Fermat’s Little Theorem - Mount Holyoke Collegejjlee/Teaching/Math 232 Introduction to RSA.pdf · Fermat’s Little Theorem Recall Fermat’s Little Theorem: Theorem (Fermat’s

Fermat’s Little Theorem

Recall Fermat’s Little Theorem:

Theorem (Fermat’s Little Theorem)

If p is prime and a is an integer not divisible by p (that is, gcd(a, p) = 1),then

ap−1 ≡ 1 (mod p).

Furthermore, for every integer a we have

ap ≡ a (mod p).

Fermat’s Little Theorem extends to a more general case.

Discrete Mathematics Introduction to RSA




If p is prime and a is an integer not divisible by p

(that is, gcd(a, p) = 1),then

ap−1 ≡ 1 (mod p).


ap ≡ a (mod p).






If p is prime and a is an integer not divisible by p (that is, gcd(a, p) = 1),

thenap−1 ≡ 1 (mod p).


ap ≡ a (mod p).







ap−1 ≡ 1 (mod p).


ap ≡ a (mod p).







ap−1 ≡ 1 (mod p).


ap ≡ a (mod p).







ap−1 ≡ 1 (mod p).


ap ≡ a (mod p).



Euler’s φ-function

Recall Euler’s φ-function:

φ(n) = the number of positive integers less than or equal to n

that are relatively prime to n

= n− the number of integers less than or equal to n

that are not relatively prime to n.

Example

1 φ(5) = 4, because 1, 2, 3, 4 are all relatively prime to 5 and 5 is not.

2 To compute φ(10), consider {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. It is easy tosee that φ(10) = 4.








Example










Example










Example

1 φ(5) = 4,

because 1, 2, 3, 4 are all relatively prime to 5 and 5 is not.









Example










Example


2 To compute φ(10), consider {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.

It is easy tosee that φ(10) = 4.








Example


2 To compute φ(10), consider {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. It is easy tosee that φ(10) =

4.








Example




Properties of Euler’s φ-function

Remark

Let p, q be primes.

Then

1 φ(p) = p− 1.

2 φ(pk) = pk − pk−1 for any positive integer k.

3 φ(pq) = (p− 1)(q − 1).

Example

1 {1, 2, 3, 4, 5, 6, 7}, so φ(7) = 7− 1.

2 {1, 2, 3, 4, 5, 6, 7, 8}, so φ(23) = 23 − 22.

3 {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}, soφ(3 · 5) = (3− 1)(5− 1).



Remark

Let p, q be primes. Then

1 φ(p) = p− 1.


3 φ(pq) = (p− 1)(q − 1).

Example

1 {1, 2, 3, 4, 5, 6, 7}, so φ(7) = 7− 1.

2 {1, 2, 3, 4, 5, 6, 7, 8}, so φ(23) = 23 − 22.

3 {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}, soφ(3 · 5) = (3− 1)(5− 1).



Remark


1 φ(p) = p− 1.


3 φ(pq) = (p− 1)(q − 1).

Example

1 {1, 2, 3, 4, 5, 6, 7}, so φ(7) = 7− 1.

2 {1, 2, 3, 4, 5, 6, 7, 8}, so φ(23) = 23 − 22.

3 {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}, soφ(3 · 5) = (3− 1)(5− 1).



Remark


1 φ(p) = p− 1.


3 φ(pq) = (p− 1)(q − 1).

Example

1 {1, 2, 3, 4, 5, 6, 7}, so φ(7) = 7− 1.

2 {1, 2, 3, 4, 5, 6, 7, 8}, so φ(23) = 23 − 22.

3 {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}, soφ(3 · 5) = (3− 1)(5− 1).



Remark


1 φ(p) = p− 1.


3 φ(pq) = (p− 1)(q − 1).

Example

1 {1, 2, 3, 4, 5, 6, 7},

so φ(7) = 7− 1.

2 {1, 2, 3, 4, 5, 6, 7, 8}, so φ(23) = 23 − 22.

3 {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}, soφ(3 · 5) = (3− 1)(5− 1).



Remark


1 φ(p) = p− 1.


3 φ(pq) = (p− 1)(q − 1).

Example

1 {1, 2, 3, 4, 5, 6, 7}, so φ(7) = 7− 1.

