Download Math 2201 Sheet 1

Math 2201 Sheet 1
Minhyong Kim (following Richard Hill)
December 12, 2008
Questions 1, 2, and 4 are assessed. The other questions are optional. The assessed
homework is to be handed in on Thursday 9th October in class.
If you find a mistake in any question then please email me ([email protected]).
1. Using Euclid’s algorithm calculate hcf(207, 21). Find h, k ∈ Z such that
hcf(207, 21) = 207h + 21k.
2. Find the inverse of 43 modulo 125. Hence solve
43x ≡ 3 mod 125.
n
3. Prove that if pn is the n-th prime then pn < 22 . [Hint: use induction on n and try
to modify the proof of Euclid’s theorem.]
4. Prove that there exist infinitely many primes p such that p ≡ 3 mod 4? [Hint:
note that if a, b ≡ 1 mod 4 then ab ≡ 1 mod 4. Then try something similar to the
proof of Euclid’s theorem.]
5. Prove that there exist infinitely many primes p such that p ≡ 2 mod 3.
6. Prove that there exist infinitely many primes p such that p ≡ 5 mod 6.
7. Why are there no more questions like the last 3 questions?
8. Prove that for any prime p we have (p − 1)! ≡ −1 mod p. [Hint: consider pairing
off the elements of {1, 2, . . . , p − 1} as pairs (a, b) where ab ≡ 1 mod p.] This result
is known as “Wilson’s Theorem”.
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Definitions
• We say that a ∈ Z divides b ∈ Z iff there exists c ∈ Z such that b = ac. This is
written a|b.
• An integer p ≥ 2 is prime iff the only divisors of p are ±1 and ±p.
• A common divisor of a and b is an integer that divides both a and b.
• The highest common factor of a and b is a common factor d of a and b such that
any other common factor is smaller than d. This is written d = hcf(a, b).
• We define a ≡ b mod m iff m|(a − b). We say a is congruent to b modulo m.
• The congruency class of a is the set of integers congruent to a modulo m. This is
written [a].
• Every integer is congruent to exactly one of the numbers 0, 1, . . . , m − 1, so the set
of all congruency classes is {[0], . . . , [m − 1]}. This is written Z/m.
• An inverse if a modulo m is an integer b such that ab ≡ 1 mod m. The set of
elements of Z/m with inverses is written (Z/m)× .
Corollary 1 (Bezout’s Lemma) Let d = hcf(a, b). Then there are integers h, k ∈ Z
such that d = ha + kb.
Theorem 1 If p is a prime and p|ab then p|a or p|b.
Theorem 2 (Unique factorisation) If a ≥ 2 is an integer then there are primes pi > 0
such that a = p1 p2 · · · ps . Moreover if a = q1 q2 · · · qt for primes qj > 0 then s = t and
(after reordering if necessary) pi = qi for i = 1, . . . , s.
Theorem 3 There are infinitely many primes.
Lemma 1 An integer a has an inverse modulo m if and only if a and m are coprime.
Corollary 2 Z/p is a field.
Corollary 3 The set (Z/p)× = {1, 2, . . . , p − 1} is a group with the operation of multiplication modulo p.
Theorem 4 (Fermat’s Little Theorem) If p is prime and a ∈ Z then ap ≡ a mod p.
Theorem 5 (Chinese Remainder Theorem) Suppose m and n are coprime; let [x] ∈
Z/n and [y] ∈ Z/m. Then there is a unique [z] ∈ Z/nm such that z ≡ x mod m and
z ≡ y mod n.
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2201 Sheet 1 Solutions
1. Using Euclid we calculate gcd(207, 21) = 3 as follows.
207 = 9 × 21 + 18
21 = 1 × 18 + 3
18 = 6 × 3.
3 = 21 − 1 × 18
= 21 − (207 − 9 × 21)
= 10 × 21 − 207.
2. We need to find the inverse of 43 modulo 125.
125
43
39
4
1
=
=
=
=
=
=
=
=
=
2 × 43 + 39
1 × 39 + 4
9×4+3
1×3+1
4−3
4 − (39 − 9 × 4) = 10 × 4 − 39
10 × (43 − 39) − 39 = 10 × 43 − 11 × 39
10 × 43 − 11(125 − 2 × 43)
32 × 43 − 11 × 125.
Therefore 43−1 ≡ 32 (125).
x ≡ 32 × 3 = 96 (125).
1
3. We use induction on n. The result holds for n = 1 since p1 = 2 < 22 = 4. So
k+1
suppose the result holds for n ≤ k we need to show that pk+1 < 22 . Consider
Q = p1 p2 · · · pk + 1,
this is not divisible by any of the first k primes and so has a prime factor p ≥ pk+1 .
Hence pk+1 ≤ p ≤ Q. But now using our inductive hypothesis for n ≤ k we have
1
2
k
k
k+1 −2
pk+1 ≤ Q < 22 22 · · · 22 + 1 = 22+4+···+2 + 1 = 22
Thus the result follows by induction.
k+1
+ 1 < 22
.
2
3
4. Suppose, for a contradiction, that there are only finitely many primes p ≡ 3 mod 4,
say p1 , p2 , . . . , pk , with p1 = 3. Note that if a, b ≡ 1 mod 4 then ab ≡ 1 mod 4. So
any number M ≡ 3 mod 4 has at least one prime factor q satisfying q ≡ 3 mod 4.
Consider
M = 4p2 p3 · · · pk + 3.
Then M ≡ 3 mod 4 and M is not divisible by 3 (since p1 = 3). But M has a
factorisation into (odd) primes and by the argument above at least one of these
must be congruent to 3 modulo 4. Thus there is 2 ≤ i ≤ k such that pi | M but
then pi | 3, a contradiction, since pi > 3.
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5. This is similar to question 4, except we take
M = 3p2 p3 · · · pn + 2,
where p1 , . . . , pn are the primes congruent to 2 modulo 3 and p1 = 2.
6. This follows from the previous question together with the Chinese Remainder Theorem.
7. The numbers n = 3, 4, 6 are the only natural numbers for which (Z/n)× has exactly
2 elements. For this reason, we can’t use the same method to prove more general
results. However the following is still true (proved in the 3rd year course on analytic
number theory):
Dirichlet’s Prime number Theorem Let n, a ∈ N be coprime. Then there are
infinitely many prime numbers congruent to a modulo n.
8. The result is trivial for p = 2, 3 so suppose that p ≥ 5. Each element a ∈ {1, . . . , (p−
1)} has a unique inverse b modulo p. Moreover if the inverse of a is itself then a2 ≡ 1
mod p and so a = 1 or a = p − 1.
Thus we can pair off all the elements in {2, . . . , (p − 2)} into inverse pairs and so
2 · 3 · · · (p − 2) ≡ 1 · 1 · · · 1 ≡ 1 mod p.
Hence
(p − 1)! ≡ p − 1 ≡ −1 mod p.
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4