MATH 3070 Theory of Numbers
Assignment 2 Solutions
Problem 1. Find the prime-power factorizations of 5500 and 1683.
Solution. We have 5500 = 22 · 53 · 11 and 1683 = 32 · 11 · 17.
Problem 2. Let k ∈ N. Prove that a positive integer n is a k-th power if and only
if every exponent appearing in the prime-power factorization of n is a multiple of k.
Then use this to find a positive integer n for which n/3 is a square and n/7 is a cube.
Solution. Suppose first that every exponent that appears in the prime-power factorization of n is a multiple of k. If the prime-power factorization of n is given
by
n = pe11 . . . pet t ,
then this assumption means that for all i we have ei = fi k for some fi . But then, we
have
n = pe11 . . . pet t
= pf11 k . . . pft t k
= (pf11 . . . pft t )k
= mk
for m = pf11 . . . pft t . We conclude that n = mk for some integer m so that it is a
perfect k-th power. Conversely, suppose that n is a perfect k-th power. Then
n = mk
for some integer m. If m has the prime-power factorization
m = pf11 . . . pft t ,
1
we then see that
n = mk
= (pf11 . . . pft t )k
= pf11 k . . . pft t k .
We conclude that the exponents appearing in the prime-power factorization of n are
f1 k, . . . , ft k. Since each of these is a multiple of k we conclude that every exponent
appearing in the prime-power factorization of n is a multiple of k as required. Now,
in order to find a positive integer n for which n/3 is a square and n/7 is a cube, we
investigate integers n divisible by 3 and 7 and figure out what exponents must appear
in the prime-power factorization of such an n. Suppose then that the prime-power
factorization of n is given by
n = 3a · 7b
for positive integers a and b. We see that
n
= 3a−1 · 7b ,
3
n
= 3a · 7b−1 .
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Using the first part, we conclude that
2 | (a − 1),
2 | b,
3|a
3 | (b − 1)
(1)
(2)
We can satisfy (1) by taking a = 3 and (2) by taking b = 4. This gives
n = 33 · 74 .
(3)
Problem 3. Use the Fundamental Theorem of Arithmetic to prove the following
statements:
(i) If p is a prime, every positive integer n ∈ N can be written uniquely in the form
n = pr m where r ∈ N0 and m ∈ N is relatively prime to p.
(ii) Every positive integer n ∈ N can be written uniquely in the form n = ab where
a, b ∈ N, a is square-free (not divisible by the square of any prime) and b is a
perfect square. Show then that b is the largest perfect square dividing n.
(iii) If m, n ∈ N are relatively prime and mn is a perfect cube then both of m and
n are perfect cubes. Give an example to show that this can fail if (m, n) 6= 1.
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Solution. For (i), let n be a positive integer. Suppose that
n = pe11 . . . pet t
is the prime-power factorization of n where p1 , . . . , pt are distinct primes and e1 , . . . , et ∈
N. If n is divisible by p, then one of the pi is equal to p and we can write
e
e
i−1
i+1
n = pei (pe11 . . . pi−1
pi+1
. . . pet t ) = pr m,
e
e
i−1
i+1
where r = ei ∈ N and m = pe11 . . . pi−1
pi+1
. . . pet t is relatively prime to p. On the
other hand, if n is relatively prime to p, we can write
n = pr m
for r = 0 and m = n which is relatively prime to p. In any case, there exist integers
r ∈ N0 and m ∈ N such that m is relatively prime to p and n = pr m. For (ii),
assume once more that
n = pe11 . . . pet t
is the prime-power factorization of n where p1 , . . . , pt are distinct primes and e1 , . . . , et ∈
N. Relabeling the primes if necessary, we may assume that e1 , . . . , e` are even while
e`+1 , . . . , et are odd. Writing
(
2fj
for 1 ≤ j ≤ `;
ej =
2fj + 1 for ` + 1 ≤ j ≤ t,
we find that n = ab for
a = p`+1 . . . pt ,
2
b = pf11 . . . pft t .
