Curve25519: new Diffie-Hellman speed records
Daniel J. Bernstein

?

djb@cr.yp.to

Abstract. This paper explains the design and implementation of a highsecurity elliptic-curve-Diffie-Hellman function achieving record-setting
speeds: e.g., 832457 Pentium III cycles (with several side benefits: free
key compression, free key validation, and state-of-the-art timing-attack
protection), more than twice as fast as other authors’ results at the same
conjectured security level (with or without the side benefits).
Keywords: Diffie-Hellman, elliptic curves, point multiplication, new
curve, new software, high conjectured security, high speed, constant time,
short keys

1

Introduction

This paper introduces and analyzes Curve25519, a state-of-the-art elliptic-curveDiffie-Hellman function suitable for a wide variety of cryptographic applications.
This paper uses Curve25519 to obtain new speed records for high-security DiffieHellman computations.
Here is the high-level view of Curve25519: Each Curve25519 user has a 32byte secret key and a 32-byte public key. Each set of two Curve25519 users has
a 32-byte shared secret used to authenticate and encrypt messages between the
two users.
Medium-level view: The following picture shows the data flow from secret
keys through public keys to a shared secret.
Alice’s secret key a

Public string 9

Bob’s secret key b

Public function
Curve25519
AA
 
 z
%  
AA
}}
}
A
}
Alice’s public key
Bob’s
public key
AA
}}
A
}
Curve25519(a, 9) U }}
AA Curve25519(b, 9)
U}UUU
iAiA
i
i
}
i
U
i
U
AA
}
i
U
i
U
}
UUUiUiii
AA
i
U
}}
i
U
i
U
AA
}
i
UUUU
i
}
i
i
A
}
UU*
"
~}
tiii
|
{Bob, Alice}’s shared secret
{Alice, Bob}’s shared secret
=
Curve25519(b, Curve25519(a, 9))
Curve25519(a, Curve25519(b, 9))
?

Thanks to Tanja Lange for her extensive comments. Date of this document:
2006.02.09. Permanent ID of this document: 4230efdfa673480fc079449d90f322c0.
This document is final and may be freely cited.

A hash of the shared secret Curve25519(a, Curve25519(b, 9)) is used as the key
for a secret-key authentication system (to authenticate messages), or as the key
for a secret-key authenticated-encryption system (to simultaneously encrypt and
authenticate messages).
Low-level view: The Curve25519 function is Fp -restricted x-coordinate scalar
multiplication on E(Fp2 ), where p is the prime number 2255 − 19 and E is the
elliptic curve y 2 = x3 + 486662x2 + x. See Section 2 for further details.
Conjectured security level. Breaking the Curve25519 function—for example,
computing the shared secret from the two public keys—is conjectured to be
extremely difficult. Every known attack is more expensive than performing a
brute-force search on a typical 128-bit secret-key cipher.
The general problem of elliptic-curve discrete logarithms has been attacked
for two decades with very little success. Generic discrete-logarithm algorithms
break prime groups that are not sufficiently large, but the prime group used
in this paper has size above 2252 . Elliptic curves with certain special algebraic
structures can be broken much more quickly by non-generic algorithms, but
E(Fp2 ) does not have those structures. See Section 3 of this paper for more
detailed comments on the security of the Curve25519 function.
If large quantum computers are built then they will break Curve25519 and
all other short-key discrete-logarithm systems. See [56] for details of a general
elliptic-curve-discrete-logarithm algorithm. The ramifications of this observation
are orthogonal to the topic of this paper and are not discussed further.
Efficiency. My public-domain Curve25519 software provides several efficiency
features, thanks in large part to the choice of the Curve25519 function:
• Extremely high speed. My software computes Curve25519 in just 832457
cycles on a Pentium III, 957904 cycles on a Pentium 4, 640838 cycles on a
Pentium M, and 624786 cycles on an Athlon. Each of these numbers is a
new speed record for high-security Diffie-Hellman functions. I am working
on implementations for the UltraSPARC, PowerPC, etc.; I expect to end up
with similar cycle counts.
• No time variability. Most speed reports in the cryptographic literature are
for software without any protection against timing attacks. See [12], [51],
and [50] for some successful attacks. Adding protection can dramatically
slow down the computation. In contrast, my Curve25519 software is already
immune to timing attacks, including hyperthreading attacks and other cachetiming attacks. It avoids all input-dependent branches, all input-dependent
array indices, and other instructions with input-dependent timings.
• Short secret keys. The Curve25519 secret key is only 32 bytes. This is
typical for high-security Diffie-Hellman functions.
• Short public keys. The Curve25519 public key is only 32 bytes. Typical
elliptic-curve-Diffie-Hellman functions use 64-byte public keys; those keys
can be compressed to half size, as suggested by Miller in [46], but the time
for decompression is quite noticeable and usually not reported.
• Free key validation. Typical elliptic-curve-Diffie-Hellman functions can be
broken if users do not validate public keys; see, e.g., [14, Section 4.1] and [3].

The time for key validation is quite noticeable and usually not reported. In
contrast, every 32-byte string is accepted as a Curve25519 public key.
• Short code. My software is very small. The compiled code, including all
necessary tables, is around 16 kilobytes on each CPU, and can easily fit
alongside other networking tools in the CPU’s instruction cache.
The new speed records are the highlight of this paper. Sections 4 and 5 explain
the computation of Curve25519 in detail from the bottom up.
One can improve speed by choosing functions at lower security levels; for
example, dropping from 255 bits down to 160 bits. But—as discussed in Section
3—I can easily imagine an attacker with the resources to break a 160-bit elliptic
curve in under a year. Users should not expose themselves to this risk; they
should instead move up to the comfortable security level of Curve25519.
Of course, when users exchange large volumes of data, their bottleneck is a
secret-key cryptosystem, and the Curve25519 speed no longer matters.
Comparison to previous work. There is an extensive literature analyzing the
speed of various implementations of various Diffie-Hellman functions at various
conjectured security levels.
In particular, there have been some reports of high-security elliptic-curve
scalar-multiplication speeds: [17, Table 8] reports 1920000 cycles on a 400 MHz
Pentium II for field size 2256 − 2224 + 2192 + 296 − 1; [33, Table 7] reports 1740000
cycles on a 400 MHz Pentium II for field size 2283 using a subfield curve; [4,
Table 4] reports 3086000 cycles on a 1000 MHz Athlon for a random 256-bit
prime field. At a lower security level: [7, Table 3] reports 2650000 cycles on a
233 MHz Pentium MMX for field size (231 − 1)6 ; [58, Table 4] reports 4500000
cycles on a 166 MHz Pentium Pro for field size (231 − 19)6 ; [26, Table 6] reports
1720000 cycles on an 800 MHz Pentium III for field size 2233 .
The Curve25519 timings are more than twice as fast as the above reports. The
comparison is actually even more lopsided than this, because the Curve25519
timings include free key compression, free key validation, and state-of-the-art
timing-attack protection, while the above reports do not.
I have previously reported preliminary implementation work achieving about
half of this speedup using a standard NIST curve. The other half of the speedup
relies on switching to a better-designed curve. This paper covers both halves of
the speedup.
At a lower level, designing and implementing an elliptic-curve-Diffie-Hellman
function means making many choices that affect speed. Making a few bad choices
can destroy performance. In the design and implementation of Curve25519 I have
tried to globally optimize the entire sequence of choices:
• Use large characteristic, not characteristic 2.
• Use curve shape y 2 = x3 + Ax2 + x, with (A − 2)/4 small, rather than
y 2 = x3 − 3x + a6 .
• Use x as a public key, not (x, y).
• Use a secure curve that also has a secure twist, rather than taking extra time
to prohibit keys on the twist.

