A First Course in Number Theory
1.4 The Fundamental Theorem of Arithmetic
We shall prove that every positive integer can be written as the product of prime numbers (with the convention that the empty product is equal to 1), and that this representation is unique except for the order in which the prime factors are written. This result is called thefundamental theorem of arithmetic.
Theorem 1.7 (Euclid’s lemma) Let a, b, c be integers. If a divides bc and(a, b) = 1, thena dividesc.
Proof. Sincea divides bc, we have bc=aq for some integer q. Sincea andbare relatively prime, Theorem 1.5 implies that there exist integers x andy such that
1 =ax+by.
Multiplying byc, we obtain
c=acx+bcy=acx+aqy=a(cx+qy), and so adivides c. This completes the proof.2
Theorem 1.8 Letk≥2, and let a, b1, b2, . . . , bk be integers. If(a, bi) = 1 for alli= 1, . . . , k,then(a, b1b2· · ·bk) = 1.
Proof. The proof is by induction onk. Letk= 2 andd= (a, b1b2). We must show that d = 1. Since d divides a and (a, b1) = 1, it follows that (d, b1) = 1. Sinceddividesb1b2,Euclid’s lemma implies thatddividesb2. Therefore,dis a common divisor ofaandb2, but (a, b2) = 1 and sod= 1.
Let k≥3, and assume that the result holds for k−1. Let a, b1, . . . , bk
be integers such that (a, bi) = 1 fori= 1, . . . , k. The induction assumption implies that (a, b1· · ·bk−1) = 1. Since we also have (a, bk) = 1, it follows from the case k= 2 that (a, b1· · ·bk−1bk) = 1.This completes the proof.
2
Theorem 1.9 If a prime number p divides a product of integers, then p divides one of the factors.
Proof. Letb1, b2, . . . , bk be integers such thatpdividesb1· · ·bk. By The- orem 1.8, we have (p, bi)>1 for somei. Sincepis prime, it follows thatp dividesbi.2
Theorem 1.10 (Fundamental theorem of arithmetic) Every positive integer can be written uniquely (up to order) as the product of prime num- bers.
1.4 The Fundamental Theorem of Arithmetic 27 Proof. First we prove that every positive integer can be written as a product of primes. Since an empty product is equal to 1, we can write 1 as the empty product of primes. Let n ≥ 2. Suppose that every positive integer less than n is a product of primes. Ifn is prime, we are done. If n is composite, then n = dd, where 1 < d ≤ d < n. By the induction hypothesis, d and d are both products of primes, and so n = dd is a product of primes.
Next we use induction to prove that this representation is unique. The representation of 1 as the product of the empty set of primes is unique.
Let n ≥ 2 and assume that the statement is true for all positive inte- gers less than n. We must show that if n = p1· · ·pk = p1· · ·p, where p1, . . . , pk, p1, . . . , p are primes, then k =and there is a permutation σ of 1, . . . , k such thatpi =pσ(i) fori= 1, . . . , k. By Theorem 1.9, since pk
dividesp1· · ·p, there exists an integerj0∈ {1, . . . , }such thatpk divides pj0, and sopk =pj0 sincepj0 is prime. Therefore,
n
pk =p1· · ·pk−1=
j=1 j=j0
pj < n.
It follows from the induction hypothesis that k−1 =−1,and there is a one-to-one map σ from {1, . . . , k−1} into {1, . . . , k} \ {j0} such that pi=pσ(i) fori= 1, . . . , k−1. Letσ(k) =j0.This defines the permutation σ, and the proof is complete.2
For any nonzero integer n and prime numberp, we definevp(n) as the greatest integer r such that pr divides n. Then vp(n) is a nonnegative integer, andvp(n)≥1 if and only ifpdividesn. Ifvp(n) =r, then we say that the prime power pr exactly divides n, and write pr n. The standard factorizationofnis
n=
p|n
pvp(n).
Since every positive integer is divisible by only a finite number of primes, we can also write
n=
p
pvp(n),
where the product is an infinite product over the set of all prime numbers, and vp(n) = 0 and pvp(n) = 1 for all but finitely many primes p. The function vp(n) is called thep-adic value of n. It is completely additive in the sense that vp(mn) = vp(m) +vp(n) for all positive integers mand n (Exercise 13). For example, sincen! = 1·2·3· · ·n, we have
vp(n!) = n m=1
vp(m).