2 {1, 2, 3, 4, 5, 6, 7, 8}, so φ(23) = 23 − 22.

3 {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}, soφ(3 · 5) = (3− 1)(5− 1).



Remark


1 φ(p) = p− 1.


3 φ(pq) = (p− 1)(q − 1).

Example

1 {1, 2, 3, 4, 5, 6, 7}, so φ(7) = 7− 1.

2 {1, 2, 3, 4, 5, 6, 7, 8}

, so φ(23) = 23 − 22.

3 {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}, soφ(3 · 5) = (3− 1)(5− 1).



Remark


1 φ(p) = p− 1.


3 φ(pq) = (p− 1)(q − 1).

Example

1 {1, 2, 3, 4, 5, 6, 7}, so φ(7) = 7− 1.

2 {1, 2, 3, 4, 5, 6, 7, 8}, so φ(23) = 23 − 22.

3 {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}, soφ(3 · 5) = (3− 1)(5− 1).



Remark


1 φ(p) = p− 1.


3 φ(pq) = (p− 1)(q − 1).

Example

1 {1, 2, 3, 4, 5, 6, 7}, so φ(7) = 7− 1.

2 {1, 2, 3, 4, 5, 6, 7, 8}, so φ(23) = 23 − 22.

3 {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}

, soφ(3 · 5) = (3− 1)(5− 1).



Remark


1 φ(p) = p− 1.


3 φ(pq) = (p− 1)(q − 1).

Example

1 {1, 2, 3, 4, 5, 6, 7}, so φ(7) = 7− 1.

2 {1, 2, 3, 4, 5, 6, 7, 8}, so φ(23) = 23 − 22.

3 {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}, soφ(3 · 5) = (3− 1)(5− 1).


Euler’s Theorem

Theorem (Euler’s Theorem)

Let a and n be positive integers with gcd(a, n) = 1. Then

aφ(n) ≡ 1 (mod n).

Corollary

As a special case, when n = p is a prime, then we get Fermat’s LittleTheorem:

ap−1 ≡ 1 (mod p).


Euler’s Theorem


Let a and n be positive integers with gcd(a, n) = 1.

Then

aφ(n) ≡ 1 (mod n).

Corollary


ap−1 ≡ 1 (mod p).


Euler’s Theorem



aφ(n) ≡ 1 (mod n).

Corollary


ap−1 ≡ 1 (mod p).


Euler’s Theorem



aφ(n) ≡ 1 (mod n).

Corollary


ap−1 ≡ 1 (mod p).


Before We Get Started...Part I

Cryptography has a lot of applications .

We will focus on sending amessage securely. For example, suppose Alice wishes to send the message“STOP” to Bob.

1 They use the following rule to get the numerical equivalent of eachletter in alphabet: A = 01, B = 02, · · · , Z = 26. For example, thenumerical equivalent of “CAT” is 030120. Conversely, thealphabetical equivalent of 13012008 is “MATH”.

2 Alice translates “STOP” into 19201516 and sends this number toBob, and Bob decodes 19201516 using the rule to get “STOP” back.

3 Problem is, in practice, it is very likely that a potential code breaker,say Charlie, also knows this rule. So if the code breaker Charlieintercepts the number 19201516, then he also knows what theoriginal plain text was.



Cryptography has a lot of applications . We will focus on sending amessage securely.

For example, suppose Alice wishes to send the message“STOP” to Bob.






Cryptography has a lot of applications . We will focus on sending amessage securely. For example, suppose Alice wishes to send the message“STOP” to Bob.







1 They use the following rule to get the numerical equivalent of eachletter in alphabet: A = 01, B = 02, · · · , Z = 26.

For example, thenumerical equivalent of “CAT” is 030120. Conversely, thealphabetical equivalent of 13012008 is “MATH”.






1 They use the following rule to get the numerical equivalent of eachletter in alphabet: A = 01, B = 02, · · · , Z = 26. For example, thenumerical equivalent of “CAT” is 030120.

Conversely, thealphabetical equivalent of 13012008 is “MATH”.













2 Alice translates “STOP” into 19201516 and sends this number toBob,

and Bob decodes 19201516 using the rule to get “STOP” back.