•
•
•
•
•
•
•
•
•

Use x/z inside scalar multiplication, not (x/z, y/z) or (x/z 2 , y/z 3 ).
Convert variable array indexing into arithmetic.
Use a fixed position for the leading 1 in the secret key.
Multiply the secret key by a small power of 2 to account for cofactors in the
curve group and the twist group.
Use a prime field, not an extension field.
Use a prime extremely close to 2b for some b.
Use radix 2b/w for some w, even if b/w is not an integer.
Allow coefficients slightly larger than the radix, rather than reducing each
coefficient as soon as possible.
Put coefficients into floating-point registers, not integer registers. Choose w
accordingly.

See Sections 4 and 5 for details and credits. Beware that these choices interact
across many levels of design and implementation: for example, there are other
curve shapes and prime shapes for which (x/z 2 , y/z 3 ) is better than x/z. This
type of interaction makes the optimal sequence of choices difficult to identify
even when all possible choices are known.

2

Specification

This section defines the Curve25519 function. Readers not familiar with rings,
fields, and elliptic curves should consult Appendix A for definitions and for a
proof of Theorem 2.1.
Theorem 2.1. Let p be a prime number with p ≥ 5. Let A be an integer such
that A2 − 4 is not a square modulo p. Define E as the elliptic curve y 2 = x3 +
Ax2 + x over the field Fp . Define X0 : E(Fp2 ) → Fp2 as follows: X0 (∞) = 0;
X0 (x, y) = x. Let n be an integer. Let q be an element of Fp . Then there exists
a unique s ∈ Fp such that X0 (nQ) = s for all Q ∈ E(Fp2 ) such that X0 (Q) = q.
In particular, define p as the prime 2255 − 19. Define Fp as the prime field
Z/p = Z/(2255 −
Fp2 as the field
√ 19). Note that 2 is not a square in Fp ; define
255
2
(Z/(2 − 19))[ 2]. Define A = 486662. Note that 486662 − 4 is not a square in
Fp . Define E as the elliptic curve y 2 = x3 + Ax2 + x over Fp . Define a function
X0 : E(Fp2 ) → Fp2 as follows: X0 (∞) = 0; X0 (x, y) = x. Define a function
X : E(Fp2 ) → {∞} ∪ Fp2 as follows: X(∞) = ∞; X(x, y)
 = x.
At this point I could say that, given n ∈ 2254 + 8 0, 1, 2, 3, . . . , 2251 − 1
and q ∈ Fp , the Curve25519 function produces s in Theorem 2.1. However, to
match cryptographic reality and to catch the types of design error explained by
Menezes in [45], I will instead define the inputs and outputs of Curve25519 as
sequences of bytes.
The set of bytes is, by definition, {0, 1, . . . , 255}. The encoding of a byte as
a sequence of bits is not relevant
to this document. Write s 7→ s for the standard

little-endian bijection from 0, 1, . . . , 2256 − 1 to theset {0, 1, . . . , 255}32 of 32byte strings: in other words, for each integers ∈ 0, 1, . . . , 2256 − 1 , define
s = (s mod 256, bs/256c mod 256, . . . , s/25631 mod 256).

The set of Curve25519
public keys is, by definition, {0, 1, . . . , 255}32 ; in

other words, q : q ∈ 0, 1, . . . , 2256 − 1 . The set of Curve25519 secret keys
is, by definition,{0, 8, 16, 24, . . . ,248} × {0, 1, . . . , 255}30 × {64, 65, 66, . . . , 127};
in other words, n : n ∈ 2254 + 8 0, 1, 2, 3, . . . , 2251 − 1 .
Now Curve25519 : {Curve25519 secret keys} × {Curve25519
public keys} →

{Curve25519public keys} is defined as follows. Fix q ∈ 0, 1, . . . , 2256 − 1 and
n ∈ 2254 + 8 0, 1, 2, 3, . . . , 2251 − 1 . By Theorem 2.1, there is a unique integer
s ∈ 0, 1, 2, . . . , 2255 − 20 with the following property: s = X0 (nQ) for all
Q ∈ E(Fp2 ) such that X0 (Q) = q mod 2255 − 19. Finally, Curve25519(n, q) is
defined as s. Note that Curve25519 is not surjective: in particular, its final output
bit is always 0 and need not be transmitted.

3

Security

This section discusses attacks on Curve25519. The bottom line is that all known
attacks are extremely expensive.
Responsibilities of the user. The legitimate users are assumed to generate
independent uniform random secret keys. A user can, for example, generate 32
uniform random bytes, clear bits 0, 1, 2 of the first byte, clear bit 7 of the last
byte, and set bit 6 of the last byte.
Large deviations from uniformity can eliminate all security. For example, if
the first 16 bytes of the secret key n were instead chosen as a public constant,
then a moderately large computation would deduce the remaining bytes of n
from the public key Curve25519(n, 9). This is not Curve25519’s fault; the user
is responsible for putting enough randomness into keys.
Legitimate users are also assumed to keep their secret keys secret. This means
that a secret key n is not used except to compute the public key Curve25519(n, 9)
and to compute the shared-secret hash H(Curve25519(n, q)) given q.
Users are not assumed to throw n away after a single q. Diffie-Hellman secret
keys can—and, for efficiency, should—be reused with many public keys, as in
[23, Section 3]. Each user’s secret key n is combined with many other users’
public keys q1 , q2 , q3 , . . ., producing shared-secret hashes H(Curve25519(n, q1 )),
H(Curve25519(n, q2 )), H(Curve25519(n, q3 )), . . . .
Choice of key-derivation function. There are no theorems guaranteeing the
safety of any particular key-derivation function H with, e.g., 512-bit output.
Some silly choices of H are breakable. As an extreme example, if H outputs just
64 bits followed by all zeros, then an attacker can perform a brute-force search
for those 64 bits.
On the other hand, from the perspective of a secret-key cryptographer, it
seems very easy to design a safe function H. A small amount of mixing, far less
than necessary to make a safe secret-key cipher, stops all known attacks.
For concreteness I will define H(x0 , x1 , x2 , x3 , x4 , x5 , x6 , x7 ) as the 64-byte
string Salsa20(c0 , x0 , 0, x1 , x2 , c1 , x3 , 0, 0, x4, c2 , x5 , x6 , 0, x7 , c3 ). Here Salsa20 is