The standard factorizations of the first 60 integers are
1 = 1 21 = 3·7 41 = 41
2 = 2 22 = 2·11 42 = 2·3·7
3 = 3 23 = 23 43 = 43
4 = 22 24 = 23·3 44 = 22·11
5 = 5 25 = 52 45 = 32·5
6 = 2·3 26 = 2·13 46 = 2·23
7 = 7 27 = 33 47 = 47
8 = 23 28 = 22·7 48 = 24·3
9 = 32 29 = 29 49 = 72
10 = 2·5 30 = 2·3·5 50 = 2·52
11 = 11 31 = 31 51 = 3·17
12 = 22·3 32 = 25 52 = 22·13
13 = 13 33 = 3·11 53 = 53
14 = 2·7 34 = 2·17 54 = 2·33 15 = 3·5 35 = 5·7 55 = 5·11 16 = 24 36 = 22·32 56 = 23·7
17 = 17 37 = 37 57 = 3·19
18 = 2·32 38 = 2·19 58 = 2·29
19 = 19 39 = 3·13 59 = 59
20 = 22·5 40 = 23·5 60 = 22·3·5.
Let a1, . . . , ak be nonzero integers. An integer m is called a common multiple of a1, . . . , ak if it is a multiple of ai for all i = 1, . . . , k, that is, every integer ai divides m. Theleast common multiple of a1, . . . , ak is a positive integermsuch thatmis a common multiple ofa1, . . . , ak, andm divides every common multiple ofa1, . . . , ak. For example, 910 is a common multiple of 35 and 91, and 455 is the least common multiple. We shall show that there is a unique least common multiple for every finite set of nonzero integers. We denote by [a1, . . . , ak] the least common multiple ofa1, . . . , ak. Theorem 1.11 Let a1, . . . , ak be positive integers. Then
(a1, . . . , ak) =
p
pmin{vp(a1),...,vp(ak)}
and
[a1, . . . , ak] =
p
pmax{vp(a1),...,vp(ak)}.
Proof. This follows immediately from the fundamental theorem of arith- metic.2
Letxbe a real number. Recall that theinteger part ofxis the greatest integer not exceeding x, that is, the unique integern such that n ≤x <
1.4 The Fundamental Theorem of Arithmetic 29 n+1. We denote the integer part ofxby [x]. For example,4
3
= 1, [√ 7] = 2, and
−43
=−2.Thefractional partofxis the real number {x}=x−[x]∈[0,1).
Thus, 4
3
= 13 and
−43
= 23.We can use the greatest integer function to compute the standard factorization of factorials.
Theorem 1.12 For every positive integer nand primep,
vp(n!) = [loglognp]
r=1
n pr
.
Proof. Let 1 ≤ m ≤ n. If pr divides m, then pr ≤ m ≤ n and r ≤ logn/logp. Sinceris an integer, we haver≤[logn/logp] and
vp(m) = [loglognp]
pr|mr=1
1.
The number of positive integers not exceedingnthat are divisible bypris exactly [n/pr], and so
vp(n!) = n m=1
vp(m) = n m=1
[loglognp]
pr|mr=1
1
= [loglognp]
r=1
n
m=1 pr|m
1 = [loglognp]
r=1
n pr
.
This completes the proof.2
We shall use Theorem 1.12 to compute the standard factorization of 10!.
The primes not exceeding 10 are 2,3,5,and 7,and v2(10!) =
10 2
+
10 4
+
10 8
= 5 + 2 + 1 = 8,
v3(10!) = 10
3
+ 10
9
= 4, v5(10!) =
10 5
= 2, v7(10!) =
10 7
= 1.
Therefore,
10! = 2834527.
For every nonzero integerm, theradical ofm, denoted by rad(m), is the product of the distinct primes that dividem,that is,
rad(m) =
p|m
p=
vp(m)≥1
p.
For example, rad(15) = rad(−45) = rad(225) = 15 and rad(pr) =pfor p prime and r≥1.
Theorem 1.13 Let m anda be nonzero integers. There exists a positive integer ksuch that mdivides ak if and only if rad(m)divides rad(a).
Proof. We know that m divides ak if and only if vp(m) ≤ vp(ak) = kvp(a) for every prime p (Exercise 14). If there exists an integer k such thatmdividesak, thenvp(a)>0 whenevervp(m)>0, and so every prime that dividesmalso dividesa. This implies that rad(m) divides rad(a).
Conversely, if rad(m) divides rad(a), then vp(a) >0 for every prime p such thatvp(m)>0. Since only finitely many primes dividem, it follows that there exists a positive integer k such that vp(ak) = kvp(a)≥ vp(m) for all primesp, and somdivides ak.2
Exercises
1. Factor 51,948 into a product of primes.
2. Factor 10k+ 1 into a product of primes fork= 1,2,3,4,5.
3. Find the greatest common divisor and least common multiple ofa= 2338712132 andb= 365511213.
4. Compute the least common multiple of the integers 1,2,3, . . . ,15.
5. Compute the standard factorization of 15!.
6. Prove thatn, n+ 2, n+ 4 are all primes if and only ifn= 3.
7. Prove thatn, n+ 4, n+ 8 are all primes if and only ifn= 3.
8. Letn≥2. Prove that (n+ 1)! +k is composite fork= 2, . . . , n+ 1.
This shows that there exist arbitrarily long intervals of composite numbers.
