Download Math 5330 Spring 2016 Exam 2 Solutions In class questions 1. (15

Math 5330
Spring 2016
Exam 2
Solutions
In class questions
1. (15 points) Solve the following congruences. Put your answer in the form of a congruence. I usually find it easier to go from largest to smallest modulus.
x ” 4 pmod 23q
(a)
x ” 6 pmod 25q
Solution: Starting with x ” 6 pmod 25q, we have x “ 6 ` 25k ” 4 pmod 23q,
which reduces to 2k ” ´2 pmod 23q. Using k “ ´1 ` 23m, x “ 6 ` 25p´1 `
23mq “ ´19 ` 575m. Thus, x ” ´19 pmod 575q, or x ” 556 pmod 575q.
(b)
x ” 8 pmod 9q
x ” 5 pmod 10q
x ” 4 pmod 11q
Solution: Again, I’ll start with the largest congruence. We have x “ 4 ` 11k ”
5 pmod 10q, or k ” 1 pmod 10q. This means x “ 4`11p1`10jq “ 15`110j ” 8
pmod 9q. Reducing, 2j ” 2 pmod 9q, or j “ 1`9m. Thus, x “ 15`110p1`9mq “
125 ` 990m. We have x ” 125 pmod 990q.
As an aside, given an answer, you can check it. That is, you can check that
x “ 125 is congruent to 8 modulo 9, 5 modulo 10, and 4 modulo 11.
2. (20 points) This problem concerns numbers n with φpnq “ 90. Recall that if p n, then
p ´ 1 φpnq. Note also that 91 is not prime.
(a) By the above, if φpnq “ 90, what are the concevable prime divisors of n? Hint:
Write down the 12 divisors of 90, and ask which of these can be p ´ 1.
Solution: The divisors of 90 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90. Adding
1 gives 2, 3, 4, 6, 7, 10, 11, 16, 19, 31, 46, 91. Selecting the primes gives
p “ 2, 3, 7, 11, 19, 31 as primes with p ´ 1 dividing 90.
(b) Show that for all but two of these primes, we can’t have p2 n.
Solution: If P k is the highest power of p dividing n then pk´1 pp ´ 1q divides
n. In particular, if k ě 2 then we need p to divide φpnq. Of our primes, 2, 3, 7,
11, 19, 31, only 2 and 3 divide 90 so only these could have p2 n.
(c) What is the highest power of 2 that might divide n? Explain.
Solution: Since φp4q “ 2, φp8q “ 4, and 4 does not divide 90, the 2 is the
highest allowable exponent on 2.
Page 2
(d) If φpnq “ 90, why can’t n be divisible by more than one odd prime?
Solution: If distinct primes p and q divide n then both p ´ 1 and q ´ 1 divide
φpnq “ 90. If p and q are odd, then p ´ 1 and q ´ 1 are even, meaning that 90
would have to be divisible by 4, but it is not.
(e) Use parts (a)-(d) to prove that there is no n for which φpnq “ 90. Hint: We have
reduced the problem to checking n “ 2m or 2m pk for odd primes p from part (a),
m is small and k “ 1 for most p.
Solution: Here is a solution from the class. φp33 q “ 18, a divisor of 90, but
φp34 q “ 54 is not. This means that based on parts (b) and (d), the only odd
possibilities for n are 1, 3, 9, 27, 7, 11, 19, 31. Applying φ gives something
less than 90. Next, we might have twice each of these: 2, 6, 18, 27, 14, 22,
38, 62, but again applying φ gives something less than 90. Finally, by part (c),
the maximum exponent on 2 is 2 so we check n “ 4, 12, 36, 108, 28, 44, 76, 124,
which have φpnq “ 2, 4, 12, 36, 12, 20, 36, 60, and again none of these is 90.
3. (15 points) Consider the number n “ 1111 “ 11 ˆ 101.
(a) How many steps of the p ´ 1 method are needed to factor n? Why?
Solution: What I was expecting: If p “ 11 then p ´ 1 “ 10, a divisor of 5! so
it should take 5 steps. Some people actually did the calculation: 21! “ 1, 22! “
4, 23! “ 64, 24! ” 5, 25! ” 55 ” 903. If we subtract 1 from each of these and
find the greatest common divisor of the result with 1111 we get 1, 1, 1, 1, 11,
so we factor the number in 5 steps.