the function defined in [13, Section 8]; (c0 , c1 , c2 , c3 ) is “Curve25519output” in
ASCII; and each xi has 4 bytes.
If fewer than 64 bytes are needed then the Salsa20 output can simply be
truncated. If more than 64 bytes are needed then Salsa20 can be invoked again
with (c0 , x0 , 1, x1 , . . .) to produce another 64 bytes.
Powers of the attacker. An attacker sees public keys q1 = Curve25519(n1 , 9),
q2 = Curve25519(n2 , 9), . . . generated from the legitimate users’ independent
uniform random secret keys n1 , n2 , . . ..
The attacker also sees messages protected by a secret-key cryptosystem C
where the keys for C are the shared-secret hashes H(Curve25519(ni , qj )) =
H(Curve25519(nj , qi )) for various sets {i, j}. The attacker’s goal is to decrypt
or forge these messages.
The attacker can also compute a public key q 0 ∈
/ {q1 , q2 , . . .} and—by using
0
q in the Diffie-Hellman protocol—see messages protected by C where the keys
for C are H(Curve25519(n1 , q 0 )), H(Curve25519(n2 , q 0 )), . . . . This would be
pointless if the attacker generated q 0 in the normal way, but the attacker is not
required to generate q 0 in the normal way; legitimate users are not assumed to
check that q 0 was generated from a secret key, let alone a secret key known to
the attacker. The attacker might take q 0 = 1, for example, or q 0 = q1 ⊕ 1. The
attacker can adaptively generate many public keys q 0 .
Of course, security depends on the choice of secret-key cryptosystem C. One
could make a poor choice of C, allowing messages to be decrypted or forged
without any weakness in Curve25519. But standard choices of C are conjectured
to be safe. Further discussion of the choice of C is outside the scope of this
document.
Simplified attack notions. There are many papers using simpler models of
Diffie-Hellman attackers, and proving theorems of the form “a fast attack in
complicated-security-model implies a fast attack in simplified-security-model.”
The reader might wonder why I am not using one of these simplified notions.
Example: Bentahar in [10], improving an algorithm by Muzereau, Smart,
and Vercauteren in [48] based on an idea by Maurer in [44], showed that one
can evaluate discrete logarithms on typical elliptic curves using roughly 213 calls
to a reliable oracle for the function (mQ, nQ) 7→ mnQ. Bentahar then repeated
the standard conjecture that computing discrete logarithms on a typical 256-bit
elliptic curve costs at least 2128 (never mind the question of exactly what “cost”
means), and deduced the conjecture that computing (mQ, nQ) 7→ mnQ costs at
least 2115 . Why, then, should one make a conjecture regarding the difficulty of
computing (mQ, nQ) 7→ mnQ, rather than a simplified conjecture regarding the
difficulty of computing discrete logarithms?
Answer: A standard conjecture says that computing (mQ, nQ) 7→ mnQ costs
at least 2128 . This conjecture is quantitatively stronger than anything that can
be obtained by applying Bentahar’s theorem to a simplified conjecture.
Similar comments apply to other theorems of this type; see, e.g., [39, Section
3.2]. Often the theorems are so weak that they say nothing about any real-world
system. To focus attention on the security properties that applications actually

need, I have chosen to make a complicated but strong conjecture about security,
rather than a simplified but weak conjecture.
Generic discrete logarithms by the rho and kangaroo methods. The
attacker can expand Curve25519(n, 9) into a point (x, y) on E(Fp2 ), namely the
nth multiple of the base point (9, . . .). The attacker can then use Pollard’s rho
method or Pollard’s kangaroo method to compute the discrete logarithm of this
point, namely n. The main cost in either method is the cost of performing a huge
number of additions of elliptic-curve points; both methods are almost perfectly
parallelizable, with negligible communication costs. See [63], [55], [61], and [60].
The number of additions here is about the square root of the length of the
n interval: in this case, about 2125 . The computation can finish after far fewer
additions, but the success chance is at most (and conjecturally at least) about
a2 /2251 after a additions.
How many elliptic-curve additions can an attacker perform? The traditional
estimate is roughly 270 elliptic-curve additions: a modern CPU costs about 26
dollars; a modern CPU cycle is about 2−31 seconds; each elliptic-curve addition
in the rho or kangaroo method costs about 210 CPU cycles for roughly 22 field
multiplications that each cost 28 cycles; the attacker is willing to spend a year,
i.e., 225 seconds; the attacker can afford to spend 230 dollars.
I don’t agree with the traditional estimate. I agree that modern circuitry
takes about 2−21 seconds for a single rho/kangaroo step; but it is a huge error to
assume that this circuitry costs as much as 26 dollars. One can fit many parallel
rho/kangaroo circuits into the same amount of circuitry as a modern CPU. A
reasonable estimate for “many” is 210 ; see [28] for a fairly detailed chip design,
and [28, Section 5.2] for the estimate. By switching to this chip, the attacker
can perform roughly 280 elliptic-curve additions. The attacker has an excellent
chance of computing a 160-bit discrete logarithm, but only about a 2−90 chance
of computing a 251-bit discrete logarithm.
Of course, one must adjust these estimates as chip technology improves. It
is not enough to account for increases in cycle speed and for decreases in chip
cost; one must also account for increases in chip size. However, the Curve25519
security level will remain comfortable for the foreseeable future.
Batch discrete logarithms. Silverman and Stapleton observed, and Kuhn
and Struik proved in [41, Section 4] assuming standard conjectures,
that the rho
√
method can compute u discrete logarithms using about u times as much effort
as computing a single discrete logarithm.
For example, given public keys Curve25519(n1 , 9), . . . , Curve25519(nu , 9), √
the
125
attacker can discover most √
of the secret keys n1 , . . . , nu using only about 2
u
125
additions, i.e., about 2 / u additions per key.
This does √
not mean, however, that one of the keys will be found within
125
the first 2 / u additions. On the contrary: the attacker is√likely to wait for
2125 additions before finding the first key, then another 2125 ( 2 − 1) additions
before finding the second key, etc. Curve25519 is at a comfortable security level
where finding the first key is, conjecturally, far out of reach, so the reduced cost
of finding subsequent keys is not a threat. The attacker can perform only 2125 

