MATH 118 HW 1
LINDSAY CROSS, KELLY DOUGAN, KEVIN LIU
1.2
Problem 2. Find the greatest common divisor g of the numbers 1819
and 3574, and then find integers x and y to satisfy 1819x + 3587y = g.
Use the Euclidean Algorithm to find gcd(3587, 1819):
3587 = 1819(1) + 1768
1819 = 1768(1) + 51
1768 = 51(34) + 34
51 = 34(1) + 17
34 = 17(2) + 0,
so g = 17. Rearranging these equations gives:
17 = 51 − 34(1)
34 = 1768 − 51(34)
51 = 1819 − 1768(1)
1768 = 3587 − 1819(1)
Substituting to solve for x and y we get:
17 = 51 − (1768 − 34(51))
= 35(51) − 1(1768)
= 35(1819 − 1(1768)) − 1(1768)
= 35(1819) − 36(1768)
= 35(1819) − 36(3587 − 1819)
= 71(1819) − 36(3587)
Therefore (x, y) = (71, −36) is a solution.
1
2
LINDSAY CROSS, KELLY DOUGAN, KEVIN LIU
Problem 29. Let g and l be given positive integers. Prove that integers x and y exist satisfying (x, y) = g and [x, y] = l if and only if
g | l.
“⇒” Let x, y ∈ Z such that (x, y) = g and [x, y] = l. Then g | x,
and x | l, so g | l.
“⇐” Suppose g | l. (g, l) = g since there is no greater divisor of g.
[g, l] = l since there is no lesser multiple of l. So x = g and y = l satisfy
the proposition.
a
Problem 43. Prove that a | bc if and only if (a,b)
| c.
“⇒” Suppose that a|bc. Let (a, b) = d, and write a = da0 and
b = db0 . Then (a0 , b0 ) = 1. By substitution, a|bc ⇒ da0 |db0 c ⇒ a0 |b0 c.
(The second implication here follows because there exists x ∈ Z such
that da0 x = db0 c; cancelling d we obtain a0 x = b0 c, so a0 | b0 c.) Since
(a0 , b0 ) = 1, we have a0 | b0 c ⇒ a0 |c by the fundamental theorem of
a
arithmetic, as desired, since a0 = a/(a, b). “⇐” Suppose (a,b)
|c. Then
a|c(a, b) Since (a, b)|b by definition of GCD, the above statement implies
a|cb as desired.
Problem 46. Prove that there are no positive integers a, b, n > 1 such
that (an − bn )|(an + bn )
For the sake of a contradiction, assume (an − bn )|(an + bn ) holds for
a, b, n > 1. Let d = gcd(a, b). Then we may write a = Ad, b = Bd,
where A and B are coprime. Then
(an − bn )|(an + bn ) ⇒ ((Ad)n − (Bd)n )|((Ad)n + (Bd)n )
⇒ dn (An − B n )|dn (An + B n )
⇒ (An − B n )|(An + B n )
⇒ (An − B n )|(An + B n ) + (An − B n )
⇒ (An − B n )|2An
Claim: (An − B n , An ) = 1
Proof of claim: If p|An , for p a prime, and p|(An −B n ), then p divides
their difference, so p|B n . So then p|An and p|B n , which implies p|A
and p|B, since p is a prime. This is a contradiction since (A, B) = 1.
Thus (An − B n , An ) = 1 as claimed.
Therefore, since (An − B n )|2An and (An − B n , An ) = 1, we have
(An − B n )|2 by the Fundamental Theorem of Arithmetic. But since
a 6= b (since otherwise an − bn = 0), we have A 6= B, and A, B ≥ 1, we
have |An − B n | is at least (2)2 − (1)2 = 3. So (An − B n ) - 2. This is
MATH 118 HW 1
3
a contradiction, so there are no positive integers a, b, n > 1 such that
(an − bn )|(an + bn ).
1.3
Problem 5. Prove that an integer is divisible by 11 if and only if the
difference between the sum of the digits in the odd places and the sum
of the digits in the even places is divisible by 11.
Let n ∈ Z. Let nk , nk−1 , . . . , n1 , n0 be the digits of n, so that
n = nk 10k + nk−1 10k−1 + · · · + n1 10 + n0 .
Note that 10 ≡ −1 (mod 11). Thus
n ≡ nk (−1)k + nk−1 (−1)k−1 + · · · − n1 + n0
(mod 11)
= (n0 + n2 + n4 + · · · ) − (n1 + n3 + n5 + · · · )
(mod 11).
We have 11 | n if and only if n ≡ 0 (mod 11), i.e. if and only if
n0 + n2 + n4 + · · · ≡ n1 + n3 + n5 + · · ·
(mod 11).
In other words, 11 | n if and only 11 divides the difference of the sums
of the even and odd place digits of n.