?
(b) How many steps of Fermat’s method are need to factor n? (Use x0 “ r 1111s “ 34.)
Solution: What I expected: We need px ´ yqpx ` yq “ 11 ˆ 101 so x ´ y “
11, x`y “ 101 Ñ 2x “ 112. Thus, x “ 56. We check x “ 34, x “ 35, . . . , x “ 56,
so it takes 23 steps. More than one person actually applied Fermat’s method,
factoring 1111 in 23 steps.
(c) How many steps of Pollard’s rho method are needed to factor n? Hint: Apply the
rho-method not to 1111 but to 11, and see when you get the gcd to be 11.
Solution: Following the hint,
k
1 2 3 4 5
ak 2 5 4 6 4
ak{2 - 2 - 5 -
6 7 8
6 4 6
4 - 6
Page 3
so we should uncover 11 as a factor in 8 steps. Some people applied the rho
method:
k
1 2 3
4
5
6
7
8
ak 2 5 26 677 598 974 994 358
ak{2 - 2 5
26
677
and gcd(1111, 358 - 677) = 11.
Extra credit.
4. (5 points) Find the largest number m for which p4 ” 1 pmod mq for all primes p ą 5.
Solution: Several people noted that 74 ´1 “ 2400, 114 ´1 “ 14640 “ 240ˆ61. This
puts 240 as the largest possibility for m. To actually prove that m “ 240 we need
to prove that p4 ” 1 pmod 240q for every prime p ą 5. To that end, note that for
p ą 5, p5´1 ” 1 pmod 5q by Fermat’s Little Theorem. This means 5 m. Similarly,
p3´1 ” 1 pmod 3q, so P 4 ” 1 pmod 3q, and 3 m. Finally, p4 ´1 pp´1qpp`1qpp2 `1q,
the product of three even numbers. Moreover, one of two consecutive even numbers is
divisible by 4, giving another factor of 2. This means 16 m. Putting these together,
m is divisible by 16, 3, and 5, so m is divisible by 240, showing that m “ 240.
5. (5 points) We call b the inverse of a modulo n if ab ” 1 pmod nq.
(a) If a is relatively prime to n prove that the inverse of a is aφpnq´1 pmod nq.
Solution: One of the easiest extra credit problems you will find! Let b “
aφpnq´1 . Then ab “ aφpnq ” 1 pmod nq, as desired.
(b) Use the binary squaring method and part a to find the inverse of 23 modulo 55.
Solution: φp55q “ 4 ˆ 10 “ 40, so by part (a), we want 2339 pmod 55q. The
calculation goes like this:
exponent 1 2 4 8 16 32
value
23 34 1 1 1 1
so 2339 “ 231 ˆ 232 ˆ 234 ˆ 2332 ” 23 ˆ 34 ˆ 1 ˆ 1 ” 12 pmod 55q. That is, the
inverse of 23 is 12.
Page 4
6. (10 points) Find the three digit prime which could be found as a factor of n in the fewest
number of steps using:
(a) The p ´ 1 method
Solution: One can ask for the smallest k for which k! is divisible by p ´ 1 for
the right 3-digit prime p. Since 121 is not prime, we reject k “ 5 and look for
k “ 6. What 3-digit primes have p ´ 1 dividing 720? 181 and 241 work. As
I mentioned in class, it is possible to uncover a prime before p ´ 1 divides k!.
One way to look for them: factor 2k! ´ 1. If p 2k! ´ 1 then p can be found in
k steps. It turns out that 241 224 ´ 1 so 241 can be found in 4, not 6 steps!
Since 224 ´ 1 “ 32 ˆ 5 ˆ 7 ˆ 13 ˆ 17 ˆ 241, these are the primes that can be
found in 4 steps. At 5 steps, we pick up additional primes 11, 31, 41, 61, 151,
331, 1321, 61681 and 4562284561. The list of 3-digit primes actually found at
stage 6 is 109, 151, 181, 241, 257, 331, 433, 577, 631, 673.
(b) The rho method.