additions for small , so the attacker’s chance of success—of finding any keys—is
only about 2 .
Generic discrete logarithms are often claimed to be about as difficult as
brute-force search for a half-size key. But brute-force search computes a batch of
u keys with about the same effort as computing a single key. Furthermore, bruteforce search has probability roughly u of finding some key after the first  of
the computation, whereas discrete logarithms have only an 2 chance. Evidently
generic discrete logarithms are more difficult than brute-force search for a halfsize key: u is much larger than 2 , except in the extreme case where u and  are
both close to 1.
Small-subgroup attacks. If the subgroup of E(Fp2 ) generated by the base
point (9, . . .) has non-prime order then the attacker can use the Pohlig-Hellman
method to save time in computing discrete logarithms. See, e.g., [5, Section 19.3].
This attack fails against Curve25519. The order of the base point is a prime,
namely 2252 + 27742317777372353535851937790883648493.
An active attacker has more options. Say there is a point (x, y) ∈ E(Fp2 )
of order b, with x ∈ Fp and with b not very large. The attacker can issue a
public key x. The legitimate user will then authenticate and encrypt data under
H(Curve25519(n, x)) = H(X0 (n(x, y))) = H(X0 ((n mod b)(x, y))); the attacker
can compare the results to all possibilities for n mod b, presumably determining
n mod b.
The active attack also fails against Curve25519. The group {∞} ∪ (E(F p2 ) ∩
(Fp × Fp )) has size 8p1 , where p1 = 2252 + · · · √is the prime number displayed
above. The “twist” group {∞}∪(E(Fp2 )∩(Fp × 2Fp )) has size 2(p+1)−8p1 =
4p2 , where p2 is the prime 2253 − 55484635554744707071703875581767296995.
Consequently, the only possibilities for b below 2252 are 1, 2, 4, 8. Secret keys n
by definition have n mod 8 = 0 and thus n mod b = 0.
History: Lim and Lee in [42] pointed out active attacks on Diffie-Hellman
in the group F∗p . They recommended in [42, Section 4] that, rather than taking
the time to test that public keys are in a particular subgroup of prime order q,
one choose a prime p such that “each prime factor of (p − 1)/2q is larger than
q.” Biehl, Meyer, and Müller in [14, Section 4.1] pointed out analogous attacks
on elliptic curves when public keys are represented as pairs (x, y); they did not
propose any workaround other than testing keys. In a November 2001 sci.crypt
posting I wrote “You can happily skip both the y transmission and the square
root. In fact, if both the curve and its twist have nearly prime order, then you
can even skip square testing.”
Other attacks. The kangaroo method actually searches
simultaneously for n/8

and p1 − n/8 in an interval. Therange of n/8 is 2251 , . . . , 2252 − 1 , so either
n/8 or p1 − n/8 is in the range (p1 + 1)/2, . . . , 2252 − 1 . However, p1 is only
marginally above 2252 , so this range has length only marginally below 2251 .
More generally, when a group G has an easily computed automorphism ϕ of
small order
method to the orbits of ϕ, using only
p b, one can apply the kangaroo
√
about #G/b steps rather than #G steps. See, e.g., [5, Section 19.5.5]. But
my elliptic curve has no structure of this type other than negation. In fact, it

has no complex endomorphisms of small norm. To prove this, compute the trace
t = p + 1 − 8p1 , and observe that t2 − 4p is not a small multiple of a square: it
is divisible once by the prime 8312956054562778877481, for example.
My elliptic curve also resists the transfer attacks surveyed in [30, Chapter
22]. The primes p1 and p2 do not equal the field characteristic p. The order of p
modulo p1 is not small: in fact, it is (p1 − 1)/6. The order of p modulo p2 is not
small: in fact, it is p2 − 1. Weil descent simply splits E(Fp2 ) into the subgroup
E(Fp ), of order 8p1 , and the twist, of order 4p2 ; there are no proper subfields of
Fp to exploit.

4

Fast arithmetic modulo 2255 − 19

This section explains one way to use common CPU instructions, specifically
floating-point instructions, to quickly multiply and add in the field Fp where
p = 2255 − 19. I will focus on the Pentium M for concreteness, but the same
techniques work well for a wide variety of CPUs. This section also discusses the
choice of field structure and the choice of prime.
In this section, “floating-point” is abbreviated “fp.”
Representing
integers modulo 2255 −19. Define R as the ring of polynomials
P
i
d25.5ie
. One way to see that R is a
i ui x where ui is an integer multiple of 2
ring is to observe that it is the intersection of the subrings Z[x] and Z[225.5 x] of
Z[x], where Z is the ring of algebraic integers in C.
Elements of R represent elements of Z/(2255 −19): each polynomial represents
its value at 1. Often a polynomial is chosen to meet two restrictions:
• The polynomial degree is small, to limit the number of coefficients that need
to be multiplied as part of polynomial multiplication. Specifically, reduceddegree polynomials have degree at most 9.
• Each coefficient ui is a small multiple of 2d25.5ie , to limit the effort of
multiplying coefficients.
Specifically, reduced-coefficient polynomials have

ui /2d25.5ie ∈ −225 , −225 + 1, . . . , −1, 0, 1, . . . , 225 − 1, 225 .

To summarize: A reduced-degree reduced-coefficient polynomial is a polynomial
u0 +u1 x+· · ·+u9 x9 with u0 /20 , u1/226 , u2 /251 , u3 /277 , u4 /2102 , u5 /2128 , u6 /2153 ,
u7 /2179 , u8 /2204 , u9 /2230 all in −225 , −225 + 1, . . . , −1, 0, 1, . . . , 225 − 1, 225 .
This polynomial represents the integer u0 + u1 + · · · + u9 .
Note that integers are not converted to a unique “smallest” representation
until the end of the Curve25519 computation. Producing reduced representations
is generally much faster than producing “smallest” representations.
Representing coefficients inside CPUs. The Pentium M has eight “fp
registers,”
which holds a real number 2e f for integers e and f with
 64 each of
f ∈ −2 , . . . , 264 and with e in an adequate range for all the computations
discussed here. My computations hold polynomial coefficients in fp registers to
the extent possible, as in [11, Section 4].

The Pentium M has many
“L1-cache doublewords” that can hold 2e f
 more
with f limited to the range −253 , . . . , 253 ; e.g., reduced coefficients. To perform
arithmetic on numbers in L1-cache doublewords, the Pentium M must take time
to copy (“load”) the numbers into registers; but this is not a big problem, because
these loads can be overlapped with arithmetic if they are not too frequent.
Why split 255-bit integers into ten 26-bit pieces, rather than nine 29-bit
pieces or eight 32-bit pieces? Answer: The coefficients of a polynomial product
do not fit into the Pentium M’s fp registers if pieces are too large. The cost of
handling larger coefficients outweighs the savings of handling fewer coefficients.
The overall time for 29-bit pieces is sufficiently competitive to warrant further
investigation, but so far I haven’t been able to save time this way. I’m sure that
32-bit pieces, the most common choice in the literature, are a bad idea.
Of course, the same question must be revisited for each CPU. The Pentium 1,
Pentium MMX, Pentium Pro, Pentium II, Pentium III, Pentium 4, Athlon, and
Athlon XP work well with 26-bit pieces; on the Athlon 64 and Opteron, 32-bit
pieces
be slightly better. On
UltraSPARC and PowerPC, fp registers
 might
 the
53
53
64
use −2 , . . . , 2
rather than −2 , . . . , 264 , and I recommend twelve 22bit pieces. The UltraSPARC and PowerPC can overlap fp additions with fp
multiplications, so I expect them to end up with comparable cycle counts to the
Pentium M despite the larger number of pieces.
Given that there are 10 pieces, why use radix 225.5 rather than, e.g., radix
225 or radix 226 ? Answer: My ring R contains 2255 x10 − 19, which represents
0 in Z/(2255 − 19). I will reduce polynomial products modulo 2255 x10 − 19 to
eliminate the coefficients of x10 , x11 , etc. With radix 225 , the coefficient of x10
could not be eliminated. With radix 226 , coefficients would have to be multiplied
by 25 · 19 rather than just 19, and the results would not fit into an fp register.
Using floating-point operations. The Pentium M has circuits for three fast
operations on numbers stored in fp registers: sum, difference, and product. These
are exact operations if the results fit into the 64-bit fp precision; otherwise the
results are rounded to the nearest fp numbers.
The Pentium M can perform, at best, one fp operation per cycle. About 92%
of the cycles in my Curve25519 computation (589825 out of 640838) are occupied
by fp operations. One can understand the cycle counts fairly well by simply
counting the fp operations. Similar comments apply to other CPUs, although
the details depend on the CPU.
Warning: Writing an fp program in the C programming language, and feeding
the result to a C compiler, often produces machine language that takes 3 or more
Pentium M cycles for each fp operation. Further discussion of this phenomenon
is outside the scope of this paper. My Curve25519 software is actually written
in qhasm, a new programming language designed for high-speed computations.
Beware that a few CPUs have input-dependent fp timings. An old example
is the Sun microSPARC-IIep. A newer example is the IBM PowerPC RS64 IV,
which takes an extra cycle to multiply by 0. Fast constant-time computations
on these CPUs need extra effort.

