est. How many existing implementations will fail to inter-operate ?
$endgroup$
– Pierre
2 hours ago






$begingroup$
It looks to be a typo in RFC. When BB is used (as in EFD and original P.L. Montgomery paper), the test vectors can be reproduced. Submitted a review comment to RFC. Errare humanum est. How many existing implementations will fail to inter-operate ?
$endgroup$
– Pierre
2 hours ago












2 Answers
2






active

oldest

votes


















4












$begingroup$

This is not a bug: it arises from different choice of sign in the definition of a24 := (a ± 2)/4; the RFC uses - while the EFD uses +.



RFC, following the Curve25519 paper:




The constant a24 is (486662 - 2) / 4 = 121665 for curve25519/X25519 and (156326 - 2) / 4 = 39081 for curve448/X448.




EFD, following Montgomery's paper (paywall-free):




Assumptions: 4*a24=a+2.




This apparent discrepancy was raised by Paul Lambert on the CFRG mailing list during discussion on the draft. It doesn't really matter which one you choose, as long as you're consistent about it!






share|improve this answer









$endgroup$













  • $begingroup$
    Thanks for the explanation. I didn't spot the little difference on a24 definition between the RFC and the EFD.
    $endgroup$
    – Pierre
    13 mins ago



















2












$begingroup$

This is not a typo; it is a difference in how the Montgomery doubling formula was derived between the original paper and the curve25519 paper. Both are correct.



To double a point on a Montgomery curve
$$
y^2 = x^3 + Ax^2 + x,,
$$
one has the identity relating the doubled point $(x_3, cdot)$ and the source point $(x_1, cdot)$:
$$
x_3 4x_1(x_1^2 + Ax_1 + 1) = (x_1^2 - 1)^2,.
$$
The doubled point $x_3$ can thus be computed as the fraction
$$
frac{(x_1^2 - 1)^2}{4x_1(x_1^2 + Ax_1 + 1)},.
$$
But to minimize the operation number, and obtain several common subexpressions, we can write $(x_1^2 - 1)^2$ as $(x_1+1)^2(x_1-1)^2$, $4x_1$ as $(x_1 + 1)^2 - (x_1 - 1)^2$, and $x_1^2 + Ax_1 + 1$ as either $(x_1-1)^2 + ((A+2)/4)4x_1$ or $(x_1+1)^2 + ((A-2)/4)4x_1$. It is this latter somewhat arbitrary choice that results in there being two almost identical Montgomery doubling formulas.






share|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

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    active

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    4












    $begingroup$

    This is not a bug: it arises from different choice of sign in the definition of a24 := (a ± 2)/4; the RFC uses - while the EFD uses +.



    RFC, following the Curve25519 paper:




    The constant a24 is (486662 - 2) / 4 = 121665 for curve25519/X25519 and (156326 - 2) / 4 = 39081 for curve448/X448.




    EFD, following Montgomery's paper (paywall-free):




    Assumptions: 4*a24=a+2.




    This apparent discrepancy was raised by Paul Lambert on the CFRG mailing list during discussion on the draft. It doesn't really matter which one you choose, as long as you're consistent about it!






    share|improve this answer









    $endgroup$













    • $begingroup$
      Thanks for the explanation. I didn't spot the little difference on a24 definition between the RFC and the EFD.
      $endgroup$
      – Pierre
      13 mins ago
















    4












    $begingroup$

    This is not a bug: it arises from different choice of sign in the definition of a24 := (a ± 2)/4; the RFC uses - while the EFD uses +.



    RFC, following the Curve25519 paper:




    The constant a24 is (486662 - 2) / 4 = 121665 for curve25519/X25519 and (156326 - 2) / 4 = 39081 for curve448/X448.




    EFD, following Montgomery's paper (paywall-free):




    Assumptions: 4*a24=a+2.




    This apparent discrepancy was raised by Paul Lambert on the CFRG mailing list during discussion on the draft. It doesn't really matter which one you choose, as long as you're consistent about it!






    share|improve this answer









    $endgroup$













    • $begingroup$
      Thanks for the explanation. I didn't spot the little difference on a24 definition between the RFC and the EFD.
      $endgroup$
      – Pierre
      13 mins ago














    4












    4








    4





    $begingroup$

    This is not a bug: it arises from different choice of sign in the definition of a24 := (a ± 2)/4; the RFC uses - while the EFD uses +.



    RFC, following the Curve25519 paper:




    The constant a24 is (486662 - 2) / 4 = 121665 for curve25519/X25519 and (156326 - 2) / 4 = 39081 for curve448/X448.




    EFD, following Montgomery's paper (paywall-free):




    Assumptions: 4*a24=a+2.




    This apparent discrepancy was raised by Paul Lambert on the CFRG mailing list during discussion on the draft. It doesn't really matter which one you choose, as long as you're consistent about it!






    share|improve this answer









    $endgroup$



    This is not a bug: it arises from different choice of sign in the definition of a24 := (a ± 2)/4; the RFC uses - while the EFD uses +.



    RFC, following the Curve25519 paper:




    The constant a24 is (486662 - 2) / 4 = 121665 for curve25519/X25519 and (156326 - 2) / 4 = 39081 for curve448/X448.




    EFD, following Montgomery's paper (paywall-free):




    Assumptions: 4*a24=a+2.




    This apparent discrepancy was raised by Paul Lambert on the CFRG mailing list during discussion on the draft. It doesn't really matter which one you choose, as long as you're consistent about it!