Solution: The way to do this without trial-and-error is similar to the above:
If a1 “ 2, an “ a2n´1 ` 1, factor ak ´ ak{2 for even k, looking for the first
appearance of a 3-digit prime. I get a4 ´ a2 “ 672, with no 3-digit prime
divisors, a6 ´ a3 “ 210066388875 “ 33 ˆ 53 ˆ 7 ˆ 31 ˆ 97 ˆ 2957, not giving any
3-digit primes. For a8 ´ a4 , a large number, I get 3-digit primes 113 and 163.
Page 5
Take Home Exam Questions
5n ´ 1
1. (15 points) This problem concerns numbers of the form Tn “
. Have a look at
4
the notes related to Mersenne numbers and the notes on congruences for help with this
question because it shares many properties with the Mersenne sequence Mn “ 2n ´ 1.
(a) How many vanes of n can you find for which Tn is prime?
Solution: I checked values of n ď 10, 000 and found the values n: 3, 7, 11,
13, 47, 127, 149, 181, 619, 929, 3407. A student in the class said they checked
n ď 1, 000, 000 and added 10949, 13241, 13873, 16519. The must have stopped
their program early (or meant 100,000 rather than 1,000,000) because 201359
and 396413 also work.
(b) Prove that Tn is prime only when n is prime.
Solution: Since xmk ´ 1 “ pxm ´ 1qpxpk´1qm ` ¨ ¨ ¨ ` xm ` 1q, it follows that
5mk ´ 1
5m ´ 1
Tmk “
“
j for some integer j and j ą 5m ą 2 if m and k are
4
4
larger than 1. Thus, if n is not prime and m n then Tm properly divides Tn ,
so Tn is not prime if n is not prime.
(c) Use the binary squaring algorithm to show that T5 is not prime but that it is a base
5 pseudoprime.
Solution: We have T5 “ 781 “ 11 ˆ 71, so T5 is not prime. We need to show
that 5780 ” 1 pmod 781q. To that end, we have 780 = 512 + 256 + 8 + 4 so
k
1 2
4
8 16 32 64 128 256 512
5k pmod 781q 5 25 625 125 5 25 625 125 5
25
giving 5780 ” 25 ˆ 5 ˆ 125 ˆ 625 ” 5 ˆ 625 “ 3125 ” 1 pmod 781q, as desired.
(d) Without using a symbolic manipulation package (ie no Maple, Mathematica, Alpha,
Sage), show that for each prime divisor, p of T17 , that 517 ” 1 pmod pq. Use this to
show that T17 is a base 5 pseudoprime.
517 ´ 1
“ 190734863281 “ 409 ˆ 466344409. If
4
p T 17 then p 517 ´ 1, or 517 ” 1 pmod pq, so the first part of the question is
done. For the second part, we must show that 5T17 ” 5 pmod T17 q, or 5T17 ´1 ” 1
pmod T17 q. I will work with the second of these. By the Chinese Remainder
Theorem, it is enough to show that 5T17 ´1 ” 1 pmod pq for each prime divisor
of T17 . To use the hint, note that T17 ´ 1 “ 190734863280 “ 17 ˆ 11219697840,
say T17 ´ 1 “ 17m. If p is one of the divisors of T17 then 5T17 ´1 “ 517m “
Solution: We have T17 “
Page 6
p517 qm ” 1m ” 1 pmod pq. Since this works for both primes, it is true modulo
their product, T17 , as desired.
2. (20 points) This problem deals with the congruence x2 ” 1 pmod nq.
(a) If n is an odd prime, prove that this congruence has exactly two solutions (mod n.)
Solution: The key property is that if p is a prime and p mk then p m or
p k. In our case, if x2 ” 1 pmod nq then n x2 ´ 1, so n px ´ 1qpx ` 1q.
Thus, n x ´ 1 or n x ` 1. In the first case, x ” 1 pmod nq, in the second,
x ” ´1 ” n ´ 1 pmod nq, and we have exactly two solutions.
(b) Use the Chinese Remainder Theorem and the fact that 305 “ 5 ˆ 61 to solve
the congruence x2 ” 1 pmod 305q. That is, use solutions to x2 ” 1 pmod 5q and
x2 ” 1 pmod 61q to build solutions modulo 305. You should get 4 solutions, Mathematica, Maple, Excel, etc are not allowed.