Adding integers modulo 2255 − 19. If two integers are represented by two
polynomials u and v then the sum of the two integers is represented by u + v.
Similarly, the difference of the two integers is represented by u − v.
If u and v are reduced-degree reduced-coefficient polynomials then computing
u+v (or u−v) involves 10 additions (or subtractions) of fp numbers. Note that the
sum is reduced-degree but usually not reduced-coefficient. In a long chain of sums
one would occasionally have to take extra time to reduce the coefficients. This
is never necessary in the Curve25519 computation: every sum (and difference)
is used solely as input to products, as Appendix B illustrates.
Statistics: Each addition or subtraction takes 10 fp operations. There are
8 additions and subtractions, totalling 80 fp operations, in each iteration of
the Curve25519 main loop. There are 2040 additions and subtractions, totalling
20400 fp operations, in the entire Curve25519 computation.
Multiplying integers modulo 2255 − 19. If two integers are represented by
polynomials u and v then their product is represented by the polynomial product
uv. If u and v are reduced-degree reduced-coefficient polynomials, or sums of
two such polynomials, then computing uv in the simplest way involves 100 fp
multiplications and 81 fp additions; I am experimenting with other polynomialmultiplication algorithms and expect to end up with slightly better results. The
product uv is then replaced by a reduced-degree reduced-coefficient polynomial:
• The coefficients of x10 , x11 , . . . , x18 in uv are eliminated by reduction modulo
2255 x10 −19. For example, the coefficient of x18 is multiplied by 19·2−255 and
added to the coefficient of x8 . Each reduction involves 1 fp multiplication
and 1 fp addition.
• The “high” part of each coefficient is subtracted from that coefficient and
added (“carried”) to the next coefficient. The high part is, by definition, the
nearest multiple of the power of 2 for the next coefficient. One carry involves
4 fp additions: 2 to identify the high part (by a rounded addition and then
subtraction of a large constant), 1 to subtract, and 1 to add.
Starting from uv, I carry from x8 to x9 , then from x9 to x10 ; then I eliminate
coefficients of x10 , x11 , . . . , x18 ; then I carry from x0 to x1 , from x1 to x2 , . . . ,
from x7 to x8 , and once more from x8 to x9 . Note that the coefficient of x9 is a
multiple of 2230 , and is between −2254 and 2254 after subtraction of its original
high part, so the final carry from x8 to x9 produces reduced coefficients. Overall
there are 18 fp operations to eliminate 9 coefficients, and 44 fp operations for
11 carries. There are many other reasonable carry sequences; on some CPUs it
might be a good idea to have two parallel carry chains, decreasing latency at the
expense of an extra carry.
Squaring is easier than general multiplication, because polynomial squaring
is easier than general polynomial multiplication. Overall a squaring eliminates
92 + 9 coefficient multiplications at the expense of 9 initial coefficient doublings;
note that doubling coefficients at the beginning is slightly better than doubling
products later. Multiplication by a small constant is also easier than general
multiplication, because the constant is represented by a polynomial of degree 0.

Statistics: Each multiplication by a small constant takes 55 fp operations.
Each squaring takes 162 fp operations. Each general multiplication takes 243 fp
operations. Each iteration of the Curve25519 main loop has 1 multiplication by a
small constant, using 55 fp operations; 4 squarings, using 648 fp operations; and
5 general multiplications, using 1215 fp operations; in total 10 multiplications,
using 1918 fp operations. The Curve25519 computation has 255 multiplications
by small constants, using 14025 fp operations; 1274 squarings, using 206388 fp
operations; and 1286 general multiplications, using 312498 fp operations; in total
2815 multiplications, using 532911 fp operations.
Note that the squaring-to-multiplication floating-point-operation ratio is only
162/243 = 2/3, far below the 0.8 ratio often used in the literature for estimating
the costs of elliptic-curve operations.
Selecting integers. Consider the problem of computing x[b], where x[0], x[1]
are integers modulo 2255 −19 and b is an input-dependent bit. Using b as an array
index—without taking extra time for preloads, interrupt elimination, etc.—could
allow hyperthreading attacks and other cache-timing attacks; see [12, Sections
8–15]. I instead compute x[b] as (1−b)x[0]+bx[1]. Similarly, if I need to compute
the pair (x[b], x[1 − b]), I compute (x[0] − b(x[0] − x[1]), x[1] + b(x[0] − x[1])).
Statistics: Each iteration of the Curve25519 main loop has 2 fp operations
inside computing b and 1 − b; 2 paired selections, taking 80 fp operations; and
2 more selections, taking 60 more fp operations. The total is 142 fp operations.
The entire Curve25519 computation spends 36210 fp operations, about 6% of
the total, on selection. Of course, these operations could be eliminated if timing
attacks were not a concern.
Why this field? CPUs include fast integer-multiplication circuits (usually
buried inside fp-multiplication circuits aimed at the large fp market) but not
circuits for fast multiplication of polynomials modulo 2. Characteristic-2 fields
allow several other speedups—see, e.g., [35, Section 3.4] and [25, Section 15.1]—
but I can’t see any way for them to set speed records on existing CPUs.
“Optimal extension fields,” such as degree-10 extensions of prime fields of
size around 226 , are advertised in [7] and [6] as allowing faster multiplication and
much faster inversion, perhaps so fast as to make affine-coordinate elliptic-curve
computations faster than projective-coordinate elliptic-curve computations. My
current assessment is that these fields have some slight advantages: there are no
carry chains, so operations are easier to reorder; there are 10 reductions modulo
a prime, rather than 11 carries, although one reduction is usually slightly more
expensive than one carry; inversion is faster, although not fast enough to make
affine coordinates worthwhile; and, most importantly, degree 9 might fit into
64-bit fp. Unfortunately, these fields have a huge disadvantage: even if they are
slightly faster on some CPUs, they are much slower on other CPUs. A 255-bit
integer can be split into 4 or 8 or 10 or 12 pieces to accommodate the capabilities
of different processors; an “optimal extension field” is tied to a particular number
of pieces.

