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MATH 324 Summer 2011
Elementary Number Theory
Solutions to Practice Problems for Final Examination
Thursday August 11, 2011
Question 1.
Find integers x and y such that314x + 159y = 1.
Solution: Using the Euclidean algorithm to find the greatest common divisor of a = 314 and b = 159, wehave
314 = 1 · 159 + 155
159 = 1 · 155 + 4
155 = 38 · 4 + 3
4 = 1 · 3 + 1
3 = 3 · 1 + 0
and the last nonzero remainder is (314, 159) = 1.
Working from the bottom up, we have
1 = 4 − 3
= 4 − (155 − 38 · 4) = 39 · 4 − 155
= 39 · (159 − 155)− 155 = 39 · 159− 40 · 155
= 39 · 159− 40(314− 159) = (−40) · 314 + 79 · 159
and therefore,1 = (−40) · 314 + 79 · 159,
and one solution is x = −40, y = 79.
Question 2.
Let n be a composite positive integer and let p be the smallest prime divisor of n. Show that if p > n1/3,then n/p is prime.
Solution: Suppose p > n1/3 and that n/p is composite, so that n = p · a · b, where 1 < a, b < n. Let q andr be any prime divisors of a and b, respectively, then q and r are also prime divisors of n, so that
q ≥ p and r ≥ p.
This implies thatp3 ≤ p · q · r ≤ p · a · b = n,
that is, p ≤ n1/3, which is a contradiction. Therefore, n/p is prime.
Question 3.
Find all positive solutions in integers to the system of linear equations
x + y + z = 31
2x + 2y + 3z = 41.
Solution: Solving the first equation, we get the two-parameter family of solutions
x = 31− t, y = t − s, z = s
for s, t ∈ Z. Since this must also be a solution of the second equation, we must have
2(31 − t) + 2(t − s) + 3s = 41,
that is, s = −21. Therefore, there are no positive solutions to the given system.
Question 4.
Prove or disprove that if a2 ≡ b2 (mod m), then a ≡ b (mod m) or a ≡ −b (mod m).
Solution: The statement is false, as can be seen by taking m = 24, a = 8 and b = 4, then
a2 ≡ 82 ≡ 64 ≡ 16 ≡ 42 ≡ b2 (mod 24),
but 8 6≡ 4 (mod 24).
Question 5.
Find all positive integers m such that 1066 ≡ 1776 (mod m).
Solution: We only need find all positive divisors of 1776− 1066 = 710, and these are
1, 2, 5, 10, 71, 142, 355, 710.
Question 6.
Show that the difference of two consecutive cubes is never divisible by 5.
Solution: For any integer n, we have
(n + 1)3 − n3 = 3n(n + 1) + 1,
and it is easily seen that n(n + 1) is congruent to 0, 1, or 2 modulo 5, so that (n + 1)3 − n3 can only becongruent to 1, 2, or 4 modulo 5.
Question 7.
Find the smallest odd integer n, with n > 3, such that 3∣∣n, 5
∣∣n + 2, and 7
∣∣n + 4.
Solution: We have to solve the simultaneous congruences
n ≡ 3 (mod 3)
n ≡ 3 (mod 5)
n ≡ 3 (mod 7)
and since 3, 5, and 7 are pairwise relatively prime, their smallest common multiple is 3 · 5 · 7 = 105, and wemay take n = 3 + 105 = 108.
Question 8.
(a) Find a positive integer n such that 32∣∣n, 42
∣∣ n + 1, and 52
∣∣n + 2.
(b) Can you find a positive integer n such that 22∣∣n, 32
∣∣ n + 1, and 42
∣∣n + 2.
Solution:
(a) We solve the following simultaneous congruences using the Chinese remainder theorem:
n ≡ 0 (mod 9)
n ≡ −1 ≡ 15 (mod 16)
n ≡ −2 ≡ 23 (mod 25).
Here
a1 = 0, a2 = 15, a3 = 23
m1 = 9, m2= 16, m3 = 25,
andM1 = 16 · 25 = 400, M2 = 9 · 25 = 225, M3 = 9 · 16 = 144.