Problem 11. If x and y are odd, prove that x2 +y 2 cannot be a perfect
square.
If x is odd then x ≡ 1 (mod 4) or x ≡ 3 (mod 4). Hence x2 ≡ 1
(mod 4) or x2 ≡ 9 ≡ 1 (mod 4), i.e. in both cases x2 ≡ 1 (mod 4).
Similarly y 2 ≡ 1 (mod 4) and hence x2 + y 2 ≡ 1 + 1 ≡ 2 (mod 4).
Now suppose that x2 + y 2 = n2 for some n ∈ Z. Then in particular
we have n2 ≡ 2 (mod 4) by our calculation above. However, we can
check all possible cases for n (mod 4) and see that this is impossible:
02 ≡ 0, 12 ≡ 1, 22 ≡ 4 ≡ 0, and 32 ≡ 9 ≡ 1 (mod 4). This is a
contradiction, so we cannot have x2 + y 2 = n2 .
Problem 23. Given integers a, b, c, d, m, n, u and v satisfying ad−bc =
±1, u = am + bn, v = cm + dn, prove that (m, n) = (u, v).
Let g = (m, n). Then g|m and g|n and hence g | (am + bn) = u
and g | (cm + dn) = v. Therefore g = (m, n)|(u, v). To get the reverse
divisibility, we would like to express a and b as a linear combination
of u and v and then apply the same logic. To do this, we consider the
original equations as a matrix equation:
u
a b
m
=
∗
v
c d
n
4
LINDSAY CROSS, KELLY DOUGAN, KEVIN LIU
−1 a b
u
m
⇒
∗
=
c d
v
n
1
d −b
u
m
∗
=
⇒
v
n
ad − bc −c a
Since ad − bc = ±1, we obtain:
m = ±(du − bv)
n = ±(−cu + av).
The argument above then implies that (u, v)|(m, n). Since (m, n)|(u, v)
and (u, v)|(m, n), we have (m, n) = (u, v).
Problem 26. Prove that there are infinitely many primes of the form
4n + 3; of the form 6n + 5.
Suppose there exist only finitely many primes of the form 4n + 3, say
(4n1 + 3), (4n2 + 3), . . . , (4nk + 3). Define
N = 4(4n1 + 3)(4n2 + 3) · · · (4nk + 3) − 1.
Since a prime number is by definition greater than 1, and all the
above primes divide N + 1, none of them divide N . Furthermore, N
is clearly odd and hence not divisible by 2. Therefore, all the primes
that divide N are of the form 4n + 1, (i.e. they are all congruent to 1
(mod 4)). Thus, N is a product of 1’s (mod 4) and so N is congruent
to 1 (mod 4), and cannot be congruent to 3 (mod 4).
But since N is one less than a multiple of 4, N ≡ −1 ≡ 3 (mod 4).
This is a contradiction.
Therefore, there must be infinitely many primes of the form 4n + 3.
For 6n + 5, the exact same argument as above works by considering
N = 6p1 p2 · · · pk − 1
where p1 , . . . , pk is the complete list of primes congruent to 5 mod
6. Note that N is not divisible by 2, 3, or any of the pi . Since any
prime not equal to 2 or 3 is congruent to 1 or 5 mod 6 (check that
you understand why!), this implies that N is only divisible by primes
congruent to 1 mod 6. But then N would be congruent to 1 mod
6, which is a contradiction since it is patently congruent to 5 mod 6.
Therefore the number of primes congruent to 5 mod 6 must be infinite.
MATH 118 HW 1
Problem √
47. Prove that 2 +
the ring Z[ −6].
5
√
√
−6 and 2 − −6 are irreducibles in
√
For a, b, c, d ∈ Z, we have N (a + b −6) = a2 + 6b2 . Therefore
√
N (2 + −6) = 10.
√
If 2 + −6 could be factored into the form
√
√
(a + b −6)(c + d −6)
Then,
√
√
√
N (2 + −6) = N (a + b −6)N (c + d −6)
and thus
10 = (a2 + 6b2 )(c2 + 6d2 )
The only possible ways to factor 10 are (1 ∗ 10) and (2 ∗ 5) and the
reverses.
2
2
For the first factorization (1 ∗ 10), It is
√ easy to see that a + 6b = 1
implies that a = ±1 and b = 0, so a + b −6 = ±1 is a unit.
For the second factorization (2 ∗ 5) we obtain that 2 = a2 + 6b2 and
5 = c2 + 6d2 . However, there are no integer solutions to either of these
two equations. For example, if b = 0 then a2 = 2 and hence a 6∈ Z.
If b 6= 0, then a2 + 6b2 > 2. (This examination also holds true for
5 = c2 + 6d2 .)
√
Therefore, in any factorization
of 2 + −6, one of the factors must
√
be a unit and thus 2 + √ −6 is irreducible by definition. The same
argument applies to 2 − −6.