2
Solution: Since 305 “ 5 ˆ 61, by the"Chinese Remainder
* Theorem, x ” 1
2
x ” 1 pmod 5q
pmod 305q is equivalent to the system
. By part (a), each
x2 ” 1 pmod 61q
of these congruences has two solutions, so we get four systems:
x ” 1 pmod 5q x ” ´1 pmod 5q x ” 1 pmod 5q
x ” ´1 pmod 5q
,
,
,
.
x ” 1 pmod 61q x ” 1 pmod 61q x ” ´1 pmod 61q x ” ´1 pmod 61q
The first system has solution x ” 1 pmod 305q and the last one has solution
x ” ´1 ” 304 pmod 305q. This leaves us with the middle two. For the second,
if we start with x ” 1 pmod 61q, then we have x “ 1 ` 61k ” ´1 pmod 5q, or
k ” ´2 ” 3 pmod 5q. Thus, x “ 1 ` 61p3 ` 5mq “ 184 ` 305m so x ” 184
pmod 305q. A cute trick for getting the last solution: If x is a solution, then
´x must be as well, so the other solution is x ” ´184 ” 121 pmod 305q.
Without the trick, x “ ´1 ` 61k ” 1 pmod 5q gives k ” 2 pmod 5q, so x “
´1 ` 61p2 ` 5mq “ 121 ` 305m, as expected. There are four solutions: 1, 121,
184, 304.
(c) How many solutions does the congruence x2 ” 1 pmod 1525q have? Justify your
answer. Again, you are not allowed to use symbolic calculators or Excel.
Solution: As with* the previous problem, this one is equivalent to a system
"
x2 ” 1 pmod 25q
. The bottom congruence has exactly two solutions, we need
x2 ” 1 pmod 61q
Page 7
to know how many solutions the top one has. If x2 ” 1 pmod 25q, then 25 x2 ´ 1, or 25 px ´ 1qpx ` 1q. This means 5 divides the product, so 5 x ´ 1
or 5 x ` 1. Now 5 cannot divide both terms (if it did, then 5 would have to
divide px ` 1q ´ px ´ 1q “ 2.) This means that which ever term is divisible by
5 is also divisible by 25. Thus, we have only two solutions to the congruence
x2 ” 1 pmod 25q, and the original congruence will have 2 ˆ 2 “ 4 solutions.
(d) Prove that if n is divisible by at least two odd primes, then x2 ” 1 pmod nq has at
least 4 solutions. Hint: Write n “ pi q j m, where p, q are distinct primes relatively
prime to m, and use the Chinese Remainder Theorem.
Solution: If we write n “ pi q j m, where p, q are distinct primes relatively prime
to m, and use the Chinese
then x2 ” 1 pmod nq is equiv$ 2 Remainderi Theorem,
,
& x ” 1 pmod p q .
alent to the system x2 ” 1 pmod q j q . Since every odd prime is larger than
% 2
x ” 1 pmod mq
2, the first two congruences have at least two solutions: x “ ˘1. Certainly the
last congruence has at least the solution x ” 1, so all totaled, we have at least
2 ˆ 2 ˆ 1 “ 4 solutions.
(e) If n “ 2p, where p is an odd prime, how many solutions does x2 ” 1 pmod nq have?
What if n “ 4p?
Solution: Since x2 ” 1 pmod 2q has a single solution x ” 1 pmod 2q (that is,
x must be odd) and x2 ” 1 pmod pq has two, the Chinese Remainder Theorem
says we have exactly 1 ˆ 2 “ 2 solutions.
For x2 ” 1 pmod 4pq, we get two solutions from p and two more from 4: just
trying x “ 0, 1, 2, 3, we see that only the first and last are solutions to x2 ” 1
pmod 4q, so we are back to 4 solutions. I probably should have asked about
x2 ” 1 pmod 8pq, which has 8 solutions, and x2 ” 1 pmod 2k pq for k ą 3, which
also has 8 solutions. That is, as k increases, we have two solutions for k “ 0, 1,
four solutions for k “ 2 and 8 solutions for all k ě 3.
