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Project Euler Solutions

Project Euler 182 Solution

Problem Description

The RSA encryption is based on the following procedure:

Generate two distinct primes p and q.
Compute n=pq and φ=(p-1)(q-1).
Find an integer e, 1<e<φ, such that gcd(e,φ)=1.

A message in this system is a number in the interval [0,n-1].
A text to be encrypted is then somehow converted to messages (numbers in the interval [0,n-1]).
To encrypt the text, for each message, m, c=me mod n is calculated.

To decrypt the text, the following procedure is needed: calculate d such that ed=1 mod φ, then for each encrypted message, c, calculate m=cd mod n.

There exist values of e and m such that me mod n=m.
We call messages m for which me mod n=m unconcealed messages.

An issue when choosing e is that there should not be too many unconcealed messages.
For instance, let p=19 and q=37.
Then n=19*37=703 and φ=18*36=648.
If we choose e=181, then, although gcd(181,648)=1 it turns out that all possible messages
m (0≤mn-1) are unconcealed when calculating me mod n.
For any valid choice of e there exist some unconcealed messages.
It’s important that the number of unconcealed messages is at a minimum.

Choose p=1009 and q=3643.
Find the sum of all values of e, 1<e<φ(1009,3643) and gcd(e,φ)=1, so that the number of unconcealed messages for this value of e is at a minimum.

Analysis

At the heart of this problem is the simple formula: [gcd(e-1,p-1)+1][gcd(e-1,q-1)+1] as outlined in many texts such as Recent Advances in RCA Cryptography in the chapter Properties of the RSA cryptosystem. Specifically Lemma 5.1 on page 69 shows a nice proof of this equation.

So encoding this into a program is straightforward and simple. Just sum all values of e where gcd(e,phi) = 1 and gcd(e-1,q-1) and gcd(e-1,p-1) are at a minimum (equal 2).

Project Euler 182 Solution

Runs in under 1 second in Python.

from Euler import gcd
p, q, sum, e = 1009, 3643, 0, 3
phi = (p-1) * (q-1)
 
while (e<phi):
    if gcd(e, phi)==1 and gcd(e-1, q-1)==2 and gcd(e-1, p-1)==2:
        sum += e
    e += 4
 
print "Answer to PE 182 =", sum

Comments

Christner discovered the every 4 rule. See the gcd subroutine here.

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