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1. (1)WhatistheciphertextthatisproducedwhenRSAencryptionwithpublickey (e, n) = (3, 2669) is used
to encrypt the message BESTWISHES? Use the protocol A = 00, B = 01, ..., Z = 25 and break your message
up into blocks of length 4. (If a cipherthext block has fewer than four digits, add enough zeros at the beginning
to make it have four digits.)
Solution: The numerical equivalent of the message, broken up into blocks of 4, is 0104, 1819, 2208, 1807, 0418.
We encrypt by raising each block to the power e = 3 and reducing modulo n = 2669 to get
1043
≡ 1215 (mod 2669);
18193
≡ 1224 (mod 2669);
22083
≡ 1471 (mod 2669);
18073
≡ 23 (mod 2669);
4183
≡ 116 (mod 2669).
The ciphertext is thus 12151224147100230116.
(2)UseMathematicatosolvethefollowingproblem.AlicehasaslowcomputerandpublishesanRSA
public key
(e, n) = (952859796010804402537, 137515498286936942046107746703624480207957)
that is not very secure since the numbers aren’t very large. Bob sends her an urgent ciphertext message using
this public key, broken up into the following blocks:
113403756589260093949278318603930681886413,
125987468369297610213653477519525084178728,
43944784289127338075793492865014310001164,
135739830937229599489426895906452425083187,
109850539078433810931837688030802520077173,
112469132477024143792191502981504997455475,
94610945754028346269684292682885558347040.
You intercepted the message. Assuming you know that the plaintext was broken into blocks of eight digits,
decipher the message. (Remember that you need to add zeros at the beginning of those plaintext blocks that
come out with fewer than eight numbers. Also remember to load the crypto notebook and use the command
alph1 to convert your numerical plaintext to letters.)
Solution: See the last two pages.
(3) Recall that if we know the factorization of n = pq, then φ(n) = (p − 1)(q − 1) is easy to compute. In this
problem, you will show that knowing n and φ(n) leads to the factorization of n. Thus factoring n is a problem
of the same complexity as finding φ(n).
(a) (5 pts) Show that p + q = n − φ(n) + 1.
(b) (3 pts) By using the fact that q = n/p, show that p satisfies the quadratic equation
p2
+ (φ(n) − n − 1)p + n = 0.
CRYPTOGRAPHY
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2. (c) (3 pts) Deduce that p and q are
p =
(n − φ(n) + 1) + (n − φ(n) + 1)2 − 4n
2
q =
(n − φ(n) + 1) − (n − φ(n) + 1)2 − 4n
2
Solution:
(a) There are n integers up to and including n. Since n = pq, the following are the integers that are not
relatively prime to n:
p, 2p, 3p, ..., qp,
q, 2q, 3q, ..., (p − 1)q.
Note that we do not include pq in the second line since this number is listed in the first line already. Thus
there are p + q − 1 integers up to and including n that are not relatively prime to n. This leaves φ(n)
integers. Thus (p + q − 1) + φ(n) = n as desired.
(b) Plugging in q = n/p in the equation from part (a), we get
p + n/p − 1 = n − φ(n)
p2
+ n − p = np − φ(n)p
p2
+ φ(n)p − np − p + n = 0
p2
+ (φ(n) − n − 1)p + n = 0.
(c) Solving the quadratic equation from (b) gives
p =
(n − φ(n) + 1) ± (n − φ(n) + 1)2 − 4n
2
.
Switching the roles of p and q in part (b) gives the same quadratic equation with the same solutions. Since
p and q are distinct, one of p or q thus must be
(n − φ(n) + 1) + (n − φ(n) + 1)2 − 4n
2
and the other
(n − φ(n) + 1) − (n − φ(n) + 1)2 − 4n
2
.
RSA security
(4) Notice that the RSA decryption requires the use of Euler’s Formula with base P, where P is a plaintext block,
and enciphering modulus n = pq, but the condition gcd(P, n) = 1 is never in fact checked (and this a hypothesis
for the Euler’s Formula).
(a) (5 pts) Show that it is extremely unlikely that this condition is not satistied by showing that the probability
that P and n are not relatively prime is 1
p + 1
q − 1
pq . Thus if both p and q are larger than 10100, the
probability that gcd(P, n) = 1 is less than 10−99. (Recall that the probability of an event occurring is the
number of ways it can occur divided by the total number of possible events.)
(b) (3 pts) Suppose a cryptanalyst discovers a plaintext block P that is not relatively prime to n. Show that
the cryptanalyst can factor n. (Hint: Recall that P < n.)
Solution:

3. (a) From an earlier problem, there are q + p − 1 integers (number of ways the event we’re interested in can
occur) up to including n that are not relatively prime to n (total number of outcomes). The probability is
then
q + p − 1
n
=
q
n
+
p
n
−
1
n
=
q
pq
+
p
pq
−
1
pq
=
1
p
+
1
q
−
1
pq
.
(b) Since a block P is less than n, if gcd(P, n) = 1, we must have gcd(P, n) = p or gcd(P, n) = q since p and
q are the only factors of n. Euclidean algorithm then can be used for finding gcd(P, n) quickly. This gives
one factor of n, and dividing n by this factor immediately gives the other.
(5) (3 pts) Suppose the length of each block in an RSA cipher is precisely the length of the numerical equivalent
of each letter. How could this cipher be broken?
Solution: Since each letter is being encrypted the same way throughout a message, the RSA in this case reduces
to a simple monoalphabetic substitution cipher which can be broken by frequency analysis.
(6) (3 pts) The exponent e = 2 should never be used in an RSA public key. Why?
Solution: To find the private key d from the public key (e, n), we need gcd(e, φ(n)) = gcd(e, (p−1)(q−1)) = 1.
However, (p − 1)(q − 1) is necessarily even, so if e = 2, gcd(e, φ(n)) = 2, and private key does not exist. (Note
that this shows that e should more generally never be even.)
(7) (5 pts) One instance of how RSA can be subverted is when there is a common modulus protocol failure, which
means that two parties are using the same modulus n but different exponents e for encryption. Show that the
plaintext of a message sent to each of these two parties can be recovered from the ciphertext messages if the
exponents are relatively prime.
Solution: Suppose Alice1 and Alice2 are using same modulus n and exponents e1 and e2 for their public keys.
Suppose Bob encrypts the same message P using (e1, n) and (e2, n) and sends the ciphertexts to Alice1 and
Alice2. Eve intercepts both ciphertexts Pe1 and Pe2 and first uses Euclidean algorithm to solve ue1 + ve2 = 1
fast (she can do this since gcd(e1, e2) = 1). Note that this step can be performed without the ciphertexts.
Then she computes (Pe1 )u and (Pe2 )v mod n. She can then recover P by multiplying these two numbers since
(Pe1
)u
(Pe2
)v
≡ Pe1u
Pe2v
(mod n) ≡ Pe1u+e2v
(mod n) ≡ P (mod n).

4. H* Problem 2 solution *L
H*The following few lines is how e and n were computed. You did not need to know this.*L
p = NextPrime@Random@Integer, 810^20, 10^21<DD
Out[170]= 906 819 166 412 721 858 083
In[171]:= q = NextPrime@Random@Integer, 810^20, 10^21<DD
Out[171]= 151 645 998 872 005 891 879
e = NextPrime@Random@Integer, 810^20, 10^21<DD
Out[172]= 952 859 796 010 804 402 537
H*So far, we have randomly choosen large Hbut not too largeL p, q, and e.*L
In[173]:= n = p * q
Out[173]= 137 515 498 286 936 942 046 107 746 703 624 480 207 957
In[148]:= P = num1@"youreallyneedafastercomputer"D
Out[148]= 25 152 118 050 112 122 514 050 504 010 601 192 005 180 315 131 621 200 518
H*This is the plaintext, converted to numbers*L
In[191]:= c1 = PowerMod@25 152 118, e, nD
c2 = PowerMod@05 011 212, e, nD
c3 = PowerMod@25 140 505, e, nD
c4 = PowerMod@04 010 601, e, nD
c5 = PowerMod@19 200 518, e, nD
c6 = PowerMod@03 151 316, e, nD
c7 = PowerMod@21 200 518, e, nD
Out[191]= 113 403 756 589 260 093 949 278 318 603 930 681 886 413
Out[192]= 125 987 468 369 297 610 213 653 477 519 525 084 178 728
Out[193]= 43 944 784 289 127 338 075 793 492 865 014 310 001 164
Out[194]= 135 739 830 937 229 599 489 426 895 906 452 425 083 187
Out[195]= 109 850 539 078 433 810 931 837 688 030 802 520 077 173
Out[196]= 112 469 132 477 024 143 792 191 502 981 504 997 455 475
Out[197]= 94 610 945 754 028 346 269 684 292 682 885 558 347 040
H*This is the ciphertext,
made out of plaintext blocks of length 8. This is what you intercepted. *L
H*To break the message, do the following*L

5. In[188]:= Phi = EulerPhi@nD
Out[188]= 137 515 498 286 936 942 045 049 281 538 339 752 457 996
H*The cipher is weak since Mathematica can figure out PhiHnL fast*L
In[190]:= d = PowerMod@e, -1, PhiD
Out[190]= 15 837 563 809 419 715 974 639 737 748 420 536 708 137
H*Since you have PhiHnL, you can figure out the private key d fast*L
In[198]:= p1 = PowerMod@c1, d, nD
p2 = PowerMod@c2, d, nD
p3 = PowerMod@c3, d, nD
p4 = PowerMod@c4, d, nD
p5 = PowerMod@c5, d, nD
p6 = PowerMod@c6, d, nD
p7 = PowerMod@c7, d, nD
Out[198]= 25 152 118
Out[199]= 5 011 212
Out[200]= 25 140 505
Out[201]= 4 010 601
Out[202]= 19 200 518
Out[203]= 3 151 316
Out[204]= 21 200 518
H* These are the plaintext blocks. Remember to add zeros
at the beginning of each block whose size is less than eight. *L
H*Finally we can convert back to text.*L
alph1@25 152 118 050 112 122 514 050 504 010 601 192 005 180 315 131 621 200 518D
Out[208]= youreallyneedafastercomputer
2 RSA homework problem.nb
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