3. (15 points) This problem introduces a suped up version of Fermats Little theorem. It
is the basis for the Miller-Rabin test (given in part c). Let p be an odd prime, and
suppose that p “ 2k m ` 1, where m is odd. For example, if p “ 47, then p “ 21 ¨ 23 ` 1,
if p “ 61, then p “ 22 ¨ 15 ` 1, and if p “ 127, then p “ 27 ¨ 1 ` 1. In these three cases,
k “ 1, m “ 23 for 47, k “ 2, m “ 15 for 61, k “ 7, m “ 1 for 127.
Page 8
Given an integer a, with p ffl a, consider the sequence
am ,
a2m ,
a4m ,
km
¨ ¨ ¨ , a2
km
(a) What is the sequence am , a2m , a4m , ¨ ¨ ¨ , a2
pmod pq.
pmod pq if a “ 2, p “ 97?
Solution: 97 “ 1 ` 25 ˆ 3, so k “ 5, m “ 3. We have 23 ” 8 pmod 37q, 26 ”
64, 212 ” 22, 224 ” 96 ” ´1 pmod 97q and 248 ” 296 ” 1 pmod 97q. The
sequence is 8, 64, 22, -1, 1, 1.
km
(b) What is the sequence am , a2m , a4m , ¨ ¨ ¨ , a2
pmod pq if a “ 6, p “ 97?
Solution: Here, 63 ” 22, 612 ” ´1 pmod 97q, so we get 22, -1, 1, 1, 1, 1.
(c) For general a and p with p ffl a, prove that either the sequence consists of all 1’s, or
that the sequence has the form x0 , x1 , . . . , xj , ´1, 1, . . . , 1.
k
What this says is that the sequence am , a2m , a4m , ¨ ¨ ¨ , a2 m pmod pq must
end with 1, and if the sequence is not all 1’s, then it must contain -1. Hint:
If we let the sequence be x0 , x1 , . . . , xk , then Fermat’s Little Theorem tells us
that xk “ 1. (Why?) Since each xi is x2i´1 , we can use problem 2a and work
backwards from xk . An alternative might be to use an extension of a factorization
like a40 ´1 “ pa5 ´1qpa5 `1qpa10 `1qpa20 `1q, which is divisible by 41 if gcdpa, 41q “ 1.
k
Solution: Following the second hint, if A “ am then ap´1 “ A2 so ap´1 ´ 1 “
k
k´1
A2 ´ 1 “ pA ´ 1qpA ` 1qpA2 ` 1q ¨ ¨ ¨ pA2 ` 1q. Since p ap´1 ´ 1, p has to
divide one of these terms. If it is the first term, then am ” 1 pmod pq, and the
j
sequence is all 1’s. If p A2 ` 1, then xj ” ´1 pmod pq, and the sequence
becomes 1’s after that.
k
Alternatively, xk “ 22 ¨m “ 2p´1 ” 1 pmod pq, and x2k´1 ” xk pmod pq, so
x2k´1 ” 1 pmod pq. By problem 2 of the take home, this means xk´1 ” ˘1
pmod pq. If xk´1 ” ´1, then our sequence is x0 , . . . , xk´2 , ´1, 1. The alternative
is that xk´1 ” 1 pmod pq. Now x2k´2 ” 1 pmod pq, so xk´2 ” ˘1 pmod pq. One
possibility is that the sequence of x’s is all 1’s. If this is not the case, then some
last xj is 1, and the previous xj´1 is not 1. In this case, x2j´1 ” xj ” 1 pmod pq,
so xj´1 ” ˘1 pmod pq. Since we have said xj´1 ı 1 pmod pq, we are left with
xj´1 ” ´1.
(d) Any number n which does not have the property above for some a is not prime. For
example, if n “ 341 “ 22 ¨ 85 ` 1, then 285 ” 32 pmod 341q, 22¨85 ” 1 pmod 341q,
so the sequence is 32, 1, 1, which does not pass through -1 to get to 1. This means
Page 9
341 is not prime, even though it is a base 2 pseudoprime. It can be shown that for
every composite number n, there is an a for which either xk ı 1 pmod nq or the
sequences of x’s does not pass through -1 to get to 1. Find the smallest values of a
for the numbers n “ 561, a Carmichael number, and n “ 1373653.
Solution: For n “ 561 “ 24 ˆ 35 ` 1, we have 235 ” 263, 270 ” 166, 2140 ”
67, 2280 ” 1 pmod 561q. The sequence is 263, 166, 67, 1, 1, which does not pass
through -1 so 561 can’t be prime by this test, even though it is a base 2 pseudoprime.
