HOMEWORK # 4
MARIA SIMBIRSKY
SANDY ROGERS
MATTHEW WELSH
1. Section 2.1, Problems 5, 8, 28, and 48
Problem. (2.1.5) Write a single congruence that is equivalent to the pair of congruences
x ≡ 1 (mod 4) and x ≡ 2 (mod 3).
Solution. x ≡ 1 (mod 4) and x ≡ 2 (mod 3) if and only if x ≡ 5 (mod 12).
Proof. Note: (3, 4) = 1, so the Chinese Remainder Theorem applies. We have
φ : Z/12Z ∼
= Z/3Z × Z/4Z,
where the map φ is given by φ(x) = (x (mod 3), x (mod 4)). We need to find φ−1 (x).
Since 4 ≡ 1 (mod 3), we note that 2 · 4 ≡ 2 (mod 3) and 2 · 4 ≡ 0 (mod 4). Similarly,
3 · 3 ≡ 1 (mod 4) and 3 · 3 ≡ 0 (mod 3).
Therefore, if we let x = 2 · 4 + 1 · 3 · 3, we have x ≡ 2 (mod 3) and x ≡ 1 (mod 4).
Reducing x modulo 12, we have, x = 2(4) + (3)(3) = 17 ≡ 5 (mod 12), so φ(5) = (2, 1) as
desired.
Problem. (2.1.8) Prove that any number that is a square must have one of the following
for its units digit:
0, 1, 4, 5, 6, 9.
Proof. By Theorem 2.2 in the book, we know a ≡ b (mod 10) implies a2 ≡ b2 (mod 10).
Every x ∈ Z is equivalent (mod 10) to some a ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. It follows that
x2 ≡ a2 (mod 10). The set formed by a2 is: {0, 1, 4, 9, 6, 5, 6, 9, 4, 1} = {0, 1, 4, 5, 6, 9}. (Note
that we only have to check 0, . . . , 5 since a2 ≡ (10 − a)2 (mod 10).)
Problem. (2.1.28) What is the last digit in the ordinary decimal representation of 3400 ?
Solution. The answer is 1 because 3400 ≡ 1 (mod 10).
Proof. Theorem 2.2 states that a ≡ b (mod m) implies an ≡ bn (mod m). Since 34 = 81 ≡ 1
(mod 10), It follows 3400 = (34 )100 ≡ 1100 ≡ 1 (mod 10).
Problem. (2.1.48) If r1 , r2 , . . . , rp and r10 , r2 ,0 . . . , rp0 are any two complete residue systems
modulo a prime p > 2, prove that the set r1 r10 , r2 r20 , . . . , rp rp0 cannot be a complete residue
system modulo p.
Proof. Any complete residue system modulo p is a permutation of the equivalence classes
0, 1, 2, . . . , p − 1 in Z/pZ. Let i and j be the indices such that ri ≡ rj0 ≡ 0 (mod p). If i 6= j,
then the elements ri ri0 and rj rj0 are both equivalent to 0 (mod p), so they cannot be part of
a complete residue system mod p. Therefore, we can assume i = j. After removing ri and
1
ri0 from these lists, we thus see that the remaining elements, which we relabel r1 , . . . , rp−1
0
and r10 , . . . , rp−1
, are complete residue systems for (Z/pZ)∗ , i.e. they represent the nonzero
classes mod p in some permuted order.
Therefore, by Wilson’s Theorem,
r1 r2 · · · rp−1 ≡ (p − 1)! ≡ −1
(mod p)
0
r10 r20 · · · rp−1
≡ (p − 1)! ≡ −1
(mod p)
Hence,
0
r1 r10 · r2 r20 · · · rp−1 rp−1
≡ (−1)(−1) ≡ 1
r1 r10
0
· · · rp−1 rp−1
(mod p)
∗
0
If
was a complete residue system for (Z/pZ) , we would have r1 r10 · · · rp−1 rp−1
≡
0
0
−1 (mod p). But 1 6≡ −1 (mod p) since p > 2. Therefore, r1 r1 · · · rp−1 rp−1 is not a complete
residue system for (Z/pZ)∗ , and hence the original r1 r10 · · · rp rp0 is not a complete residue
system mod p as desired.