Also, solving the congrences
M1y1 ≡ 1 (mod m1)
M2y2 ≡ 1 (mod m2)
M3y3 ≡ 1 (mod m3)
for the inverses y1, y2, and y3, we have
y1 ≡ 7 (mod 9), y2 ≡ 1 (mod 16), y3 ≡ 4 (mod 25),
and the unique solution modulo 9 · 16 · 25 is given by
n = a1M1y1 + a2M2y2 + a3M3y3 = 15 · 225 · 1 + 23 · 144 · 4 = 3375 + 13248 = 16623.
(b) There does not exist a positive integer n such that
22∣∣ n, 32
∣∣ n + 1, and 42
∣∣ n + 2,
since if such an n existed, then we would have
4∣∣n and 16
∣∣n + 2,
however, this implies that 4∣∣n + 2 − n = 2, which is a contradiction.
Question 9.
What is the last digit of 7355 ?
Solution: We have
72 ≡ 9 (mod 10)
73 ≡ 3 (mod 10)
74 ≡ 1 (mod 10)
and 355 = 88 · 4 + 3, so that
7355 ≡(74
)88· 73 ≡ 3 (mod 10),
and the last digit of 7355 is a 3.
Question 10.
What is the remainder when 314162 is divided by 165 ?
Solution: We have the following prime power decompositions:
314 = 157 · 2 and 165 = 3 · 5 · 11
and the following congruences:
22 ≡ 1 (mod 3)
24 ≡ 1 (mod 5)
28 ≡ 1 (mod 11)
so that 3∣∣ 28 − 1, 5
∣∣ 28 − 1, and 11
∣∣28 − 1, and since the integers 3, 5, and 11 are pairwise relatively prime,
then 165∣∣28 − 1 also, that is,
28 ≡ 1 (mod 165).
We also have the congruences:
157 ≡ 1 (mod 3)
1574 ≡ 1 (mod 5)
1575 ≡ 1 (mod 11)
and again, since 3∣∣ 15720 − 1, 5
∣∣ 15720 − 1, and 11
∣∣15720 − 1, and the integers 3, 5, and 11 are pairwise
relatively prime, then 165∣∣15720 − 1 also, that is,
15720 ≡ 1 (mod 165).
Therefore, 240 ≡ 1 (mod 165) and 15740 ≡ 1 (mod 165), so that 31440 ≡ 1 (mod 165), and since162 = 4 · 40 + 2, then
314162 ≡(31440
)4· 3142 ≡ 3142 ≡ 91 (mod 165).
Question 11.
Prove that an − bn is divisible by the prime n + 1 if neither a nor b is.
Solution: Note that if p is a prime and (a, p) = 1 and (b, p) = 1, then from Fermat’s theorem we have
ap−1 ≡ 1 (mod p) and bp−1 ≡ 1 (mod p),
so thatap−1 − bp−1 ≡ 0 (mod p).
Now, if n + 1 a prime and (a, n + 1) = (b, n + 1) = 1, then from the above we have
an − bn ≡ 0 (mod n + 1),
that is, n + 1∣∣an − bn.
Question 12.
Suppose that p is an odd prime.
(a) Show that1p−1 + 2p−1 + · · · + (p − 1)p−1 ≡ −1 (mod p).
(b) Show that1p + 2p + · · · + (p − 1)p ≡ 0 (mod p).
Solution:
(a) Since p is an odd prime and (k, p) = 1 for 1 ≤ k ≤ p − 1, then Fermat’s theorem implies that
kp−1 ≡ 1 (mod p)
for 1 ≤ k ≤ p − 1, and therefore
1p−1 + 2p−1 + · · · + (p − 1)p−1 ≡ 1 + 1 + · · · + 1︸ ︷︷ ︸
p−1 times
≡ p − 1 ≡ −1 (mod p).
