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Difference on montgomery curve equation between EFD and RFC7748

How is the curve equation used in ECC?Montgomery Ladder vs Double-and-AddWhat is the difference between order of base point and curve order in EC?Inversion Free Direct Conversion between Twisted Edwards (X,Y,Z) and Montgomery (X,Z)Differential addition on Montgomery curveHow Elliptic Curve equation is chosen?What is the difference between regular and “twisted” ECC curves?Understanding the elliptic curve equation by exampleDiscrete logarithm on Montgomery curve twistCurve 25519 (X25519, Ed25519) Convert coordinates between Montgomery curve and twisted Edwards curve

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

edited 5 hours ago

puzzlepalace

2,8701133

asked 6 hours ago

PierrePierre

36718

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  4 hours ago
  
  

  
  
  
  

add a comment | 

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

edited 5 hours ago

puzzlepalace

2,8701133

asked 6 hours ago

PierrePierre

36718

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  4 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

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

edited 5 hours ago

puzzlepalace

2,8701133

asked 6 hours ago

PierrePierre

36718

$endgroup$

 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)

 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)

share|improve this question

edited 5 hours ago

puzzlepalace

2,8701133

asked 6 hours ago

PierrePierre

36718

share|improve this question

edited 5 hours ago

puzzlepalace

2,8701133

asked 6 hours ago

PierrePierre

36718

edited 5 hours ago

puzzlepalace

2,8701133

edited 5 hours ago

puzzlepalace

2,8701133

asked 6 hours ago

PierrePierre

36718

asked 6 hours ago

PierrePierre

36718

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

answered 3 hours ago

Squeamish OssifrageSqueamish Ossifrage

19.3k12883

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

4

$begingroup$

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
$$
frac(x_1^2 - 1)^24x_1(x_1^2 + Ax_1 + 1),.
$$
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

answered 2 hours ago

Samuel NevesSamuel Neves

7,6402641

$endgroup$

add a comment | 

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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

answered 3 hours ago

Squeamish OssifrageSqueamish Ossifrage

19.3k12883

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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

answered 3 hours ago

Squeamish OssifrageSqueamish Ossifrage

19.3k12883

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

$begingroup$

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

answered 3 hours ago

Squeamish OssifrageSqueamish Ossifrage

19.3k12883

$endgroup$

share|improve this answer

answered 3 hours ago

Squeamish OssifrageSqueamish Ossifrage

19.3k12883

answered 3 hours ago

Squeamish OssifrageSqueamish Ossifrage

19.3k12883

answered 3 hours ago

Squeamish OssifrageSqueamish Ossifrage

19.3k12883

4

$begingroup$

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
$$
frac(x_1^2 - 1)^24x_1(x_1^2 + Ax_1 + 1),.
$$
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

answered 2 hours ago

Samuel NevesSamuel Neves

7,6402641

$endgroup$

add a comment | 

4

$begingroup$

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
$$
frac(x_1^2 - 1)^24x_1(x_1^2 + Ax_1 + 1),.
$$
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

answered 2 hours ago

Samuel NevesSamuel Neves

7,6402641

$endgroup$

add a comment | 

$begingroup$

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
$$
frac(x_1^2 - 1)^24x_1(x_1^2 + Ax_1 + 1),.
$$
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

answered 2 hours ago

Samuel NevesSamuel Neves

7,6402641

$endgroup$

share|improve this answer

answered 2 hours ago

Samuel NevesSamuel Neves

7,6402641

answered 2 hours ago

Samuel NevesSamuel Neves

7,6402641

answered 2 hours ago

Samuel NevesSamuel Neves

7,6402641