So I selected a prime field. Prime fields also have the virtue of minimizing
the number of security concerns for elliptic-curve cryptography; see, e.g., [29]
and [22].
I chose my prime 2255 − 19 according to the following criteria: primes as close
as possible to a power of 2 save time in field operations (as in, e.g, [9]), with no
effect on (conjectured) security level; primes slightly below 32k bits, for some
k, allow public keys to be easily transmitted in 32-bit words, with no serious
concerns regarding wasted space; k = 8 provides a comfortable security level. I
considered the primes 2255 + 95, 2255 − 19, 2255 − 31, 2254 + 79, 2253 + 51, and
2253 + 39, and selected 2255 − 19 because 19 is smaller than 31, 39, 51, 79, 95.

5

Fast Curve25519 computation

This section explains fast x-coordinate point addition on my elliptic curve y 2 =
x3 + 486662x2 + x; explains fast x-coordinate scalar multiplication, i.e., fast
computation of Curve25519; and compares this curve to other elliptic curves.
Recall that Section 2 defines two x-coordinate functions. One function X0
maps ∞ to 0; the other function X maps ∞ to ∞. Curve25519 is defined using
X0 , but inside the computation it is convenient to use X until the last moment.
Addition. Montgomery in [47, Section 10.3.1] published formulas to compute
X(2Q) given X(Q), and to compute X(Q + Q0 ) given X(Q), X(Q0), X(Q − Q0 ),
assuming that Q 6= ∞, Q0 6= ∞, Q − Q0 6= ∞, Q + Q0 6= ∞. It turns out that
Montgomery’s formulas also work for ∞, provided that Q − Q0 ∈
/ {∞, (0, 0)}, so
the Curve25519 computation can avoid checking for ∞. See Appendix B of this
paper.
Montgomery’s formulas represent each X value as a fraction x/z, replacing
divisions with multiplications. Montgomery commented that, when d is large,
one can perform d divisions in Fp at about the same cost as 4d multiplications
in Fp , so dividing x by z may be a good idea when there are many separate
elliptic-curve computations to perform at once; I have not implemented this
option yet.
The formula for X(2Q) involves 2 squarings, 1 multiplication by 121665 =
(486662−2)/4, and 2 more multiplications. The formula for X(Q+Q0 ) involves 2
squarings and 3 more multiplications when z1 in Theorem B.2, the denominator
of X(Q − Q0 ), is known to be 1; otherwise it involves 2 squarings and 4 more
multiplications. The Curve25519 computation always has z1 = 1.
Scalar multiplication. Montgomery suggested using his formulas to obtain
X(nQ + Q), X(nQ), X(Q) given X(bn/2cQ + Q), X(bn/2cQ), X(Q): if n is even
then nQ = 2bn/2cQ and nQ + Q = (bn/2cQ + Q) + (bn/2cQ); if n is odd then
nQ + Q = 2(bn/2cQ + Q) and nQ = (bn/2cQ + Q) + (bn/2cQ). Either case
involves one doubling and one addition.
The formulas, repeated k times, produce
X(Q)
with k
 X(nQ
 + Q), X(nQ),


k
k
doublings and k additions starting from
X(
n/2
Q
+
Q),
X(
n/2
Q),
X(Q).
I

254
251
compute X(nQ) for any n ∈ 2 + 8 0, 1, . . . , 2 − 1 with 255 doublings and

255 additions starting from X(Q), X(0), X(Q). The first and last few iterations
could be simplified.
The final X(nQ), like other X values, is represented as a fraction x/z. I
compute X0 (nQ) = xz p−2 using a straightforward sequence of 254 squarings
and 11 multiplications. This is about 7% of the Curve25519 computation. An
extended-Euclid inversion of z, randomized to protect against timing attacks,
might be faster, but the maximum potential speedup is very small, while the
cost in code complexity is large.
Theorems B.1 and B.2 justify the above procedure if X0 (Q) 6= 0. The same
formulas also work for X0 (Q) = 0: every computed fraction has denominator 0,
so the final output is 0 as desired.
Other addition
pointed out that one can replace the
 Montgomery

 chains.
k
k
addition chain n/2
∪ n/2 + 1 with any differential addition chain (any
“Lucas chain”), i.e., any addition chain where each sum is already accompanied
by a difference. One can find such a chain with only about 384 elements, as
discussed in [59, Section 5]. On the other hand, most of the additions then
require z1 6= 1 in Theorem B.2, costing extra multiplications in Fp . It is also
not clear how easily these addition chains can be protected against cache-timing
attacks. Further investigation is required.
A more common strategy is to drop the difference requirement, compensate
by computing more coordinates of each multiple of Q (Jacobian coordinates,
for example, or Chudnovsky coordinates), and use an addition chain with only
about 320 elements. See, e.g., [17] or [4]. Unfortunately, even if A is selected
so that y 2 = x3 + Ax2 + x is isomorphic to a curve y 2 = x3 − 3x − a6 , each
doubling in known coordinate systems takes at least 8 field multiplications, and
each general addition takes even more. All of my experiments with this strategy
have ended up using more field operations, more floating-point operations, and
more cycles than the x-coordinate strategy.
One can save a large fraction of the time for computing Curve25519(n, q)
when q is fixed—in particular, for computing public keys Curve25519(n, 9)—
by precomputing various multiples of (q, . . .). An essentially optimal algorithm,
published by Pippenger in [52] in 1976, computes u public keys with only about
256/lg 8u additions per key. This speedup is negligible in the Diffie-Hellman
context (and is not provided by my current software), since each key is used
many times; but the speedup is useful for other applications of elliptic curves.
Why this curve? I chose the curve shape y 2 = x3 + Ax2 + x, as suggested
by Montgomery, to allow extremely fast x-coordinate point operations. Curves
of this shape have order divisible by 4, requiring a marginally larger prime for
the same conjectured security level, but this is outweighed by the extra speed
of curve operations. I selected (A − 2)/4 as a small integer, as suggested by
Montgomery, to speed up the multiplication by (A − 2)/4; this has no effect on
the conjectured security level.
To protect against various attacks discussed in Section 3, I rejected choices
of A whose curve and twist orders were not {4 · prime, 8 · prime}; here 4, 8 are
minimal since p ∈ 1+4Z. The smallest positive choices for A are 358990, 464586,

and 486662. I rejected A = 358990 because one of its primes is slightly smaller
than 2252 , raising the question of how standards and implementations should
handle the theoretical possibility of a user’s secret key matching the prime;
discussing this question is more difficult than switching to another A. I rejected
464586 for the same reason. So I ended up with A = 486662.
Special curves with small complex automorphisms have potential benefits,
as discussed in [31], and are worth further investigation, but so far I have not
succeeded in saving time using them.