(b) Again, from Fermat’s theorem we have
kp ≡ k (mod p)
for 1 ≤ k ≤ p − 1, and therefore
1p + 2p + · · · + (p − 1)p ≡ 1 + 2 + · · · + p − 1 ≡ p
(p − 1
2
)
≡ 0 (mod p)
sincep − 1
2∈ Z
+.
Question 13.
Show that σ(n) is odd if n is a power of 2.
Solution: If n = 2α, then
σ (2α) =∑
d | 2α
d = 1 + 2 + 22 + · · · + 2α =2α+1 − 1
2 − 1= 2α+1 − 1,
and σ (2α) = 2α+1 − 1 is odd for all integers α ≥ 0.
Question 14.
For which n is σ(n) odd?
Solution: We have seen that if n = 2α, then σ(n) is odd. Suppose now that p is an odd prime and that αis a positive integer, then
σ (pα) = 1 + p + p2 + · · · + pα =pα+1 − 1
p − 1,
and σ (pα) is odd if and only if the sum contains an odd number of terms, that is, if and only if α is an eveninteger.
Therefore, σ(n) is odd if and only if in the prime power decomposition of n, every odd prime occurs to aneven power, that is, if and only if n is perfect square or n is 2 times a perfect square.
Question 15.
Find a formula for σ2(n) =∑
d |n
d2.
Solution: If p is a prime and α is a positive integer, then
σ2 (pα) =∑
d | pα
d2 = 1 + p2 + p4 + p6 + · · · + p2α =
(p2
)α+1− 1
p2 − 1=
p2α+2 − 1
p2 − 1,
and if the prime power decomposition of n is given by
n =
r∏
k=1
pαk
k ,
where the pk’s are distinct primes, then
σ2(n) =
r∏
k=1
p2αk+2 − 1
p2k − 1
.
Question 16.
It was long thought that even perfect numbers ended alternately in 6 and 8. Show that this is not the caseby verifying that the perfect numbers corresponding to the primes
213 − 1 and 217 − 1
both end in 6.
Solution: Note that 24 ≡ 6 (mod 10) and 6k ≡ 6 (mod 10) for all k ≥ 1. Now, if k ≡ 1 (mod 4), thenk = 1 + 4m for some integer m ≥ 1, and
2k ≡ 21 · 24m ≡ 2 ·(24
)m≡ 2 · 6m ≡ 2 · 6 ≡ 2 (mod 10),
so that 2k − 1 ≡ 1 (mod 10) for all k ≥ 1, k ≡ 1 (mod 4).
Also, if k = 1 + 4m, then k − 1 = 4m and
2k−1 ≡(24
)m≡ 6m ≡ 6 (mod 10)
for all k ≥ 5, k ≡ 1 (mod 4), and therefore
2k−1(2k − 1
)≡ 6 · 1 ≡ 6 (mod 10)
for all k ≥ 5, k ≡ 1 (mod 4). The result now follows since both 13 and 17 are congruent to 1 modulo 4.
Question 17.
Show that all even perfect numbers end in 6 or 8.
Solution: We have seen that any even perfect number n has the form
n = 2k−1(2k − 1
)
where 2k − 1 is prime (and hence k is also prime).
If k ≡ 1 (mod 4), then we saw in the previous problem that the last digit of n is a 6.
Suppose now that k ≡ 3 (mod 4), then k = 3 + 4m for some integer m ≥ 0, so that
2k ≡ 23 ·(24
)m≡ 23 · 6m ≡ 8 · 6 ≡ 8 (mod 10)
and 2k − 1 ≡ 7 (mod 10).
Also,2k−1 ≡ 22
(24
)m≡ 4 · 6 ≡ 4 (mod 10),
so thatn = 2k−1
(2k − 1
)≡ 4 · 7 ≡ 8 (mod 10)
for all k ≡ 3 (mod 4).
Question 18.
If n is an even perfect number, with n > 6, show that the sum of its digits is congruent to 1 modulo 9.
Solution: Since 10k ≡ 1 (mod 9) for all k ≥ 0, then any positive integer n is congruent modulo 9 to thesum of it digits, so we only need to show that if
n = 2k−1(2k − 1
), k ≥ 3
is an even perfect number, then n is congruent to 1 modulo 9.