3 Problem is, in practice, it is very likely that a potential code breaker,say Charlie, also knows this rule.

So if the code breaker Charlieintercepts the number 19201516, then he also knows what theoriginal plain text was.








One Possible Fix

How to fix this problem?

Alice and Bob may want to change the rule, say:

A = 03, B = 04, · · · , X = 26, Y = 01, Z = 02.

But this is not long-lived. There are several ways to figure out this newrule (e.g. frequency analysis).

Another problem is: if Alice wants to send a message to Charlie, thennow Alice and Charlie must set up a different rule, because now Bobcould be a possible code breaker!


One Possible Fix

How to fix this problem? Alice and Bob may want to change the rule, say:

A = 03, B = 04, · · · , X = 26, Y = 01, Z = 02.




One Possible Fix


A = 03, B = 04, · · · , X = 26, Y = 01, Z = 02.

But this is not long-lived.

There are several ways to figure out this newrule (e.g. frequency analysis).



One Possible Fix


A = 03, B = 04, · · · , X = 26, Y = 01, Z = 02.




One Possible Fix


A = 03, B = 04, · · · , X = 26, Y = 01, Z = 02.


Another problem is: if Alice wants to send a message to Charlie, thennow Alice and Charlie must set up a different rule,

because now Bobcould be a possible code breaker!


One Possible Fix


A = 03, B = 04, · · · , X = 26, Y = 01, Z = 02.




As an idea of resolving this problem, in 1976, three researchers at M.I.T.– Ronald Rivest, Adi Shamir, and Leonard Adleman came up with anidea now called the RSA System.


Before We Get Started...Part II

1 Find φ(143).

Since 143 = 11× 13, φ(143) = 10× 12 = 120.

2 Find φ(3127). Since 3127 = 53× 59, φ(3127) = 52× 58 = 3016.

KEY OBSERVATION

1 Computing 53× 59 = 3127 is easy, but finding the primefactorization of 3127 (reverse process) is difficult!

2 If p, q are large primes (each with more than 200 digits, forexample), then calculating the product pq is not difficult with acomputer, but when the product pq is given, finding the individualprime numbers p, q is not easy, even with a computer.



1 Find φ(143). Since 143 = 11× 13,

φ(143) = 10× 12 = 120.

2 Find φ(3127). Since 3127 = 53× 59, φ(3127) = 52× 58 = 3016.

KEY OBSERVATION





1 Find φ(143). Since 143 = 11× 13, φ(143) = 10× 12 = 120.

2 Find φ(3127). Since 3127 = 53× 59, φ(3127) = 52× 58 = 3016.

KEY OBSERVATION





1 Find φ(143). Since 143 = 11× 13, φ(143) = 10× 12 = 120.

2 Find φ(3127).

Since 3127 = 53× 59, φ(3127) = 52× 58 = 3016.

KEY OBSERVATION





1 Find φ(143). Since 143 = 11× 13, φ(143) = 10× 12 = 120.

2 Find φ(3127). Since 3127 = 53× 59,

φ(3127) = 52× 58 = 3016.

KEY OBSERVATION





1 Find φ(143). Since 143 = 11× 13, φ(143) = 10× 12 = 120.

2 Find φ(3127). Since 3127 = 53× 59, φ(3127) = 52× 58 = 3016.

KEY OBSERVATION





1 Find φ(143). Since 143 = 11× 13, φ(143) = 10× 12 = 120.

2 Find φ(3127). Since 3127 = 53× 59, φ(3127) = 52× 58 = 3016.

KEY OBSERVATION

1 Computing 53× 59 = 3127 is easy,

but finding the primefactorization of 3127 (reverse process) is difficult!




1 Find φ(143). Since 143 = 11× 13, φ(143) = 10× 12 = 120.

2 Find φ(3127). Since 3127 = 53× 59, φ(3127) = 52× 58 = 3016.

KEY OBSERVATION





1 Find φ(143). Since 143 = 11× 13, φ(143) = 10× 12 = 120.

2 Find φ(3127). Since 3127 = 53× 59, φ(3127) = 52× 58 = 3016.

KEY OBSERVATION


2 If p, q are large primes (each with more than 200 digits, forexample),

then calculating the product pq is not difficult with acomputer, but when the product pq is given, finding the individualprime numbers p, q is not easy, even with a computer.



