Math 4400/6400 – Homework #3 solutions
MATH 4400 problems.
1. Prove that if a, b, c, d ∈ Z+ and ad − bc = 1, then gcd(a + b, c + d) = 1.
Proof. The equation 1 = ad − bc = (a + b)d + (c + d)(−b) shows that 1 is a linear
combination of a + b and c + d. Hence, gcd(a + b, c + d) | 1. Since the only (positive)
divisor of 1 is itself, we get gcd(a + b, c + d) = 1.
2. In class, we saw that every square mod 3 was congruent to 0 or 1. In particular, there
are no squares congruent to 2 modulo 3. Suppose now that m is a positive integer
with the property that there are squares congruent to all of 0, 1, 2, . . . , m − 1 modulo
m. Show that m = 1 or m = 2.
Proof. We show that if m > 2, then at least one of the residue classes 0, 1, 2, . . . , m − 1
mod m is not a square.
Suppose otherwise. Then 02 , 12 , 22 , . . . , (m − 1)2 leave distinct remainders when
divided by m. But 12 and (m − 1)2 leave the same remainder (namely, 1). (This proof
needs the condition m > 2 to guarantee that the integers 1 and m − 1 are distinct
modulo m.)
3. Show that there is no perfect square that ends with the decimal digits ‘79’.
Proof. Saying that the positive integer n ends with the digits 79 is equivalent to saying
that n ≡ 79 (mod 100). So we want to show that there is no square m2 with m2 ≡ 79
(mod 100).
We could solve this problem by computing all of the squares 02 , 12 , . . . , 992 , but here
is an easier way: Since 4 is a divisor of 100, we get that if m2 ≡ 79 (mod 100), then
m2 ≡ 79 (mod 4). (Unwind the definitions if you don’t see this.) Now 79 ≡ 3 (mod 4).
Thus, m2 ≡ 3 (mod 4). But modulo 4, we have 02 ≡ 0, 12 ≡ 1, 22 ≡ 0, and 32 ≡ 1, so
the only squares are 0 and 1. We conclude that no such m exists.
4. Show that 7 divides 32n+1 + 2n+2 for all natural numbers n.
Proof. We have 32n+1 = 3 · 32n = 3 · 9n ≡ 3 · 2n (mod 7). Also, 2n+2 = 4 · 2n . We
deduce from these facts that
32n+1 + 2n+2 ≡ 3 · 2n + 4 · 2n ≡ 7 · 2n ≡ 0
(mod 7).
Hence, 7 | 32n+1 + 2n+2 .
5. (a) Show that for any real numbers x and y, we have bx + yc ≥ bxc + byc.
(b) Using the result of part (a) and the formula
∞
X
vp (n!) =
bn/pk c
(1)
k=1
(which we proved in class), show that m!n! | (m + n)! for every pair of positive
integers m and n.
NOTE: The sum in (1) should start at k = 1 and not at k = 0. This was a typo in
the HW and in the original version of the solutions!
Proof. We start with (a). Since bxc ≤ x and byc ≤ y, we have bxc + byc ≤ x + y. By
definition, bx + yc is the largest positive integer not exceeding x + y. Since bxc + byc
is one integer with this property, it follows that bxc + byc ≤ bx + yc, as desired.
Now we go on to (b). To show that m!n! | (m + n)!, we will show that vp (m!n!) ≤
vp ((m + n)!) for all primes p. Now
vp ((m + n)!) =
∞ X
m+n
pk
k=1
.
(2)
On the other hand,
vp (m!n!) = vp (m!) + vp (n!) =
=
∞ X
m
pk
k=1
∞ X
k=1
+
∞ X
n
pk
m
n
+ k
.
k
p
p
k=0
(3)
To show that (2) is at least as large as (3), it is enough to prove that every term in (2)
is at least as large as the corresponding term in (3), i.e., that for each k ≥ 1, we have
m
n
m+n
≥ k + k .
k
p
p
p
But this follows from part (a) with x = m/pk and y = n/pk .
6. Find all solutions to each of the following linear congruences:
(a) 7x ≡ 3 (mod 161),
(b) 32x ≡ 17 (mod 105),
(c) 20x ≡ 4 (mod 34).
Solution. Since gcd(7, 161) = 7 - 3, there are no solutions to (a).
Now we turn to (b). We easily compute that gcd(32, 105) = 1, and so there will be a
solution. To find the solution, we use the Euclidean algorithm to find X, Y ∈ Z with
2
105X + 32Y = 1. We have
105 = 32 · 3 + 9
32 = 9 · 3 + 5
9=5·1+4
5=4·1+1
4 = 1 · 4 + 0.
Thus,
1 = 5 + 4(−1)
= 5(1) + (9 − 5(1))(−1)
= 9(−1) + 5(2)
= 9(−1) + (32 − 9(3))(2)
= 32(2) + 9(−7)
= 32(2) + (105 − 32(3))(−7)
= 105(−7) + 32(23).
Thus. 32 · 23 ≡ 1 (mod 105). It follows that x = 23 · 17 = 391 is a solution to the
congruence in (b). From the theorem discussed in class, all solutions are of the form
x = 391 + 105k,
where k ∈ Z.
(So taking k = −3, we get the smallest nonnegative solution, namely x = 76.)
Finally we turn to (c). We have gcd(20, 34) = 2, and since 2 | 34, there will be
solutions. Moreover, an integer x is a solution to the congruence in (c) if and only if x
is a solution to the congruence
10x ≡ 2
(mod 17).
(4)
We now use the Euclidean algorithm to find integers X and Y with 10X + 17Y = 1.