Let n = 2k−1(2k − 1
)be an even perfect number, n > 6, then 2k − 1 and k are both odd primes. Note that
22 ≡ 4 (mod 9), 23 ≡ 8 ≡ −1 (mod 9), 26 ≡ 1 (mod 9).
Also, since k is prime, then k ≡ ±1 (mod 6).
If k ≡ 1 (mod 6), then k = 1 + 6m for some m ≥ 1, so that
2k ≡ 2 ·(26
)m≡ 2 (mod 9),
and2k − 1 ≡ 2 − 1 ≡ 1 (mod 9).
Also,2k−1 ≡
(26
)m≡ 1 (mod 9),
so thatn = 2k−1
(2k − 1
)≡ 1 (mod 9)
if k ≡ 1 (mod 6).
If k ≡ −1 (mod 6), then k = 5 + 6m for some m ≥ 0, so that
2k ≡ 25 ·(26
)m≡ 5 (mod 9),
and2k − 1 ≡ 5 − 1 ≡ 4 (mod 9).
Also,2k−1 ≡ 24
(26
)m≡ 7 (mod 9),
so thatn = 2k−1
(2k − 1
)≡ 4 · 7 ≡ 28 ≡ 1(mod 9)
if k ≡ −1 (mod 6).
Question 19.
Show that if n is odd, then φ(4n) = 2φ(n).
Solution: If n is odd, then (4, n) = 1, so that
φ(4n) = φ(4) · φ(n) = 2φ(n).
Question 20.
Perfect numbers satisfy σ(n) = 2n. Which n satisfy φ(n) = 2n ?
Solution: If n is a positive integer and φ(n) = 2n, then since
φ(n) ≤ n − 1,
this implies that 2n ≤ n − 1, that is, n ≤ −1, which is a contradiction. Therefore, there are no positiveintegers n for which φ(n) = 2n.
Question 21.
Note that
1 + 2 =3
2· φ(3)
1 + 3 =4
2· φ(4)
1 + 2 + 3 + 4 =5
2· φ(5)
1 + 5 =6
2· φ(6)
1 + 2 + 3 + 4 + 5 + 6 =7
2· φ(7)
1 + 3 + 5 + 7 =8
2· φ(8)
Guess a theorem!
Solution: The theorem was one that we proved in class, namely,
n∑
k=1(k,n)=1
k =n
2· φ(n).
Question 22.
Find all solutions of φ(n) = 4, and prove that there are no more.
Solution: If the prime power decomposition of the positive integer n is given by
n = 2α · pα1
1 · pα2
2 · · · pαk
k ,
where the pi’s are distinct odd primes, then
φ(n) = 2α−1 · pα1−11 · pα2−1
2 · · · pαk−1k (p1 − 1)(p2 − 1) · · · (pk − 1),
and it is clear that if φ(n) = 4 then n can have no more than 2 odd prime factors and these must occur tothe first power. It is also clear that the exponent of the power of 2 must be less than or equal to 3.
Thus, if φ(n) = 4, then n must have one of the forms
n = 2α for α ≤ 3,
n = 2α · p for α ≤ 2, p an odd prime
n = 2α · p · q for α ≤ 1, p, q odd primes
In the first case, if n = 2α, then φ(n) = 2α−1 = 4 if and only if α − 1 = 2, that is, α = 3, so n = 8 is onesolution to φ(n) = 4.
In the second case, if n = 2α · p, where p is an odd prime, then φ(n) = 2α−1(p− 1), and φ(n) = 4 if and onlyif either α − 1 = 1 and p − 1 = 2, or α − 1 = 0 and p − 1 = 4, or α = 0 and p − 1 = 4. Thus, n is a solutionto φ(n) = 4 if and only if n = 4 · 3 = 12 or n = 2 · 5 = 10, or n = 1 · 5 = 5.