1 Find φ(143). Since 143 = 11× 13, φ(143) = 10× 12 = 120.

2 Find φ(3127). Since 3127 = 53× 59, φ(3127) = 52× 58 = 3016.

KEY OBSERVATION


2 If p, q are large primes (each with more than 200 digits, forexample), then calculating the product pq is not difficult with acomputer,

but when the product pq is given, finding the individualprime numbers p, q is not easy, even with a computer.



1 Find φ(143). Since 143 = 11× 13, φ(143) = 10× 12 = 120.

2 Find φ(3127). Since 3127 = 53× 59, φ(3127) = 52× 58 = 3016.

KEY OBSERVATION




Central Idea of RSA

1 Alice chooses two (large) prime numbers pA and qA.

She computesnA = pAqA and φ(nA) = (pA − 1)(qA − 1). She then chooses anumber eA so that gcd(eA, (pA − 1)(qA − 1)) = 1 (this can be donewithout difficulty). Since gcd(eA, (pA − 1)(qA − 1)) = 1, there existpositive integers sA, tA such that sAeA − tA(pA − 1)(qA − 1) = 1(Backward Euclidean Algorithm). Now she makes public the pair(nA, eA) but does NOT disclose sA.

2 Bob does the same thing: he chooses two large prime numbers pBand qB , computes nB = pBqB , chooses eB , sB , and tB so thatgcd(eB , (pB − 1)(qB − 1)) = 1 and sBeB − tB(pB − 1)(qB − 1) = 1,and makes public the pair (nB , eB) while concealing sB .

3 Charlie does the same thing (makes (nC , eC) public, keeps sC),David does the same thing (makes (nD, eD) public, keeps sD), etc.


Central Idea of RSA

1 Alice chooses two (large) prime numbers pA and qA. She computesnA = pAqA and φ(nA) = (pA − 1)(qA − 1).

She then chooses anumber eA so that gcd(eA, (pA − 1)(qA − 1)) = 1 (this can be donewithout difficulty). Since gcd(eA, (pA − 1)(qA − 1)) = 1, there existpositive integers sA, tA such that sAeA − tA(pA − 1)(qA − 1) = 1(Backward Euclidean Algorithm). Now she makes public the pair(nA, eA) but does NOT disclose sA.




Central Idea of RSA

1 Alice chooses two (large) prime numbers pA and qA. She computesnA = pAqA and φ(nA) = (pA − 1)(qA − 1). She then chooses anumber eA so that gcd(eA, (pA − 1)(qA − 1)) = 1 (this can be donewithout difficulty).

Since gcd(eA, (pA − 1)(qA − 1)) = 1, there existpositive integers sA, tA such that sAeA − tA(pA − 1)(qA − 1) = 1(Backward Euclidean Algorithm). Now she makes public the pair(nA, eA) but does NOT disclose sA.




Central Idea of RSA

1 Alice chooses two (large) prime numbers pA and qA. She computesnA = pAqA and φ(nA) = (pA − 1)(qA − 1). She then chooses anumber eA so that gcd(eA, (pA − 1)(qA − 1)) = 1 (this can be donewithout difficulty). Since gcd(eA, (pA − 1)(qA − 1)) = 1, there existpositive integers sA, tA such that sAeA − tA(pA − 1)(qA − 1) = 1

(Backward Euclidean Algorithm). Now she makes public the pair(nA, eA) but does NOT disclose sA.




Central Idea of RSA

1 Alice chooses two (large) prime numbers pA and qA. She computesnA = pAqA and φ(nA) = (pA − 1)(qA − 1). She then chooses anumber eA so that gcd(eA, (pA − 1)(qA − 1)) = 1 (this can be donewithout difficulty). Since gcd(eA, (pA − 1)(qA − 1)) = 1, there existpositive integers sA, tA such that sAeA − tA(pA − 1)(qA − 1) = 1(Backward Euclidean Algorithm).

Now she makes public the pair(nA, eA) but does NOT disclose sA.




Central Idea of RSA

1 Alice chooses two (large) prime numbers pA and qA. She computesnA = pAqA and φ(nA) = (pA − 1)(qA − 1). She then chooses anumber eA so that gcd(eA, (pA − 1)(qA − 1)) = 1 (this can be donewithout difficulty). Since gcd(eA, (pA − 1)(qA − 1)) = 1, there existpositive integers sA, tA such that sAeA − tA(pA − 1)(qA − 1) = 1(Backward Euclidean Algorithm). Now she makes public the pair(nA, eA) but does NOT disclose sA.