If n “ 1373653 “ 22 ˆ 343413 ` 1, picking various values of a we have the table
a
x0
x1
x2
2 890592
-1
1
3
1
1
1
4
-1
1
1
5 1199564 73782 1370388
so n passes the test for a “ 2, 3, 4 but fails for a “ 5.
Some extra credit problems, if you are interested.
4. (10 points) Some independent problems related to question 1.
(a) If p is prime and q is a prime divisor of Tp , prove that q ” 1 pmod pq.
(b) Prove that if p is a prime and Tp is not prime, then Tp is a base 5 pseudoprime.
p
Solution: This is easier than I indicated in problem 1: Let Tp “ 5 4´1 “ m.
We have 5p “ 1 ` 4m, so 5p ” 1 pmod mq. Also, by Fermat’s Little Theorem,
p 5p ´ 5, so p Tp ´ 1. This means that 5m´1 “ 5pk ” 1k ” 1 pmod mq. If m
is not prime this means m “ Tp is a base 5 pseudoprime.
5n ` 1
5n ` 1
(c) Suppose we define Rn “
when n is even and Sn “
, when n is odd.
2
6
How much of problem 1 can we do? That is,
i. How many n can you find with Rn prime? With Sn prime?
ii. Can you give a necessary condition on n for Rn to be prime? For Sn to be
prime?
iii. When are divisors of Rn or Sn congruent to 1 modulo n?
iv. Do Rn or Sn have to be base 3 pseudoprimes under certain conditions on n?
Solution: I will not do these problems, but mention that Sn is prime for n “
5, 67, 101, 103, 229, 347, and probably infinitely many more, and that Sn can
only be prime if n is prime. For Rn , it is prime for n “ 1, 2, 4, and for no other
Page 10
values under 2000. Probably this is the complete list. In any case, Rn can only
be prime if n is a power of 2.
5. (3 points) With regard to problem 2, prove that if p is an odd prime then for all integers
n ě 1, x2 ” 1 pmod pn q has exactly two solutions. You did the case n “ 1 in 2a.
6. (4 points) Related to problem 3 on the exam, if n “ pq, where p ą q are two odd primes,
find conditions on p, q, and a so that n does not fail the compositeness test in question
3, as described in 3d. That is, the sequences of x’s is either all 1’s or passes through -1
on the way to 1. Use this to find numbers n “ pq for a “ 2, a “ 3, a “ 5.
7. (4 points) Possibly related to the previous question:
(a) Suppose that p and q both divide 2n ´ 1 and that n p ´ 1, n q ´ 1. Prove
that m “ pq is a base 2 pseudoprime. for example, 210 ´ 1 “ 3 ¨ 11 ¨ 31, and
10 p11 ´ 1q, 10 p31 ´ 1q so 11 ˆ 31 “ 341 is a base 2 pseudoprime.
(b) How large of a base 2 pseudoprime can you find using this idea? You need m ą 341
for credit and I will give an extra point to the largest base 2 pseudoprime found by
this method.
8. (8 points) This problem is related to our factoring homework. Let ωpnq be the number
of distinct prime divisors of n. In class, I stated that on average, ωpnq « lnplnpnqq ` .261,
and this matched our results fairly well on our factoring homework.
(a) Knowing how many distinct prime factors a random number has gives information
on its largest prime divisor. Suppose that n is an “average” number with ωpnq
n
distinct prime divisors. Let p be the largest prime divisor of n and suppose that
p
´ ¯
n
is also “average.” Then we should have ωpnq “ ω p `1. If ωpnq “ lnplnpnqq`.261,
what does this tell us about the size of p? Your answer should be p « nα for some
α. I have stated in class that α « .632. What I’m looking for is the actual number.
n
ÿ
ÿ
(b) To actually find ωpnq, first, show that
ωpkq “
tn{pu.
k“1
pďn
(c) How bad of an approximation is it to replace tn{pu by n{p in the sum above?
ÿ1
(d) There is a theorem that says
“ lnplnpnqq`.261`small error. Use this theorem
p
pďn
and parts b, c to show that the average value of ωpkq for k ď n is approximately
lnplnpnqq ` .261.