Solutions to Problem Set 6
1. Section 6.2 - Problem 2
Note 45 = 9 · 5 is composite.
We have 174 ≡ 1 (mod 5) and 174 ≡ (−1)4 ≡ 1 (mod 9). Thus by the Chinese remainder
theorem, 174 ≡ 1 (mod 45) and so 1745 = 17 · 174·11 ≡ 17 (mod 45). Thus 45 is a pseudoprime
to the base 17.
We have 192 ≡ 1 (mod 5) and 192 ≡ 1 (mod 9), so 192 ≡ 1 (mod 45). Thus 1945 = 19 ·
2·22
19
≡ 19 (mod 45). Thus 45 is a pseudoprime base 19.
2. Section 6.2 - Problem 8
Let N = 2p − 1. We have 2p−1 ≡ 1 (mod p) and so N = 2p−1 − 1 = pk for some odd integer
k. Thus N − 1 = 2p − 2 = 2(2p−1 − 1) = 2pk. Note that 2p = N + 1 ≡ 1 (mod N ). Thus
2N −1 = 22pk = (2p )2k ≡ 1 (mod N ). That is, N is a pseudoprime to the base 2.
3. Section 6.2 - Problem 12
We have 24 = 23 · 3. We calculate 73 ≡ −7 (mod 25), 72·3 ≡ −1 (mod 25). Thus
724 ≡ (−1)4 ≡ 1 (mod 25),
712 ≡ (−1)2 ≡ 1 (mod 25), and
76 ≡ −1 (mod 25).
This means that 25 passes Miller’s test for the base 7. In other words, 25 is a strong pseudoprime to the base 7.
4. Section 6.2 - Problem 18(a)
Let n = (6m + 1)(12m + 1)(18m + 1). Let b be a positive integer such that (b, n) = 1.
As 6m + 1 | n, we have (b, 6m + 1) = 1. Thus b6m ≡ 1 (mod 6m + 1). Similarly we have
12m
b
≡ 1 (mod 12m + 1) and b18m ≡ 1 (mod 18m + 1).
Multiplying out (6m + 1)(12m + 1)(18m + 1), we observe that
n = 6 · 12 · 18m3 + (6 · 12 + 6 · 18 + 12 · 18)m2 + 36m + 1
Thus 6m | n − 1, 12m | n − 1 and 18m | n − 1, we have
bn−1 ≡ 1 (mod 6m + 1)
bn−1 ≡ 1 (mod 12m + 1)
bn−1 ≡ 1 (mod 18m + 1)
As 6m + 1, 12m + 1 and 18m + 1 are distinct primes we have bn−1 ≡ 1 (mod n). As b was an
arbitrary positive integer satisfying (b, n) = 1 we thus have n is a Charmichael number.
5. Section 6.3 - Problem 6
This is equivalent to finding the least nonnegative residue of 7999999 modulo 10. We have
φ(10) = 4 so 74 ≡ 1 (mod 10). We have 4 | 999996, so 7999999 ≡ 73 ≡ 3 (mod 10), thus the
last digit of 7999999 is 3.
6. Section 6.3 - Problem 8
By Fermat’s little theorem we have a7 ≡ a (mod 7). We also claim that a7 ≡ a (mod 9). If
we prove this, then the Chinese Remainder Theorem will tell us that a7 ≡ a (mod 63), and our
proof will be complete. (The Chinese Reainder Theorem applies here because 63 = 7 · 9, and
gcd(7, 9) = 1.) To prove the claim, we will consider two cases.
Case 1: a is divisible by 9. In this case, a7 ≡ 0 ≡ a (mod 9).
Case 2: a is not divisible by 3. Then (a, 9) = 1. As φ(9) = 6, by Euler’s theorem we then
have a6 ≡ 1 (mod 9) and thus a7 ≡ a (mod 9).
In both cases a7 ≡ a (mod 7). This proves the claim.
7. Section 6.3 - Problem 10
We have the congruences
aφ(b) ≡ 1 (mod b)
bφ(a) ≡ 1 (mod a)
aφ(b) ≡ 0 (mod a)
bφ(a) ≡ 0 (mod b)
Thus we have
aφ(a) + bφ(a) ≡ 1 (mod a)
aφ(a) + bφ(a) ≡ 1 (mod b)
and so aφ(a) + bφ(a) ≡ 1 (mod ab).
8. Section 6.3 - Problem 14 (Problem 12 in the 5th edition)
We have mi | Mj for each i 6= j. Also, (Mi , mi ) = 1 for each i. Thus, by Euler’s theorem
we have
φ(m1 )
a1 M1
φ(m2 )
+ a2 M2
φ(mi )
+ · · · + ar Mrφ(mr ) ≡ ai Mi
φ(m )
φ(m )
≡ ai (mod mi )
φ(m )
for each i. Thus x ≡ a1 M1 1 + a2 M2 2 + · · · + ar Mr r satisfies the congruences and by
the Chinese remainder theorem this is the only solution modulo M .