2. Section 2.2, Problems 8, 9, and 14
Problem. (2.2.8) Show that if p is an odd prime then the congruence x2 ≡ 1 (mod pα ) has
only the two solutions x ≡ 1 and x ≡ −1 (mod pα ).
Proof. x2 ≡ 1 (mod pα ) ⇒ pα |(x2 − 1) ⇒ pα |(x − 1)(x + 1)
Case 1: (pα , x − 1) > 1. Then p|(x − 1) since the only divisors of pα are powers of p. Since
p > 2, p does not divide x + 1, so (pα , x + 1) = 1. By the fundamental theorem of arithmetic,
we see that pα |(x − 1)(x + 1) ⇒ pα |(x − 1). Therefore x ≡ 1 (mod pα ).
Case 2: (pα , x − 1) = 1. Then by the fundamental theorem of arithmetic, we see that
pα |(x − 1)(x + 1) ⇒ pα |(x + 1). Therefore x ≡ −1 (mod pα ).
Problem. (2.2.9) Show that the congruence x2 ≡ 1 (mod 2α ) has one solution when α = 1,
two solutions when α = 2, and precisely four solutions 1, 2α−1 − 1, 2α−1 + 1, −1 when α ≥ 3.
Proof. When α = 1, x ≡ 0 or 1 (mod 2). Squaring both of these shows that x2 ≡ 1 if and
only if x ≡ 1 (mod 2).
Similarly, when α = 2, x ≡ 0, 1, 2, or 3 (mod 4). Squaring shows that x2 ≡ 1 implies
that x ≡ 1 or x ≡ 3 (mod 4).
Now consider the case when α ≥ 3. x2 ≡ 1 (mod 2α ) implies, by definition, that 2α |x2 − 1,
or equivalently 2α |(x − 1)(x + 1). There are three cases to consider.
Case 1: (2α , x + 1) = 1. Then 2α |x − 1 by the fundamental theorem of arithmetic, so x ≡ 1
(mod 2α ).
Case 2: (2α , x − 1) = 1. Then 2α |x + 1 by FTA, so x ≡ −1 (mod 2α ).
Case 3: Both (2α , x−1) > 1 and (2α , x+1) > 1. Since the only divisors of 2α are powers of
2, this implies that 2|(x+1) and 2|(x−1). This, along with the fact that 2α |(x+1)(x−1), im)( x−1
). Note that α > 2 so this is still well-defined. Since x+1
= x−1
+ 1,
plies that 2α−2 |( x+1
2
2
2
2
x+1
x−1
x−1
α−2
α−2
either (2 , 2 ) = 1 or (2 , 2 ) = 1, depending on whether 2 is even or odd. The
fundamental theorem of arithmetic implies that either 2α−2 | x+1
or 2α−2 | x−1
. This, in turn,
2
2
α−1
α−1
implies that 2 |(x + 1) or 2 |(x − 1). This leaves us with four possible congruences of x
modulo 2α , two of which have already been considered (specifically x ≡ 1 or −1 (mod 2α )).
The two new possibilities are x ≡ 2α−1 − 1 and x ≡ 2α−1 + 1 (mod 2α ).
2
In conclusion, x2 ≡ 1 (mod 2α ) has four solutions when α ≥ 3. These are 1, 2α−1 − 1,
2α−1 + 1, and −1.
α
Problem. (2.2.14) Show that pk ≡ 0 (mod p) for 0 < k < pα and use this to show that
α
α
α
(a + b)p ≡ ap + bp (mod p).
Proof. First note that
α α
p − 1 pα
p
=
.
k
k−1 k
Indeed, we calculate
α
α
(pα )!
p
(pα )(pα − 1)!
p − 1 pα
=
=
=
.
k
k!(pα − k)!
k(k − 1)!(pα − 1 − (k − 1))!
k−1 k
Now
α
α α
p −1 α
p
p − 1 pα
∈Z⇒k|
p
=
k−1
k
k−1 k
Let k = mpβ where (m, p) = 1 and 0 ≤ β < α. Then we have
α
p − 1 α−β
m|
p .
k−1
The fundamental theorem of arithmetic implies that
α
α
α
p −1
p −1 β
p − 1 α−1
m|
⇒ k|
p ⇒ k|
p
k−1
k−1
k−1
So
α
α
p − 1 pα
p − 1 pα−1
∈Z⇒
≡0
k−1 k
k
k−1
(mod p).