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A

Appendix: rings, fields, and curves

This appendix reviews elliptic curves at the level of generality of Theorem 2.1.
See [24, Chapter 13] for much more information about elliptic curves.
The base field. Let p be a prime number with p ≥ 5. Define Fp as the set
{0, 1, . . . , p − 1}. Define a binary operation + on Fp as addition mod p. Define
a binary operation · on Fp as multiplication mod p. Define a unary operation −
on Fp as negation mod p.
Fp is a commutative ring under 0, 1, −, +, ·. This means that it satisfies every
0, 1, −, +, · identity satisfied by Z; e.g., the identity a(b + c + 1) = ab + ac + a.
Furthermore, because p is prime, Fp is a field: every nonzero element of Fp has
a reciprocal in Fp .
Squares in the base field. Squaring is a 2-to-1 map on the nonzero elements
of Fp , so there are exactly (p − 1)/2 non-squares in Fp . Find the smallest δ ∈
{1, 2, . . . , p − 1} such that δ is not a square in Fp .
Fermat’s little theorem implies that α(p−1)/2 = 1 if α is a nonzero square
in Fp ; α(p−1)/2 = −1 if α is a non-square in Fp ; and α(p−1)/2 = 0 if α = 0.
Consequently, if α is a non-square in Fp , then α/δ is a nonzero square in Fp .
The extension field. Define Fp2 as the set Fp × Fp . Define a unary operation
− on Fp2 by −(c, d) = (−c, −d). Define a binary operation + on Fp2 by (a, b) +
(c, d) = (a + c, b + d). Define a binary operation · on Fp2 by (a, b) · (c, d) =
(ac + δbd, ad + bc).
Fp2 is a commutative ring under 0, 1, −, +, ·. Furthermore, each nonzero
(a, b) ∈ Fp2 has a reciprocal (a/(a2 − δb2 ), −b/(a2 − δb2 )) ∈ Fp2 .
The injection a 7→ (a, 0) from Fp to Fp2 is a ring morphism: it preserves
0, 1, −, +, ·. Thus (a, 0) is abbreviated
a without
risk of confusion. The element
√
√
(0, 1) of Fp2 is abbreviated δ; it satisfies δ 2 = (δ, 0) = δ.

The elliptic curve. Let A be an
A2 − 4 mod p is not a square
 integer such that
in Fp . Define E(Fp2 ) as {∞} ∪ (x, y) ∈ Fp2 : y 2 = x3 + Ax2 + x .
Define a unary operation − on E(Fp2 ) as follows: −∞ = ∞; −(x, y) =
(x, −y). Define a binary operation + on E(Fp2 ) as follows:
•
•
•
•
•

∞ + ∞ = ∞.
∞ + (x, y) = (x, y).
(x, y) + ∞ = (x, y).
(x, y) + (x, −y) = ∞.
If y 6= 0 then (x, y) + (x, y) = (x00 , y 00 ) where λ = (3x2 + 2Ax + 1)/2y,
x00 = λ2 − A − 2x = (x2 − 1)2 /4y 2 , and y 00 = λ(x − x00 ) − y. Here / refers to
division in Fp2 .
• If x0 6= x then (x, y) + (x0 , y 0 ) = (x00 , y 00 ) where λ = (y 0 − y)/(x0 − x),
x00 = λ2 − A − x − x0 , and y 00 = λ(x − x00 ) − y.

Standard (although lengthy) calculations show that E(Fp2 ) is a commutative
group under ∞, −, +. This means that every 0, −, + identity satisfied by Z is
also satisfied by E(Fp2 ) when 0 is replaced by ∞.
Note that the following three sets are subgroups of E(Fp2 ):
• {∞, (0, 0)}. Indeed, ∞ + ∞ = ∞; (0, 0) + (0, 0) = ∞; and (0, 0) + ∞ = (0, 0).
• {∞} ∪ (E(Fp2 ) ∩ (Fp × Fp )). Indeed, if x, y, x0 , y 0 ∈ Fp then the quantities
λ, x00 , y 00 defined above are
√ in Fp .
• {∞} ∪ (E(Fp2 ) ∩ (Fp × δFp )). This time λ is a ratio of an element of Fp
√
√
and an element of√ δFp , and is therefore an element of δFp , producing
x00 ∈ Fp and y 00 ∈ δFp .
Note also that if x3 + Ax2 + x = 0 in Fp then x = 0. (Otherwise A2 − 4 =
(x − 1/x)2 in Fp , so A2 − 4 mod p is a square in Fp , contradiction.) In other
words, (x, 0) ∈
/ E(Fp2 ) if x 6= 0.
Proof of Theorem 2.1. Let n be an integer. Let q be an element of Fp . Define
α = q 3 +Aq 2 +q. Define X0 : E(Fp2 ) → Fp2 as follows: X0 (∞) = 0; X0 (x, y) = x.
I will show that there are exactly two Q ∈ E(Fp2 ) such that X0 (Q) = q, that
both of them have the same value of X0 (nQ), and that the value is in Fp . Here
nQ means the nth multiple of Q under the above group operations on E(Fp2 ).
 Case 1: α = 0. Then q = 0. The only square root of 0 in Fp2 is 0, so
Q ∈ E(Fp2 ) : X0 (Q) = q is exactly the group {∞, (0, 0)}. Thus each Q ∈
E(Fp2 ) with X0 (Q) = q has nQ ∈ {∞, (0, 0)}; i.e., X0 (nQ) = 0.
Case 2: α is a nonzero square in Fp . Select a square
 root r. Now q 6= 0, and the
3
2
only square roots of q + Aq + q in Fp2 are ±r, so Q ∈ E(Fp2 ) : X0 (Q) = q =
{(q, r), (q, −r)}. Define s = X0 (n(q, r)). The group {∞} ∪ (E(Fp2 ) ∩ (Fp × Fp ))
contains (q, r), so it contains n(q, r), so s ∈ {0, 1, 2, 3, . . . , p − 1}. Furthermore
n(q, −r) = n(−(q, r)) = −n(q, r), so X0 (n(q, −r)) = X0 (n(q, r)) = s. Thus
X0 (nQ) = s for all Q ∈ E(Fp2 ) such that X0 (Q) = q.
Case 3: α is a non-square in Fp . Then α/δ is a nonzero square in Fp . Select
3
2
a square root
√ of q +√Aq + q in
√ r. Now
 q 6= 0, and the only square roots
Fp2 are ±r δ, so Q ∈ E(Fp2 ) : X0 (Q) = q = {(q, r δ), (q, −r δ)}. Define

√
√
√
s = X0 (n(q, r δ)). √
The group {∞} ∪ (E(Fp2 ) ∩ (Fp × δFp )) contains (q,√
r δ),
so s ∈ {0, 1, 2, 3, . . . ,√
p − 1}. Furthermore
so it contains
n(q, r δ), √
√
√ n(q, −r δ) =
n(−(q, r δ)) = −n(q, r δ), so X0 (n(q, −r δ)) = X0 (n(q, r δ)) = s. Thus
t
u
X0 (nQ) = s for all Q ∈ E(Fp2 ) such that X0 (Q) = q.