Central Idea of RSA


2 Bob does the same thing:

he chooses two large prime numbers pBand qB , computes nB = pBqB , chooses eB , sB , and tB so thatgcd(eB , (pB − 1)(qB − 1)) = 1 and sBeB − tB(pB − 1)(qB − 1) = 1,and makes public the pair (nB , eB) while concealing sB .



Central Idea of RSA


2 Bob does the same thing: he chooses two large prime numbers pBand qB ,

computes nB = pBqB , chooses eB , sB , and tB so thatgcd(eB , (pB − 1)(qB − 1)) = 1 and sBeB − tB(pB − 1)(qB − 1) = 1,and makes public the pair (nB , eB) while concealing sB .



Central Idea of RSA


2 Bob does the same thing: he chooses two large prime numbers pBand qB , computes nB = pBqB ,

chooses eB , sB , and tB so thatgcd(eB , (pB − 1)(qB − 1)) = 1 and sBeB − tB(pB − 1)(qB − 1) = 1,and makes public the pair (nB , eB) while concealing sB .



Central Idea of RSA


2 Bob does the same thing: he chooses two large prime numbers pBand qB , computes nB = pBqB , chooses eB , sB , and tB so thatgcd(eB , (pB − 1)(qB − 1)) = 1 and sBeB − tB(pB − 1)(qB − 1) = 1,

and makes public the pair (nB , eB) while concealing sB .



Central Idea of RSA





Central Idea of RSA



3 Charlie does the same thing (makes (nC , eC) public, keeps sC),

David does the same thing (makes (nD, eD) public, keeps sD), etc.


Central Idea of RSA



3 Charlie does the same thing (makes (nC , eC) public, keeps sC),David does the same thing (makes (nD, eD) public, keeps sD),

etc.


Central Idea of RSA





How does Alice send a message to Bob?

1 Alice first translates her plain text to its numerical equivalent, sayM .

2 She then looks up her phone book to find (nB , eB).

3 Then she computes the remainder C of MeB when divided by nB(that is MeB ≡ C (mod nB)).

4 Finally, Alice sends C to Bob.











3 Then she computes the remainder C of MeB when divided by nB

(that is MeB ≡ C (mod nB)).















Example

Let’s say pB = 53 and qB = 59.

Then nB = 53× 59 = 3127 and(pB − 1)(qB − 1) = 52× 58 = 3016. Pick eB = 271, thengcd(eB , (pB − 1)(qB − 1)) = gcd(271, 3016) = 1. Using BackwardEuclidean Algorithm, one can see that 2671 · 271− 240 · 3016 = 1, whichmeans sB = 2671. Bob makes (nB , eB) = (3127, 271) open to publicwhile concealing sB = 2671.

1 To send “GO” (which is equivalent to M = 715) to Bob, Alice looksup her phone book to find (nB , eB) = (3127, 271).

2 She then computes the remainder C of MeB = 715271 when dividedby 3127. C turns out to be 1657.

3 Alice sends C = 1657 to Bob.

http://www.mtholyoke.edu/∼mpeterso/Applets/CalculatorApplet.html


Example

Let’s say pB = 53 and qB = 59. Then nB = 53× 59 = 3127

and(pB − 1)(qB − 1) = 52× 58 = 3016. Pick eB = 271, thengcd(eB , (pB − 1)(qB − 1)) = gcd(271, 3016) = 1. Using BackwardEuclidean Algorithm, one can see that 2671 · 271− 240 · 3016 = 1, whichmeans sB = 2671. Bob makes (nB , eB) = (3127, 271) open to publicwhile concealing sB = 2671.






Example

Let’s say pB = 53 and qB = 59. Then nB = 53× 59 = 3127 and(pB − 1)(qB − 1) = 52× 58 = 3016.

Pick eB = 271, thengcd(eB , (pB − 1)(qB − 1)) = gcd(271, 3016) = 1. Using BackwardEuclidean Algorithm, one can see that 2671 · 271− 240 · 3016 = 1, whichmeans sB = 2671. Bob makes (nB , eB) = (3127, 271) open to publicwhile concealing sB = 2671.