α
α
α
This proves the first part of the problem. Now we show that (a+b)p ≡ ap +bp (mod p).
α
(a + b)
pα
α
α
α
p α
p −1 α X
X p
p
p
p
α
k pα −k
0 pα
k pα −k
=
a b
=
ab +
a b
+ α ap b 0
k
0
k
p
k=0
k=1
α
α
p
p
α
0 pα
≡
a b + 0 + α ap b0 (mod p)
0
p
α
α
Notice that p0 = 1 and ppα = 1 so
α
α
p
p
α
α
α
0 pα
a b + 0 + α ap b0 ≡ bp + ap (mod p).
0
p
3
3. Section 2.3, Problems 15, 20, and 47
Problem. (2.3.15) Solve the congruence x3 + 4x + 8 ≡ 0 (mod 15).
Proof. Let f (x) = x3 + 4x + 8. x must be in some equivalence class modulo 5, where
Z/5Z = {[−2], [−1], [0], [1], [2]}.
Trying the values x = 0, ±1, ±2, we find that f (x) ≡ 0 (mod 5) has no solutions. Since
5 | 15, it follows that f (x) ≡ 0 (mod 15) has no solutions.
Problem. (2.3.20) Let m1 and m2 be arbitrary positive integers, and let a1 and a2 be arbitrary integers. Show that there is a simultaneous solution of the congruences x ≡ a1
(mod m1 ) and x ≡ a2 (mod m2 ) if and only if a1 ≡ a2 (mod g) where g = (m1 , m2 ). Show
that if this condition is met, then the solution is unique modulo [m1 , m2 ].
Proof. If ∃x ∈ Z such that x ≡ a1 (mod m1 ) and x ≡ a2 (mod m2 ) , then ∃n1 , n2 ∈ Z such
that
x − a1 = m1 n1
x − a2 = m2 n2
and hence,
a1 + m1 n1 = a2 + m2 n2 .
Conversely, if ∃n1 , n2 ∈ Z such that
a1 + m1 n1 = a2 + m2 n2 ,
then by letting x be the common value, we see x ≡ a1 (mod m1 ) and x ≡ a2 (mod m2 ).
Therefore, a simultaneous solution x exists if and only if there exist n1 and n2 such that
a1 + m1 n1 = a2 + m2 n2 . This equation is equivalent to m1 n1 − m2 n2 = a2 − a1 . By Corollary
1.1.5 in the online notes, such n1 , n2 exist ⇔ (m1 , m2 )|a2 − a1 ⇔ a1 ≡ a2 (mod (m1 , m2 )).
To show uniqueness of x modulo [m1 , m2 ], take y ∈ Z such that x ≡ y ≡ a1 (mod m1 ) and
x ≡ y ≡ a2 (mod m2 ). By Theorem 2.3 in the book, this implies x ≡ y (mod [m1 , m2 ]). Problem. (2.3.47) Let f (x) be a polynomial with integral coefficients, let N (m) denote
the number of solutions of the congruence f (x) ≡ 0 (mod m), and let φf (m) denote the
number of integers a, 1 ≤ a ≤ m, such that (f (a), m) = 1. Show that if (m, n) = 1
then φf (mn) = φf (m)φf (n). Show that if α > 1 then φf (pα ) = pα−1 φf (p). Show that
Q
φf (p) = p − N (p). Conclude that for any positive integer n, φf (n) = n p|n (1 − Np(p) ). Show
that for an appropriate choice of f (x), this reduces to theorem 2.19.
Proof. Show that if (m, n) = 1, then φf (mn) = φf (m)φf (n).
