Never Use (M**e) for Ciphers

“Go” and do it the Montgomery Reduction Way!

Our world of public key cryptographic is build within finite fields. We take a prime number N, and then all our operations are done with (mod N). The basic security of our systems are then determine by the number of bits in N. If N has 1,024 bits or more, we are normally fairly secure. Many of our operations are then done with multiplication or with exponential, such as:

Cipher = M^e (mod N)

or with:

Cipher = ab (mod N)

In our labs I see quite a quite few students then doing this:

Cipher = (M**e % N)

And while it will work with relatively small values, it will never work in production, as the calculation of M^e will take a long time. For example, let’s take an e value of 65,637 (the most common value for e in RSA is 65,537), and M as 11, and see the size of the number we produce:

And so if we do M^e and then (mod N) it is going to be rather inefficient.

With the RSA and the Diffie-Hellman method we perform large exponential calculations, such as:

C=M^e (mod N)

and where we will continually multiply large integers by an exponent to get a result. If we were to just calculate x^y and then take (mod N) it would take a while to produce the result (possibly minutes for large numbers). Thus we use Montgomery modular multiplication, and which significantly reduces the time to compute the result. In a traditional multiplication of two value (x and y) for a modulus of N, we multiply x times y and then divide by N to find the remainder. The number of bits in the multiplication with this be the number of bits in x added to the number of bits in y. In Montgomery reduction we add multiples of N in order to simplify the multiplication.

An example of this is here, and a sample run for x=10, y=5 and N=29 is [here]:

a=10, b=5, p=29Result: 10*5 (mod 29) = 21
Traditional method result = 21
Result: 10^5 (mod 29) = 8
Traditional method result = 8

In this case we get 50 (mod 29) which is 21, and 105 (mod 29) which is 8.

The sample code is [here]:

package main

import (

// == From
type mont struct {
n uint // m.BitLen()
m *big.Int // modulus, must be odd
r2 *big.Int // (1<<2n) mod m
func newMont(m *big.Int) *mont {
if m.Bit(0) != 1 {
return nil
n := uint(m.BitLen())
x := big.NewInt(1)
x.Sub(x.Lsh(x, n), m)
return &mont{n, new(big.Int).Set(m), x.Mod(x.Mul(x, x), m)}
func (m mont) reduce(t *big.Int) *big.Int {
a := new(big.Int).Set(t)
for i := uint(0); i < m.n; i++ {
if a.Bit(0) == 1 {
a.Add(a, m.m)
a.Rsh(a, 1)
if a.Cmp(m.m) >= 0 {
a.Sub(a, m.m)
return a
// ==
func main() { p:=13
argCount := len(os.Args[1:]) if (argCount>0) {a,_= strconv.Atoi(os.Args[1])}
if (argCount>1) {b,_= strconv.Atoi(os.Args[2])}
if (argCount>2) {p,_= strconv.Atoi(os.Args[3])}
m := big.NewInt(int64(p)) fmt.Printf("a=%d, b=%d, p=%d\n\n",a,b,p)

mr := newMont(m)

x1 := big.NewInt(int64(a))
x2 := big.NewInt(int64(b))

t1 := mr.reduce(new(big.Int).Mul(x1, mr.r2))
t2 := mr.reduce(new(big.Int).Mul(x2, mr.r2))
res := mr.reduce(new(big.Int).Mul(t1, t2))
fmt.Printf("Result: %s*%s (mod %s) = %s\n",x1,x2,m,mr.reduce(res)) mul:=new(big.Int).Mul(x1, x2)
fmt.Printf("Traditional method result = %s\n\n",mul.Mod(mul,m))
prod := mr.reduce(mr.r2)
base := mr.reduce(new(big.Int).Mul(x1, mr.r2))
exp := new(big.Int).Set(x2)
for exp.BitLen() > 0 {
if exp.Bit(0) == 1 {
prod = mr.reduce(prod.Mul(prod, base))
exp.Rsh(exp, 1)
base = mr.reduce(base.Mul(base, base))
fmt.Printf("\nResult: %s^%s (mod %s) = %s\n",x1,x2,m,mr.reduce(prod)) fmt.Printf("Traditional method result = %s",new(big.Int).Exp(x1, x2, m))


And there you go. If you are into Python, don’t do this:

C = m**e % N

Do this:

C = pow(m,e,N)

and it will use the Montgomery method. In this article I’ve used Go, and you can see how I use big.Int. In Python, we automatically cast to Big Ints, when we need them.

Prof Bill Buchanan OBE

Written by

Professor of Cryptography. Serial innovator. Believer in fairness, justice & freedom. EU Citizen. Auld Reekie native. Old World Breaker. New World Creator.

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