Example

Let’s say pB = 53 and qB = 59. Then nB = 53× 59 = 3127 and(pB − 1)(qB − 1) = 52× 58 = 3016. Pick eB = 271, thengcd(eB , (pB − 1)(qB − 1)) = gcd(271, 3016) = 1.

Using BackwardEuclidean Algorithm, one can see that 2671 · 271− 240 · 3016 = 1, whichmeans sB = 2671. Bob makes (nB , eB) = (3127, 271) open to publicwhile concealing sB = 2671.






Example

Let’s say pB = 53 and qB = 59. Then nB = 53× 59 = 3127 and(pB − 1)(qB − 1) = 52× 58 = 3016. Pick eB = 271, thengcd(eB , (pB − 1)(qB − 1)) = gcd(271, 3016) = 1. Using BackwardEuclidean Algorithm, one can see that 2671 · 271− 240 · 3016 = 1,

whichmeans sB = 2671. Bob makes (nB , eB) = (3127, 271) open to publicwhile concealing sB = 2671.






Example

Let’s say pB = 53 and qB = 59. Then nB = 53× 59 = 3127 and(pB − 1)(qB − 1) = 52× 58 = 3016. Pick eB = 271, thengcd(eB , (pB − 1)(qB − 1)) = gcd(271, 3016) = 1. Using BackwardEuclidean Algorithm, one can see that 2671 · 271− 240 · 3016 = 1, whichmeans sB = 2671.

Bob makes (nB , eB) = (3127, 271) open to publicwhile concealing sB = 2671.






Example

Let’s say pB = 53 and qB = 59. Then nB = 53× 59 = 3127 and(pB − 1)(qB − 1) = 52× 58 = 3016. Pick eB = 271, thengcd(eB , (pB − 1)(qB − 1)) = gcd(271, 3016) = 1. Using BackwardEuclidean Algorithm, one can see that 2671 · 271− 240 · 3016 = 1, whichmeans sB = 2671. Bob makes (nB , eB) = (3127, 271) open to publicwhile concealing sB = 2671.






Example


1 To send “GO” (which is equivalent to M = 715) to Bob,

Alice looksup her phone book to find (nB , eB) = (3127, 271).





Example







Example



2 She then computes the remainder C of MeB = 715271 when dividedby 3127.

C turns out to be 1657.




Example







Example







Example







How does Bob decode C to get the plain text back?

1 Bob receives C.

2 He then computes the remainder of CsB when divided by nB . Notethat

CsB ≡ (MeB )sB = MeBsB = M1+tB(pB−1)(qB−1).

By Euler’s Theorem, M (pB−1)(qB−1) ≡ 1 (mod nB) and this meansthat

M tB(pB−1)(qB−1) = (M (pB−1)(qB−1))tB ≡ 1 (mod nB).

Consequently,CsB ≡M,

which means that the remainder of CsB when divided by nB equalsM .

3 With this M , Bob can easily recover the original plain text.



1 Bob receives C.

2 He then computes the remainder of CsB when divided by nB .

Notethat









1 Bob receives C.










1 Bob receives C.



By Euler’s Theorem, M (pB−1)(qB−1) ≡ 1 (mod nB)

and this meansthat







1 Bob receives C.










1 Bob receives C.










1 Bob receives C.










1 Bob receives C.









Example

1 Bob received 1657.

2 He then computes the remainder of 1657sB = 16572671 whendivided nB = 3127, which turns out to be 715.

3 Bob then recovers “GO”.


Example





Example





Why is RSA robust?

Suppose that Charlie was able to intercept C.

To get the plain text M ,he must know sB . Recall that sB was determined so thatsBeB − tB(pB − 1)(qB − 1) = 1, that is

sBeB ≡ 1 (mod (pB − 1)(qB − 1)).

Since eB is known to Charlie, if Charlie knowsφ(nB) = (pB − 1)(qB − 1), then he can actually find sB . But to get(pB − 1)(qB − 1), Charlie should know pB and qB , and this is almostimpossible even though he knows what pBqB is.