For any positive integer k, let Sk = {a ∈ Z/kZ | (f (a), k) = 1} so that φf (k) = #Sk . Now
consider Smn where (m, n) = 1. By the Chinese Remainder theorem we know there is an
isomorphism ψ : Z/mnZ → (Z/mZ) × (Z/nZ). The image of Smn under ψ is
ψ(Smn ) = {ψ(a) ∈ (Z/mZ) × (Z/nZ) | (f (a), mn) = 1}.
Let ψ(a) = (a1 , a2 ). We know that a ≡ a1 (mod m) and a ≡ a2 (mod n), and, by Theorem
2.2 in the book, f (a) ≡ f (a1 ) (mod m) and f (a) ≡ f (a2 ) (mod n). Applying Theorem 2.4
4
we get (f (a), m) = (f (a1 ), m) and (f (a), n) = (f (a2 ), n). Since (f (a), mn) = 1 if and only if
(f (a), m) = (f (a), n) = 1, we have (f (a), mn) = 1 ⇐⇒ (f (a1 ), m) = (f (a2 ), n) = 1. Hence
ψ(Smn ) = {(a1 , a2 ) ∈ (Z/mZ) × (Z/nZ) | (f (a1 ), m) = (f (a2 ), n) = 1} = Sm × Sn .
We can now restrict the domain of ψ to Smn to establish a bijection between Smn and Sm ×Sn .
It follows that #Smn = (#Sm )(#Sn ) and hence φf (mn) = φf (m)φf (n).
Next we show that if α > 1 then φf (pα ) = pα−1 φf (p). Notice that (f (a), pα ) = 1 ⇔
(f (a), p) = 1. Thus φf (pα ) = number of integers a, 1 ≤ a ≤ pα , such that (f (a), p) = 1.
Note that by Theorem 2.2 (in the book): f (xpy + z) ≡ f (z) (mod p). Theorem 2.4 states:
If b ≡ c (mod m), then (b, m) = (c, m).
Thus by Theorem 2.4 and Theorem 2.2, we have
(f (xpy + z), p) = (f (z), p).
Notice that each a such that 1 ≤ a ≤ pα can be uniquely written as a = a0 + pk where 1 ≤
a0 ≤ p and 0 ≤ k ≤ pα−1 − 1. We’ve shown (f (a), pα ) = 1 ⇔ (f (a), p) = 1 ⇔ (f (a0 ), p) = 1.
So the number of such a is the number of a0 times the number of all possible k, namely pα−1 ,
i.e.
φf (pα ) = pα−1 φf (p).
Next we show φf (p) = p − N (p).
φf (p) = number of intergers a 1 ≤ a ≤ p such that (f (a), p) = 1
(f (a), p) = 1 ⇔ f (a) 6= p · k for some k. f (a) = p · k exactly when f (a) ≡ 0 (mod p). The
number of times f (a) ≡ 0 (mod p) is by definition N (p).
Thus φf (p) = p − N (p).
α
Finally, let the prime factorization of n be n = pα1 1 pα2 2 · · · pαk k . Since (pαi i , pj j ) = 1,
φf (n) = φf (pα1 1 )φf (pα2 2 ) · · · φf (pαk k )
= pα1 1 −1 φf (p1 ) · p2α2 −1 φf (p2 ) · · · pαk k −1 φf (pk )
= pα1 1 −1 (p1 − N (p1 ))pα2 2 −1 (p2 − N (p2 )) · · · pαk k −1 (pk − N (pk ))
N (p1 ) α2
N (p2 )
N (pk )
αk
α1
= p1 1 −
p2 1 −
· · · pk 1 −
p1
p2
pk
N (p2 )
N (pk )
N (p1 )
αk
α1 α2
1−
··· 1 −
= p1 p2 · · · pk 1 −
p1
p2
pk
Y
N (p)
=n
1−
.
p
p|n
If we let f (x) = x, then φf (m) = the number of integers a, 1 ≤ a ≤ m, such that (a, m) = 1
Thus φf (m) = φ(m). N (p) = 1 since f (x) ≡ 0 (mod p) ⇔ x ≡ 0 (mod p) which occurs
exactly once, when x ≡ 0 (mod p). Thus
Y
1
φ(m) = m
1−
,
p
p|m
which is the formula we proved in class.
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