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered 1 hour ago









    Squeamish OssifrageSqueamish Ossifrage

    19.3k12883




    19.3k12883












    • $begingroup$
      Thanks for the explanation. I didn't spot the little difference on a24 definition between the RFC and the EFD.
      $endgroup$
      – Pierre
      13 mins ago


















    • $begingroup$
      Thanks for the explanation. I didn't spot the little difference on a24 definition between the RFC and the EFD.
      $endgroup$
      – Pierre
      13 mins ago
















    $begingroup$
    Thanks for the explanation. I didn't spot the little difference on a24 definition between the RFC and the EFD.
    $endgroup$
    – Pierre
    13 mins ago




    $begingroup$
    Thanks for the explanation. I didn't spot the little difference on a24 definition between the RFC and the EFD.
    $endgroup$
    – Pierre
    13 mins ago











    2












    $begingroup$

    This is not a typo; it is a difference in how the Montgomery doubling formula was derived between the original paper and the curve25519 paper. Both are correct.



    To double a point on a Montgomery curve
    $$
    y^2 = x^3 + Ax^2 + x,,
    $$
    one has the identity relating the doubled point $(x_3, cdot)$ and the source point $(x_1, cdot)$:
    $$
    x_3 4x_1(x_1^2 + Ax_1 + 1) = (x_1^2 - 1)^2,.
    $$
    The doubled point $x_3$ can thus be computed as the fraction
    $$
    frac{(x_1^2 - 1)^2}{4x_1(x_1^2 + Ax_1 + 1)},.
    $$
    But to minimize the operation number, and obtain several common subexpressions, we can write $(x_1^2 - 1)^2$ as $(x_1+1)^2(x_1-1)^2$, $4x_1$ as $(x_1 + 1)^2 - (x_1 - 1)^2$, and $x_1^2 + Ax_1 + 1$ as either $(x_1-1)^2 + ((A+2)/4)4x_1$ or $(x_1+1)^2 + ((A-2)/4)4x_1$. It is this latter somewhat arbitrary choice that results in there being two almost identical Montgomery doubling formulas.






    share|improve this answer









    $endgroup$


















      2












      $begingroup$

      This is not a typo; it is a difference in how the Montgomery doubling formula was derived between the original paper and the curve25519 paper. Both are correct.



      To double a point on a Montgomery curve
      $$
      y^2 = x^3 + Ax^2 + x,,
      $$
      one has the identity relating the doubled point $(x_3, cdot)$ and the source point $(x_1, cdot)$:
      $$
      x_3 4x_1(x_1^2 + Ax_1 + 1) = (x_1^2 - 1)^2,.
      $$
      The doubled point $x_3$ can thus be computed as the fraction
      $$
      frac{(x_1^2 - 1)^2}{4x_1(x_1^2 + Ax_1 + 1)},.
      $$
      But to minimize the operation number, and obtain several common subexpressions, we can write $(x_1^2 - 1)^2$ as $(x_1+1)^2(x_1-1)^2$, $4x_1$ as $(x_1 + 1)^2 - (x_1 - 1)^2$, and $x_1^2 + Ax_1 + 1$ as either $(x_1-1)^2 + ((A+2)/4)4x_1$ or $(x_1+1)^2 + ((A-2)/4)4x_1$. It is this latter somewhat arbitrary choice that results in there being two almost identical Montgomery doubling formulas.






      share|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        This is not a typo; it is a difference in how the Montgomery doubling formula was derived between the original paper and the curve25519 paper. Both are correct.



        To double a point on a Montgomery curve
        $$
        y^2 = x^3 + Ax^2 + x,,
        $$
        one has the identity relating the doubled point $(x_3, cdot)$ and the source point $(x_1, cdot)$:
        $$
        x_3 4x_1(x_1^2 + Ax_1 + 1) = (x_1^2 - 1)^2,.
        $$
        The doubled point $x_3$ can thus be computed as the fraction
        $$
        frac{(x_1^2 - 1)^2}{4x_1(x_1^2 + Ax_1 + 1)},.
        $$
        But to minimize the operation number, and obtain several common subexpressions, we can write $(x_1^2 - 1)^2$ as $(x_1+1)^2(x_1-1)^2$, $4x_1$ as $(x_1 + 1)^2 - (x_1 - 1)^2$, and $x_1^2 + Ax_1 + 1$ as either $(x_1-1)^2 + ((A+2)/4)4x_1$ or $(x_1+1)^2 + ((A-2)/4)4x_1$. It is this latter somewhat arbitrary choice that results in there being two almost identical Montgomery doubling formulas.






        share|improve this answer









        $endgroup$



        This is not a typo; it is a difference in how the Montgomery doubling formula was derived between the original paper and the curve25519 paper. Both are correct.



        To double a point on a Montgomery curve
        $$
        y^2 = x^3 + Ax^2 + x,,
        $$
        one has the identity relating the doubled point $(x_3, cdot)$ and the source point $(x_1, cdot)$:
        $$
        x_3 4x_1(x_1^2 + Ax_1 + 1) = (x_1^2 - 1)^2,.
        $$
        The doubled point $x_3$ can thus be computed as the fraction
        $$
        frac{(x_1^2 - 1)^2}{4x_1(x_1^2 + Ax_1 + 1)},.
        $$
        But to minimize the operation number, and obtain several common subexpressions, we can write $(x_1^2 - 1)^2$ as $(x_1+1)^2(x_1-1)^2$, $4x_1$ as $(x_1 + 1)^2 - (x_1 - 1)^2$, and $x_1^2 + Ax_1 + 1$ as either $(x_1-1)^2 + ((A+2)/4)4x_1$ or $(x_1+1)^2 + ((A-2)/4)4x_1$. It is this latter somewhat arbitrary choice that results in there being two almost identical Montgomery doubling formulas.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 42 mins ago









        Samuel NevesSamuel Neves

        7,6202641




        7,6202641






























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