Example

Suppose Charlie managed to get C = 715. To obtain the plain text M ,he needs to know sB , which is hidden. If Charlie can findφ(nB) = φ(pBqB) = (pB − 1)(qB − 1), then there is a way to find sBeasily. Charlie knows (nB , eB) = (3127, 271), but computingφ(nB) = φ(3127) is difficult without knowing the prime factorization of3127.


Why is RSA robust?

Suppose that Charlie was able to intercept C. To get the plain text M ,he must know sB .

Recall that sB was determined so thatsBeB − tB(pB − 1)(qB − 1) = 1, that is

sBeB ≡ 1 (mod (pB − 1)(qB − 1)).


Example



Why is RSA robust?

Suppose that Charlie was able to intercept C. To get the plain text M ,he must know sB . Recall that sB was determined so thatsBeB − tB(pB − 1)(qB − 1) = 1, that is

sBeB ≡ 1 (mod (pB − 1)(qB − 1)).


Example



Why is RSA robust?


sBeB ≡ 1 (mod (pB − 1)(qB − 1)).

Since eB is known to Charlie,

if Charlie knowsφ(nB) = (pB − 1)(qB − 1), then he can actually find sB . But to get(pB − 1)(qB − 1), Charlie should know pB and qB , and this is almostimpossible even though he knows what pBqB is.

Example



Why is RSA robust?


sBeB ≡ 1 (mod (pB − 1)(qB − 1)).

Since eB is known to Charlie, if Charlie knowsφ(nB) = (pB − 1)(qB − 1), then he can actually find sB .

But to get(pB − 1)(qB − 1), Charlie should know pB and qB , and this is almostimpossible even though he knows what pBqB is.

Example



Why is RSA robust?


sBeB ≡ 1 (mod (pB − 1)(qB − 1)).

Since eB is known to Charlie, if Charlie knowsφ(nB) = (pB − 1)(qB − 1), then he can actually find sB . But to get(pB − 1)(qB − 1), Charlie should know pB and qB ,

and this is almostimpossible even though he knows what pBqB is.

Example



Why is RSA robust?


sBeB ≡ 1 (mod (pB − 1)(qB − 1)).


Example

Suppose Charlie managed to get C = 715.

To obtain the plain text M ,he needs to know sB , which is hidden. If Charlie can findφ(nB) = φ(pBqB) = (pB − 1)(qB − 1), then there is a way to find sBeasily. Charlie knows (nB , eB) = (3127, 271), but computingφ(nB) = φ(3127) is difficult without knowing the prime factorization of3127.


Why is RSA robust?


sBeB ≡ 1 (mod (pB − 1)(qB − 1)).


Example

Suppose Charlie managed to get C = 715. To obtain the plain text M ,he needs to know sB , which is hidden.

If Charlie can findφ(nB) = φ(pBqB) = (pB − 1)(qB − 1), then there is a way to find sBeasily. Charlie knows (nB , eB) = (3127, 271), but computingφ(nB) = φ(3127) is difficult without knowing the prime factorization of3127.


Why is RSA robust?


sBeB ≡ 1 (mod (pB − 1)(qB − 1)).


Example

Suppose Charlie managed to get C = 715. To obtain the plain text M ,he needs to know sB , which is hidden. If Charlie can findφ(nB) = φ(pBqB) = (pB − 1)(qB − 1), then there is a way to find sBeasily.

Charlie knows (nB , eB) = (3127, 271), but computingφ(nB) = φ(3127) is difficult without knowing the prime factorization of3127.


Why is RSA robust?


sBeB ≡ 1 (mod (pB − 1)(qB − 1)).


Example

Suppose Charlie managed to get C = 715. To obtain the plain text M ,he needs to know sB , which is hidden. If Charlie can findφ(nB) = φ(pBqB) = (pB − 1)(qB − 1), then there is a way to find sBeasily. Charlie knows (nB , eB) = (3127, 271),

but computingφ(nB) = φ(3127) is difficult without knowing the prime factorization of3127.


Why is RSA robust?


sBeB ≡ 1 (mod (pB − 1)(qB − 1)).


Example



Documents

Fermat’s Little Theorem - Mount Holyoke Collegejjlee/Teaching/Math 232 Introduction to RSA.pdf · Fermat’s Little Theorem Recall Fermat’s Little Theorem: Theorem (Fermat’s