We get
17 = 10 · 1 + 7
10 = 7 · 1 + 3
7=3·2+1
3 = 1 · 3 + 0.
Thus,
1 = 7 + 3(−2)
= 7 + (10 − 7(1))(−2)
= 10(−2) + 7(3)
= 10(−2) + (17 − 10 · 1)(3)
= 17(3) + 10(−5).
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Thus, 10(−5) ≡ 1 (mod 17). Hence, x = −5(2) = −10 is a solution to (4) (equivalently, to the original congruence in c) and that all solutions are given by
x = −10 + 17k,
where k ∈ Z.
[Note that while the solution x is unique mod 17, there are two solutions mod 34,
namely −10 and 7 mod 34.]
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MATH 6400 problems. Students in MATH 6400 must turn in (at least) two of these
problems, in addition to the problems assigned for MATH 4400.
G1. Consider the sequence of primes 2, 3, 7, 43, 139, . . . defined by the following procedure.
Let q1 = 2, and assuming qj has been defined for 1 ≤ j ≤ k, let qk+1 be the largest
prime divisor of 1 + q1 q2 · · · qk . (We discussed this sequence briefly in class.) Prove
that the prime 5 does not appear anywhere in the sequence {qi }∞
i=1 .
Hint: If 5 = qk for some k, show that q1 · · · qk−1 + 1 must be a power of 5. Reach a
contradiction by considering the resulting equation modulo 4.
Proof. If qk = 5, then 5 is the largest prime factor of q1 · · · qk−1 + 1. So it must be that
q1 · · · qk−1 + 1 = 2a 3b 5c
for some integers a, b ≥ 0 and c ≥ 1. But 2 and 3 both divide q1 · · · qk−1 (since k ≥ 3)
and so cannot divide q1 · · · qk−1 + 1. So it must be that a = b = 0. Hence,
q1 · · · qk−1 + 1 = 5c .
Since 5 ≡ 1 (mod 4), the right-hand side of this equation is congruent to 1 (mod 4).
But q1 · · · qk−1 = 2q2 · · · qk−1 is twice an odd number, which implies that q1 · · · qk−1 ≡ 2
(mod 4) and q1 · · · qk−1 + 1 ≡ 3 (mod 4). This is a contradiction.
G2. (This exercise relates back to the formula (1).) Show that the sequence defined explicitly by ak = bn/pk c can alternatively be defined recursively as follows: a0 = n, and
ak+1 = bak /pc for each k = 0, 1, 2, 3, . . . . This explains an empirical observation we
made in class.
Proof. We begin by proving a lemma: For every positive real number x, we have
bbxc/pc = bx/pc.
(5)
Since bxc ≤ x, we have bxc/p ≤ x/p, and so bbxc/pc ≤ bx/pc. In other words, the
left-hand side of (5) is bounded by the right-hand side.
Next, we show that the right-hand side is bounded by the left. We have pbx/pc ≤ x.
Since pbx/pc is an integer, we get that pbx/pc ≤ bxc. Thus, bx/pc ≤ bxc/p. Since
bx/pc is an integer, we deduce that bx/pc ≤ bbxc/pc. In conjunction with the result
of the previous paragraph, this completes the proof of (5).
Let ak be the sequence defined recursively by a0 = n and ak+1 = bak /pc for each
k = 0, 1, 2, 3, . . . . Let us prove by induction that
ak = bn/pk c
(6)
for all k = 0, 1, 2, . . . . This holds by definition when k = 0. Supposing (6), we find
that
ak+1 = bak /pc = bbn/pk c/pc.
Taking x = n/pk in our lemma, we get that
bbn/pk c/pc = bn/pk+1 c.
Hence, (6) holds with k replaced by k + 1. This completes the induction.
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G3. (a) Show that if n ∈ Z is a multiple of 6, then there are integers a, b, c, d with
n = a3 + b 3 + c 3 + d 3 .
Hint: For each polynomial f , define ∆f (x) = f (x+1)−f (x). Start by computing
∆(x3 ), then ∆(∆(x3 )), etc.
Solution. We get that
∆(x3 ) = (x + 1)3 − x3 = 3x2 + 3x + 1,
∆(∆(x3 )) = ∆(3x2 + 3x + 1)
= 3((x + 1)2 − x2 ) + 3((x + 1) − x) + (1 − 1)
= 3(2x + 1) + 3 = 6x + 6.
But also
∆(∆(x3 )) = ∆((x + 1)3 − x3 )
= (x + 2)3 − (x + 1)3 − ((x + 1)3 − x3 )
= (x + 2)3 − (x + 1)3 − (x + 1)3 + x3
= (x + 2)3 + (−1 − x)3 + (−1 − x)3 + x3 .
Comparing our two evaluations of ∆(∆(x3 )), we find that 6x + 6 is a sum of four
integer cubes for every x ∈ Z. But as x ranges over Z, the expression 6x + 6
ranges over the multiples of 6. This finishes off (a).
(b) Deduce from part (a) that every n ∈ Z can be written in the form n = a3 + b3 +
c3 + d3 + e3 . Hint: First prove that n3 ≡ n (mod 6).
Proof. We can check by hand that n − n3 ≡ 0 (mod 6) for each n = 0, 1, 2, . . . , 5.
This implies that n − n3 ≡ 0 (mod 6) for every n ∈ Z. Since n − n3 is a multiple
of 6, we get from (a) that n − n3 = a3 + b3 + c3 + d3 for some integers a, b, c, and
d. Thus, n = a3 + b3 + c3 + d3 + n3 . So the assertion in (b) follows with e = n.
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