9. Section 7.1 - Problem 4
Let n be a positive integer and write n = pa11 pa22 · · · pakk for distinct primes pi . Then
φ(n) =
k
Y
pai i −1 (pi − 1)
i=1
pai i −1 (pi
a) Suppose φ(n) = 1. Then
− 1) = 1 for each prime pi dividing n. Thus pi − 1 = 1
and so 2 is the only prime that could divide n. If 2 | n then 2ai −1 = 1 and so ai = 1. If 2 does
not divide n, then no primes divide n and so n = 1. Thus φ(n) = 1 if and only if n = 1 or
n = 2.
b) Suppose φ(n) = 2. Then piai −1 (pi − 1) is 1 or 2 for each prime pi dividing n. Thus each pi
is 2 or 3, i.e., and we have n = 2a 3b for some a, b ≥ 0. We have b = 0 or b = 1 as otherwise 3
would divide φ(n).
Case 1. Suppose b = 1, i.e., n = 2a · 3. If a ≥ 2, then φ(n) is divisible by 4, hence, cannot
be 2. Thus a ≤ 1, i.e., n = 3 or 6. Both of these are solutions: φ(3) = φ(6) = 2.
Case 2. Suppose b = 0, i.e., n = 2a . If a = 0 n = 1, then φ(n) = 1 6= 2. If a ≥ 1, then
φ(n) = 2a−1 = 2, so a = 2 (i.e., n = 4) is the only solution in this case.
Answer: φ(n) = 2 if and only if n = 3, 4 or 6.
c) Suppose φ(n) = 3. Then pi − 1 is 1 or 3 for each prime pi dividing n. Thus pi = 2 is the
only possibility. That is, n = 2a is a power of 2. But then φ(n) is also a power of 2, so cannot
be 3.
Answer: There are no n satisfying φ(n) = 3.
d) Suppose φ(n) = 4. Once again, the formula for φ(n) shows that pi − 1 divides 4 for every
prime divisor pi of n. Hence, no primes, other than 2, 3 and 5 can divide n.
Thus n = 2a 3b 5c for some a, b, c ≥ 0. If b ≥ 2, then φ(n) is divisible by 3, a contradiction.
Thus b ≤ 1.
Case 1: b = 1. Then n = 2a · 3 · 5c , and thus
4 = φ(n) = φ(3)φ(2a · 5c ) = 2φ(2a · 5c ) .
We conclude that φ(2a · 5c ) = 2. By part (b), the only positive integers m such that φ(m) = 2
are m = 3, 4, and 6. Of these, only 4 is of the form 2a 5c . Thus 2a · 5c = 4, the only solution to
φ(n) = 4 in the case (with b = 1) is n = 3 · 2a · 5c = 3 · 4 = 12.
Case 2: b = 0, i.e., n = 2a 5c . Clearly c ≤ 1, since otherwise φ(n) is divisible by 5.
If c = 1, then n = 2a · 5, φ(n) = φ(22 ) · 4 = 4 and thus φ(2a ) = 1. This is only possible if
a
2 = 1 or 2. Thus the only solutions with b = 0 and c = 1 are n = 5 or 10.
If c = 0, then n = 2a . The only solution to φ(n) = 4 of the form n = 2a is n = 8.
Answer: φ(n) = 4 if and only if n = 5, 8, 10 or 12.
10. Section 7.1 - Problem 8
Suppose φ(n) = 14. Then p − 1 | 14 for each prime dividing n. Thus the only primes that
could divide n are p = 2 or 3. n = 2a 3b . Then φ(n) = φ(2a )φ(3b ). But our formula shows that
7 cannot divide φ(2a ) or φ(3b ) for any a, b ≥ 0. Thus φ(n) cannot be equal to 14.
11. Section 7.1 - Problem 18
Let n be an odd integer. Then gcd(4, n) = 1. Since φ is a multiplicative function, we have
φ(4n) = φ(4)φ(n) = 2φ(n).
12. Section 7.1 - Problem 34
√
Let n be a positive integer. If n = 1 then φ(n) = 1 > n − n. If n is prime then
√
φ(n) = n − 1 > n − n
√
√
as n > 1. It remains to prove that φ(n) ≤ n − n when n is composite. We will give two
proofs of this inequality.
√
√
Proof: If n is composite, write n = ab with a ≤ n and b ≥ n. Then there are at least b
integers
√ in {1, 2, . . . , n} that are not coprime to n, being a, 2a, 3a, . . . , ba. Thus φ(n) ≤ n−b ≤
n − n.
Alternative proof: Let n = pa11 pa22 · · · pakk be the prime decomposition of n, where p1 < p2 <
· · · < pk are the primes dividing p, arranged in increasing order, and ai ≥ 1. Since we are
assuming that n is composite,
a1 + · · · + ar ≥ 2.
We will use the formula
1
1
1
)(1 − ) · · · (1 − )
p1
pk
pk
1
1
n
n
= (n − )(1 − ) · · · (1 − ) ≤ n − .
p1
pk
pk
p1
√
n
n
We want to show that
≥ n or equivalently, √ ≥ 1. Since p1 < p2 < · · · < pk , we
p1
p1 n
have
n
a /2−1 a1 /2
a /2−1 a1 /2
a /2
(a +a +...ar )/2−1
√
= p1 1
p2 . . . par r /2 ≥ p11
p1 . . . p1r ≥ p1 1 2
.
np1
Since a1 + · · · + ar ≥ 2,
(a1 + a2 + . . . ar )/2 − 1 ≥ 0
φ(n) = n(1 −
(a +a2 +...ar )/2−1
and consequently, p1 1
≥ 1, as desired.