B

Appendix: Montgomery’s double-and-add formulas

This appendix states Montgomery’s x-coordinate double-and-add formulas, and
proves that the formulas work whenever Q − Q0 ∈
/ {∞, (0, 0)}.
The following diagram summarizes Montgomery’s formulas in the case z1 = 1.
As in Theorems B.1 and B.2, x/z and x0 /z 0 are the x-coordinates of points Q, Q0 ;
x2 /z2 is the x-coordinate of 2Q; x1 is the x-coordinate of Q − Q0 ; and x3 /z3 is
the x-coordinate of Q + Q0 .
x OOO
z
x0 OOO
z0
OOOooooo
OOoOoooo
oO
ooOO
 woooo OOO' 
 woooo OOO' 
+
+ ZZZZZZZZ − OO
−
ZZZZZZOZOO
ZZOZOZZZZ
OOO Z ZZZZZZ


ZZZZ, 
'
× 4OO
×
×
O
OOO ooo ×
ooo
44 OOOo
o
Oo
O
Oo
oO
o
O
 woo44o4o OOO' 
 woooo OOO' 
44
×
−
+
−
44
44


44 / ×
×
×
44
44
 

+
×o

(A − 2)/4

x1


×

x2


x3


z2


z3

One can see at a glance that there are 4 squarings, 1 multiplication by (A − 2)/4,
and 5 other multiplications; and that there are 8 additions/subtractions, none
of which produce input to another addition/subtraction.
Theorem B.1. Let p be a prime number with p ≥ 5. Let A be an integer such
that A2 − 4 is not a square modulo p. Define E as the elliptic curve y 2 = x3 +
Ax2 + x over the field Fp . Define X : E(Fp2 ) → {∞} ∪ Fp2 as follows: X(∞) =
∞; X(x, y) = x. Fix x, z ∈ Fp with (x, z) 6= (0, 0). Define
x2 = (x2 − z 2 )2 = (x − z)2 (x + z)2 ,
z2 = 4xz(x2 + Axz + z 2 )


A−2
2
2
= ((x + z) − (x − z) ) (x + z) +
((x + z) − (x − z) ) .
4
2

2



2

Then X(2Q) = x2 /z2 for all Q ∈ E(Fp2 ) such that X(Q) = x/z.
Here x/z means the quotient of x and z in Fp if z 6= 0; it means ∞ if x 6= 0
and z = 0; it is undefined if x = z = 0.
Proof. Case 1: z = 0. Then x2 = x4 6= 0 and z2 = 0. Also X(Q) = x/0 = ∞ so
Q = ∞ so 2Q = ∞ so X(2Q) = ∞ = x2 /0 = x2 /z2 .
Case 2: z 6= 0 and x = 0. Then x2 = z 4 6= 0 and z2 = 0. Also X(Q) = 0/z = 0
so Q = (0, 0) so 2Q = ∞ so X(2Q) = ∞ = x2 /0 = x2 /z2 .
Case 3: z 6= 0 and x 6= 0. Then Q = (x/z, y) for some y ∈ Fp2 satisfying
2
y = (x/z)3 + A(x/z)2 + (x/z) and thus 4y 2 z 4 = 4(x3 z + Ax2 z 2 + xz 3 ) = z2 .
The non-squareness of A2 − 4 implies that y 6= 0; hence z2 6= 0. Also X(2Q) =
((x/z)2 − 1)2 /4y 2 by definition of doubling; thus z2 X(2Q) = z 4 ((x/z)2 − 1)2 =
(x2 − z 2 )2 = x2 .
t
u
Theorem B.2. In the context of Theorem B.1, fix x, z, x0 , z 0 , x1 , z1 ∈ Fp with
(x, z) 6= (0, 0), (x0 , z 0 ) 6= (0, 0), x1 6= 0, and z1 6= 0. Define
x3 = 4(xx0 − zz 0 )2 z1 = ((x − z)(x0 + z 0 ) + (x + z)(x0 − z 0 ))2 z1 ,

z3 = 4(xz 0 − zx0 )2 x1 = ((x − z)(x0 + z 0 ) − (x + z)(x0 − z 0 ))2 x1 .

Then X(Q+Q0 ) = x3 /z3 for all Q, Q0 ∈ E(Fp2 ) such that X(Q) = x/z, X(Q0 ) =
x0 /z 0 , and X(Q − Q0 ) = x1 /z1 .

Proof. Case 1: Q = Q0 . Then X(Q − Q0 ) = X(∞) = ∞, so z1 = 0, contradiction.
Case 2: Q = ∞. Then z = 0 and x 6= 0; also X(Q − Q0 ) = X(−Q0 ) = X(Q0 ),
so x1 /z1 = x0 /z 0 , so x0 6= 0 and z 0 6= 0. Finally x3 = 4(xx0 )2 z1 and z3 = 4(xz 0 )2 x1
so x3 /z3 = (x0 /z 0 )2 z1 /x1 = x0 /z 0 = X(Q0 ) = X(Q + Q0 ).
Case 3: Q0 = ∞. Then z 0 = 0 and x0 6= 0; also X(Q − Q0 ) = X(Q), so
x1 /z1 = x/z, so x 6= 0 and z 6= 0. Finally x3 = 4(xx0 )2 z1 and z3 = 4(zx0 )2 x1 so
x3 /z3 = (x/z)2 z1 /x1 = x/z = X(Q) = X(Q + Q0 ).
Case 4: Q = −Q0 . Then X(Q0 ) = X(Q) so x/z = x0 /z 0 so xz 0 = zx0 so
z3 = 0.
Suppose that x3 = 0. Then (x − z)(x0 + z 0 ) + (x + z)(x0 − z 0 ) = 0 and
(x−z)(x0 +z 0 )−(x+z)(x0 −z 0 ) = 0, so (x−z)(x0 +z 0 ) = 0 and (x+z)(x0 −z 0 ) = 0.
If x + z 6= 0 then x0 − z 0 = 0 so x0 + z 0 = 2x0 6= 0 so x − z = 0; i.e., X(Q) = 1 and
X(Q0 ) = 1. Otherwise x = −z so x − z = 2x 6= 0 so x0 = −z 0 ; i.e., X(Q) = −1
and X(Q0 ) = −1. Either way X(Q − Q0 ) = X(2Q) = (X(Q)2 − 1)2 / · · · =
(1 − 1)2 / · · · = 0 by definition of doubling, so x1 = 0, contradiction.
Thus x3 6= 0, and x3 /z3 = ∞ = X(∞) = X(Q + Q0 ).
Case 5: Q 6= ∞; Q0 6= ∞; Q 6= Q0 ; and Q 6= −Q0 . Then z 6= 0, z 0 6= 0,
and x/z 6= x0 /z 0 , so z3 6= 0. Find y, y 0 ∈ Fp2 such that Q = (x/z, y) and
Q0 = (x0 /z 0 , y 0 ). Write α = x0 /z 0 −x/z and β = A+x/z+x0 /z 0 . Then X(Q+Q0 ) =
((y 0 − y)/α)2 − β and X(Q − Q0 ) = ((−y 0 − y)/α)2 − β by definition of Q ± Q0 ,
so X(Q + Q0 )X(Q − Q0 ) = β 2 − 2β((y 0 )2 + y 2 )/α2 + ((y 0 )2 − y 2 )2 /α4 . Substitute
y 2 = (x/z)3 + A(x/z)2 + (x/z) and (y 0 )2 = (x0 /z 0 )3 + A(x0 /z 0 )2 + (x0 /z 0 ) and
simplify to see that X(Q+Q0 )X(Q−Q0 ) = (xx0 −zz 0 )2 /(xz 0 −x0 z)2 ; this is what
Montgomery did. Finally X(Q+Q0 ) = (xx0 −zz 0 )2 z1 /(xz 0 −x0 z)2 x1 = x3 /z3 . t
u