Since a is a perfect square and b is square-free, we are reduced to proving that b is
the largest square dividing n. But this is clear since if c2 | n, then we can write
c2 = p12g1 . . . pt2gt
for nonnegative integers g1 , . . . , gt such that 2gj ≤ ej for each j. We conclude that
gj ≤ fj for all j so that c2 | b. We must then have c2 ≤ b as required. For
(iii), suppose that m, n ∈ N are such that (m, n) = 1. We have distinct primes
p1 , . . . , pt , q1 , . . . , qs and positive integers e1 , . . . , et , f1 , . . . , fs such that
m = pe11 . . . pet t ;
n = q1f1 . . . qsfs .
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We conclude that the prime-power factorization of mn is given by
mn = pe11 . . . pet t q1f1 . . . qsfs .
But then, if mn is a cube, all of the exponents ei , fj for 1 ≤ i ≤ t and 1 ≤ j ≤ s
are divisible by 3. Since all of the ei are divisible by 3 we conclude that m is a cube
while since all of the fj are divisible by 3 we conclude that n is a cube. The reason
that this requires (m, n) = 1 is that we need to know that the pi are distinct from the
qj . An example where this does not hold is given by taking m = 2 and n = 4 = 22 .
Neither m nor n is a cube, but mn = 2 · 22 = 23 is a cube.
Problem 4. In this problem we adapt Euclid’s proof of the infinitude of primes
(Theorem 5 from the course notes) in order to prove that there are infinitely many
primes of the form 6n + 5.
(i) Prove that every positive integer of the form 6n + 5 is divisible by a prime of
the same form. (Hint: First show that any prime divisor of an integer of the
form 6n + 5 must either be of the same form or of the form 6n + 1. Then argue
that it is impossible for every prime divisor to be of the form 6n + 1).
(ii) Using Euclid’s proof of the infinitude of primes as a guide, prove that there
are infinitely many primes of the form 6n + 5. (Hint: Assume there are only
finitely many such primes, say p1 , . . . , pn and then consider the integer N =
6p1 . . . pn − 1 = 6(p1 . . . pn − 1) + 5).
Remark 1. This is a special case of a deep theorem due to Dirichlet that says that
whenever a and b are relatively prime, there are infinitely many primes of the form
an + b.
Solution. We start with the proof of (i). First of all, every integer is of one of the
forms 6n, 6n + 1, 6n + 2, 6n + 3, 6n + 4, 6n + 5 since 0, 1, 2, 3, 4, 5 are the distinct
residue classes modulo 6. Now, since every number of the form 6n is a multiple of 6,
every number of one of the forms 6n + 2, 6n + 4 is even, while every number of the
form 6n + 3 is divisible by 3 we see that the primes distinct from 2 and 3 must be of
the form 6n + 1 or 6n + 5. Since a number of the form 6n + 5 is not divisible by 2
or 3, each of its prime divisors must be of one of the forms 6n+1, 6n+5. Finally, if
a number of the form 6n + 5 only had prime divisors of the form 6n + 1, we would
have
6n + 5 = (6n1 + 1)(6n2 + 1) . . . (6nj + 1)
for some integers n1 , . . . , nj . If we multiply out the right hand side we obtain a
number of the form 6` + 1 which is a contradiction. We conclude that every integer
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of the form 6n + 5 is divisible by a prime of the same form. We now proceed to the
proof of (ii). Suppose now, towards a contradiction, that there are only finitely many
primes p1 , . . . , pn of the form 6n + 5. Define
N = 6p1 . . . pn − 1 = 6(p1 . . . pn − 1) + 5.
Since N is of the form 6n + 5, we conclude from (i) that it must be divisible by
a prime of the same form. Since p1 , . . . , pn are the only primes of this form, we
conclude that N is divisible by one of the pi . But this is impossible since pi would
then divide N − 6p1 . . . pn = −1. This contradiction proves our claim.
