SOLUTIONS TO MATH 38B GRADED HOMEWORK 3
Exercise 1
Let ϕ be Euler’s function. Show that ϕ(nm ) = nm−1 ϕ(n) for all natural numbers n, m.
Solution. Let n = pa1 1 · · · par r be the prime decomposition for n. Since ϕ is multiplicative on coprime
1
r
). Next, we know that if q is prime and k ≥ 1
elements, we have that ϕ(nm ) = ϕ(pma
) · · · ϕ(pma
r
1
k
k−1
then ϕ(q ) = q
(q − 1). Written another way reads ϕ(q k ) = q k−1 ϕ(q). Using this with what we
have already written we get
1 −1
r −1
ϕ(nm ) = pma
· · · pma
ϕ(p1 ) · · · ϕ(pr ) =
r
1
1
r
pma
· · · pma
r
1
ϕ(n) = nm−1 ϕ(n),
p 1 · · · pr
where we have used the multiplicativity of ϕ on primes to get the second equality.
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Exercise 2(a) Let a be an integer, u, v, n, m natural numbers. We assume that m and n are
relatively prime, that au ≡ 1 mod m and that av ≡ 1 mod n. Show that alcm(u,v) ≡ 1 mod (mn).
(b) Let a be an integer relatively prime to 63. Show that a36 ≡ 1 mod 63.
(c) Using question (a), show that we can improve the result in (b), by proving that for any integer
relatively prime to 63, a6 ≡ 1 mod 63.
Solution.
(a) Since m and n are relatively prime, if x ≡ 1 mod m and x ≡ 1 mod n then x ≡ 1 mod mn. Apply
this fact to x = alcm(u,v) . Since lcm(u, v) is a multiple of u (respectively v) the congruence
alcm(u,v) ≡ 1 mod m (respectively alcm(u,v) ≡ 1 mod n) follows from assumption that au ≡
1 mod m (respectively av ≡ 1 mod n).
(b) We see 63 = 32 · 7 so that ϕ(63) = 3(3 − 1) · 6 = 36. Since a is coprime to 63, we can apply
Euler’s theorem and get a36 ≡ 1 mod 63.
(c) Again we use that 63 = 32 · 7. By Euler’s theorem again, we have a6 ≡ 1 mod 32 and a6 ≡
1 mod 7. Take m = 32 , n = 7 and u = v = 6 in part (a) to see that a6 ≡ 1 mod 63.
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Exercise 3
n
For any natural number n, let Fn = 22 + 1. The Fn are called the Fermat numbers.
(a) Show that for every k > 0, Fn | (Fn+k − 2),
(b) Show that no two Fermat numbers have a common divisor greater than 1,
(c) Let pn be the nth prime (so p1 = 2, p2 = 3, . . . ). Show that pn+1 < Fn .
Solution.
n
n+1
(a) Proof by induction on k. The base case is k = 1. We want to check that 22 + 1 | 22
− 1, but
we have
n+1
n
n
n
22
− 1 = (22 )2 − 1 = (22 − 1)(22 + 1)
so the statement is verified here. Now assume that k > 1. Then we can write
! n+k−1 "2
n+k−1
n+k−1
n+k
+ 1).
− 1)(22
− 1 = (22
− 1 = 22
22
n
By induction 22 + 1 divides the left factor of the above equation and so we are done.
(b) Suppose that d divides both Fn and Fm with n $= m. Without loss of generality we may assume
that n < m. Since d divides Fn , part (a) says that d | Fm − 2. But, since d | Fm this implies
that d | 2, i.e. d = 1 or d = 2. However, each Fermat number is evidently odd and so 2 ! Fn for
any n. Thus d = 1.
(c) By part (b) the numbers F0 < F1 < · · · < Fn are all pairwise coprime and so by considering
all the prime decompositions of the Fi we can see at least n + 1 primes which are at most
Fn , i.e. pn+1 ≤ Fn . On the other hand, 2 appears in no prime decomposition because each
1
2
SOLUTIONS TO MATH 38B GRADED HOMEWORK 3
Fermat number is odd. This gives us one more prime and thus pn+2 ≤ Fn . It follows now that
pn+1 < Fn .
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