In the third case, if n = 2α · p · q, where p and q are distinct odd primes, then φ(n) = 2α−1(p − 1)(q − 1),and since both p − 1 and q − 1 are even, then we must have α = 0. Also, since p and q are distinct oddprimes, then (p− 1)(q − 1) = 4 implies that one of p− 1 or q − 1 equals 1 and the other equals 4, which is acontradiction. Thus, in this case φ(n) 6= 4, and there are no solutions.
Therefore, the only solutions to φ(n) = 4 are 5, 8, 10, and 12.
Question 23.
Show that if (m, n) = 2, then φ(m · n) = 2 · φ(m) · φ(n).
Solution: Let m = 2r ·M and n = 2s ·N where M and N are odd and one of r or s is 1, then (M, N) = 1so that
φ(m · n) = 2r+s−1 · φ(M) · φ(N),
andφ(m) · φ(n) = 2r−1 · φ(M) · 2s−1 · φ(N) = 2r+s−2 · φ(M) · φ(N),
and therefore2 · φ(m) · φ(n) = 2r+s−1 · φ(M) · φ(N) = φ(m · n).
Question 24.
Show that if n − 1 and n + 1 are both primes, with n > 4, then φ(n) ≤n
3.
Solution: Since n > 4, then n − 1 > 3, and since one of the consecutive integers n − 1, n, and n + 1 isdivisible by 3 and n − 1 and n + 1 are primes greater than 3, then 3
∣∣n. Also, n must be even, since n − 1
and n + 1 are odd primes, so that 2∣∣n.
Since (2, 3) = 1, and n is a common multiple of 2 and 3, then 6 = 2 · 3 = [2, 3] also divides n, and we canwrite
n = 2a · 3b · N
where a ≥ 1 and b ≥ 1, and (2, N) = (3, N) = 1.
Therefore,
φ(n) = 2a · 3b−1 · φ(N) ≤ 2a · 3b−1 · N =n
3.
Question 25.
One of the primitive roots of 19 is 2. Find all of the others.
Solution: Since φ(19) = 18, and φ(18) = 6, then there are 6 incongruent primitive roots of 19, and theyare
2u, 1 ≤ u ≤ 18, with (u, 18) = 1,
that is, the incongruent primitive roots of 19 are 2, 25, 27, 211, 213, 217.
Question 26.
Suppose that p is a prime and a has order 4 modulo p. What is the least positive residue of (a + 1)4 modulop ?
Solution: Since a4 ≡ 1 (mod p), then p ≥ 5, and since
p∣∣a4 − 1 = (a2 − 1)(a2 + 1),
and p is prime with a2 6≡ 1 (mod p), then p∣∣a2 + 1, that is, a2 ≡ −1 (mod p).
Therefore,
(a + 1)4 ≡ a4 + 4a3 + 6a2 + 4a + 1 ≡ 1 − 4a − 6 + 4a + 1 ≡ 2 − 6 ≡ −4 ≡ p − 4 (mod p),
and (a + 1)4 ≡ (p − 4) (mod p), so the least positive residue of (a + 1)4 modulo p is p − 4.
Question 27.
If r is a primitive root of the prime p, show that two consecutive powers of r have consecutive least positiveresidues modulo p, that is, show that there exists an integer k ≥ 1 such that
rk+1 ≡ rk + 1 (mod p).
Solution: The set{1, r, r2, . . . , rp−1}
forms reduced residue system modulo p. Therefore, since the φ(p) = p − 1 integers
{r2 − r, r3 − r2, r4 − r3, . . . , rp − rp−1}
are all incongruent modulo p, they also form a reduced residue system modulo p, and there must be aninteger k, with 1 ≤ k ≤ p − 1, such that
rk+1 − rk ≡ 1 (mod p).
Question 28.
Show that if p is an odd prime and a is a quadratic residue modulo p, and a · b ≡ 1 (mod p), then b is alsoa quadratic residue modulo p.
Solution: There exists an integer x such that x2 ≡ a (mod p), so that
(x · b)2 ≡ x2 · b2 ≡ a · b2 ≡ (a · b) · b ≡ 1 · b ≡ b (mod p),
and b is also a quadratic residue modulo p.