Problem 5. Let a, b ∈ N. In Assignment 1, the least common multiple of a and b
was defined and a relationship between the least common multiple and greatest common divisor was proved. Here we provide another proof of this relationship using
prime-power factorizations. Recall that the least common multiple of a and b, denoted [a, b], is defined to be the unique m ∈ N satisfying the pair of conditions:
(i) a | m and b | m;
(ii) For any positive integer n such that a | n and b | n we have m ≤ n.
Prove that if a and b have prime-power factorizations given by
a = pe11 . . . pekk ;
b = pf11 . . . pfkk .
for distinct primes p1 , . . . , pk and nonnegative integers e1 , . . . , ek , f1 , . . . , fk , then
max{e1 ,f1 }
[a, b] = p1
max{ek ,fk }
. . . pk
.
Use this to provide a second proof that
(a, b)[a, b] = ab,
where (a, b) denotes the greatest common divisor of a and b.
Solution. Consider the prime-power factorization of a common multiple n of a and
b. It is clear that, for each i, the exponent of pi appearing in the prime-power factorization of n must be at least ei and fi . We can therefore write the prime-power
factorization of n as
n = pg11 . . . pgkk q1h1 . . . qshs
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for distinct primes q1 , . . . , qs , nonnegative integers h1 , . . . , hs and integers g1 , . . . , gk
satisfying ei , fi ≤ gi for all i. It follows that the least common multiple is obtained
by taking the smallest possible values for the gi and hj . We therefore take gi =
max{ei , fi } for each i and hj = 0 for all j. This yields
max{e1 ,f1 }
[a, b] = p1
max{ek ,fk }
. . . pk
as required. In particular, since we know that the greatest common divisor of a and
b is given by
min{e1 ,f1 }
min{e ,f }
(a, b) = p1
. . . pk k k ,
we obtain
min{e1 ,f1 }
min{e ,f }
max{e1 ,f1 }
max{ek ,fk }
. . . pk
. . . pk k k
p1
(a, b)[a, b] = p1
min{e1 ,f1 }+max{e1 ,f1 }
= p1
min{ek ,fk }+max{ek ,fk }
. . . pk
= pe11 +f1 . . . pkek +fk
= (pe11 . . . pekk )(pf11 . . . pfkk )
= ab.
Problem 6. Find all solutions in positive integers to the linear diophantine equation 3718x − 110y = 66.
Solution. We first apply the Euclidean algorithm to determine the greatest common
divisor of 3718 and −110. This will allow us to determine if the linear diophantine
equation in question has any solutions, and determine a particular solution provided
one exists. We compute
3718 = (−110)(−33) + 88
−110 = (88)(−2) + 66
88 = (66)(1) + 22
66 = (22)(3)
We conclude that (3718, −110) = 22 and since 22 | 66 we obtain infinitely many
solutions to the linear diophantine equation
3718x − 110y = 66.
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(4)
We run the Euclidean algorithm backwards to find a particular solution:
22 = 88 − 66
= 88 − (−110 + 2(88))
= (−1)(88) − (−110)
= −(−110) − (3718 + (33)(−110))
= (−34)(−110) − 3718.
Multiplying by 3 gives
3718(−3) − 110(−102) = 66.
We therefore obtain the particular solution x = −3 and y = −102. We conclude that
all integer solutions to (4) are parametrized by
110
t = −3 − 5t
22
3718
y = −102 −
t = −102 − 169t
22
x = −3 −
for t ∈ Z. Finally, to find all positive integer solutions, we note that
x ≥ 0 and y ≥ 0 ⇐⇒ −3 − 5t ≥ 0 and − 102 − 169t ≥ 0
⇐⇒ 5t ≤ −3 and 169t ≤ −102
3
102
⇐⇒ t ≤ − and t ≤ −
5
169
⇐⇒ t ≤ −1,
since t is an integer. Letting u = −t for notational purposes, we find that all solutions
to (4) in positive integers are given by
x = −3 + 5u
y = −102 + 169u
where u ∈ N.
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