9. Prove thatn5−nis divisible by 30 for every integern.
10. Find all primespsuch that 29p+ 1 is a square.
1.4 The Fundamental Theorem of Arithmetic 31 11. The prime numberspandqare calledtwin primesif|p−q|= 2. Let pandqbe primes. Prove that pq+ 1 is a square if and only ifpand qare twin primes.
12. Prove that ifp and q are twin primes greater than 3, then p+q is divisible by 12.
13. Letm, n,andkbe positive integers. Prove that
vp(mn) =vp(m) +vp(n) and vp(mk) =kvp(m).
14. Letdandmbe nonzero integers. Prove thatddividesmif and only ifvp(d)≤vp(m) for all primesp.
15. Let m =k
i=1prii, where p1, . . . , pk are distinct primes, k ≥2, and ri ≥1 fori= 1, . . . , k. Let mi =mp−i ki fori= 1, . . . , k. Prove that (m1, . . . , mk) = 1.
16. Leta, b,andcbe positive integers. Prove that (ab, c) = 1 if and only if (a, c) = (b, c) = 1.
17. Prove that if 6 dividesm, then there exist integersbandcsuch that m=bcand 6 divides neitherbnorc.
18. Prove the following statement or construct a counterexample: Ifdis composite and d divides m, then there exist integers b and c such thatm=bcandddivides neitherbnorc.
19. Let a andb be positive integers. Prove that (a, bc) = (a, b)(a, c) for every positive integercif and only if (a, b) = 1.
20. Letm1, . . . , mk be pairwise relatively prime positive integers, and let d divide m1· · ·mk. Prove that for each i = 1, . . . , k there exists a unique divisordi ofmi such thatd=d1· · ·dk.
21. Let n ≥ 2. Prove that the equation yn = 2xn has no solution in positive integers.
22. Letn≥2, and letxbe a rational number. Prove that √n
xis rational if and only ifx=yn for some rational number y.
23. Let m1, . . . , mk be positive integers and m = [m1, . . . , mk]. Prove that there exist positive integers d1, . . . , dk such thatdi is a divisor of mi for i = 1, . . . , k, (di, dj) = 1 for 1 ≤ i < j ≤ n, and m = [d1, . . . , dk] =d1· · ·dk.
24. Prove that for any positive integersaandb, [a, b] = ab
(a, b).
25. Letaandb be positive integers with (a, b) =d. Prove that a
d,b d
= [a, b]
d . 26. Prove that for any positive integersa, b, c, [a, b, c] = abc(a, b, c)
(a, b)(b, c)(c, a).
27. Leta1, . . . , ak be positive integers. Prove that [a1, . . . , ak] =a1· · ·ak if and only if the integersa1, . . . , ak are pairwise relatively prime.
28. Letaandbbe positive integers andpa prime. Prove that ifpdivides [a, b] andpdivides a+b, thenpdivides (a, b).
29. Letaandb be positive integers such that a+b= 57 and
[a, b] = 680.
Findaandb.
Hint:Show thataandbare relatively prime. Thena(57−a) =ab= [a, b].
30. LetaZ={ax:x∈Z}denote the set of all multiples ofa. Prove that for any integersa1, . . . , ak,
k i=1
aiZ= [a1, . . . , ak]Z.
31. A positive integer is called square-free if it is the product of dis- tinct prime numbers. Prove that every positive integer can be written uniquely as the product of a square and a square-free integer.
32. Prove that the set of all rational numbers of the form a/b, where a, b∈Zandb is square-free, is an additive subgroup ofQ.
33. A powerful number is a positive integer n such that if a prime p dividesn, thenp2 divides n. Prove that every powerful number can be written as the product of a square and a cube. Construct examples to show that this representation of powerful numbers is not unique.
34. Prove thatmis square-free if and only if rad(m) =m.
35. Prove that rad(mn) = rad(m)rad(n) if and only if (m, n) = 1.
1.5 Euclid’s Theorem and the Sieve of Eratosthenes 33 36. Let H = {1,5,9, . . .} be the arithmetic progression of all positive integers of the form 4k+ 1. Elements ofH are calledHilbert numbers.
Show thatH is closed under multiplication, that is,x, y∈H implies xy∈H. An elementxofHwill be called a Hilbert prime ifx= 1 and xcannot be written as the product of two strictly smaller elements of H. Compute all the Hilbert primes up to 100. Prove that every element ofH can be factored into a product of Hilbert primes, but that unique factorization does not hold inH.
Hint:Find two essentially distinct factorizations of 441 into a product of Hilbert primes.
37. Forn≥1,consider the rational number hn= 1 +1
2 +1
3+· · ·+ 1 n. Prove thathn is not an integer for anyn≥2.
Hint:Let 2a be the largest power of 2 not exceedingn. LetP be the product of the odd positive integers not exceeding n. Consider the number 2a−1P hn.