Download 3 Solution of Homework

Math 4161
Dr. Franz Rothe
December 3, 2013
Name:
13FALL\4161_fall13h3.tex
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Solution of Homework
10 Problem 3.1. Show that for any primes p, q ≥ 5, the number 24 is a divisor
of p2 − q 2 .
Answer. Since p and q are both odd and not divisible by 3, we know p2 ≡ q 2 ≡ 1
(mod 3). Hence p2 − q 2 is divisible by 3. For every odd square p2 ≡ q 2 ≡ 1 (mod 8).
Hence p2 − q 2 is divisible by 8. From both divisors together, we see that p2 − q 2 is
divisible by 24.
10 Problem 3.2. If a prime divides both n2 + 3 and (n + 1)2 + 3, what is the
prime. (p.59 #22)
Answer. Assume a prime p divides both n2 + 3 and (n + 1)2 + 3. Hence prime p divides
also
p | (n + 1)2 + 3 − n2 − 3 = 2n + 1
p | 4(n2 + 3) − (2n + 1)2 = 11 − 4n
p | 11 − 4n + 2(2n + 1) = 13
Hence p = 13 is the only possibility.
10 Problem 3.3. Show that 5777 6= p + 2a2 for any integer a and prime p. (p.57
#5)
Answer. I check at first the case that a is not divisible by 3. In this case r := 5777 − 2a2
is divisible by 3 and not a prime. We calculate modulo 3: If a is not divisible by 3, then
a2 ≡ 1 (mod 3) and
5777 − 2a2 ≡ 2 − 2 · 0 = 0 (mod 3)
hence r is divisible by 3. Too, r 6= 3 since (5777 − 3)/2 is not a perfect square. Hence r
is not a prime.
To check the remaining cases, I may now assume that a = 3b and put r = 5777−18b2
with b = 0, 1, 2, . . . 17. None of these numbers turn out to be prime.
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10 Problem 3.4. Show the following: ”If p and p2 + 8 are both primes, then
p3 + 4 is a prime.” For which values of p are the assumptions true? (p.58#20)
Answer. For p = 2, the sum p2 + 8 = 12 is composite. For any odd prime p > 3, we
calculate modulo 3 to obtain p2 + 8 ≡ 1 + 8 ≡ 0 (mod 3), and hence p2 + 8 is divisible
by three and composite. We see the assumption that p and p2 + 8 are both prime holds
only for p = 3. In that case, one checks that p3 + 4 = 31 is a prime.
10 Problem 3.5. Show that every p-th term of the infinite arithmetic sequence
a, a + b, a + 2b, . . .
is divisible by the prime p, provided that the prime p - b does not divide the step b. Use
pr + bs = 1 from the extended Euclidean algorithm. (p.58 #18)
Answer.
a + kb = a(pr + bs) + kb = ar · p + (as + k)b
For every p-th term a+kb in the sequence, it happens that k ≡ −as (mod p), and hence
the term a + kb is divisible by p.
10 Problem 3.6. Prime factor the numbers 2n +n2 for n = 3, 9, 15, 21, 27, 33, 39.
Give the result in a readable table.
Answer.
n
n2 + 2n factored
3
9
15
21
27
33
39
17
593
32 993
2 097 593
73 · 521 · 3529
8 589 935 681
17 · 43 · 752 059 939
One gets two cases with a composite number, the remaining ones are primes.
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10 Problem 3.7. Prime factor the number 50 ! and explain the method, credited
to Legendre.
Answer. Legendre notes the exponent r of the prime p in the factorial N ! to be
r=b
N
N
N
c + b 2c + b 3c + ...
p
p
p
Clearly that is always a finite sum. For example, 2r k50! for
r=b
50
50
50
50
50
c + b c + b c + b c + b c = 25 + 12 + 6 + 3 + 1 = 47
2
4
8
16
32
For primes p = 3, 5, . . . 23 one gets the table below. The primes 29 ≤ p ≤ 47 appear
only once in 50!. Hence
50! = 247 · 322 · 512 · 78 · 114 · 133 · 172 · 192 · 232 · 29 · 31 · 37 · 41 · 43 · 47
prime
2
exponent
calculation of sum
50
47 b 50
c + b 50
c + b 50
c + b 50
c + b 32
c
2
4
8
16
3
5
7
11
16 + 5 + 1 = 22
10 + 2 = 12
7+1=8
4
13
17
19
23
3
2
2
2
b 50
c + b 50
c + b 50
c
3
9
27
50
c
+
b
c
b 50
5
25
50
b 50
c
+
b
c
7
49
50
b 11 c
c
b 50
13
b 50
c
17
50
b 19 c
b 50
c
23
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10 Problem 3.8. Find two adjacent primes 100 < a < b such that all numbers
a + 1, a + 2, . . . , b − 1 between them are composite, and the gap between has a length at
least three times the average gap. This can be checked by confirming:
(3.1)
a
b
−
>3
ln b ln a
Answer. A simple way to find long gaps between primes is the remark that the numbers
n! + 2, n! + 3, . . . , n! + n
are all composite. A slight improvement of this idea is to replace n! by the product of
all primes less or equal n. Let
Y
n] = {
q : q ≤ n is a prime }
All the numbers
n] + 2, n] + 3, . . . , n] + n
are composite. Similarly, all the numbers
n] − 2, n] − 3, . . . , n] − n
are composite. Finally, if n is odd and n] + 1 as well as n] − 1 both turn out to be
composite, we get the even longer string
n] − n − 1, n] − n, . . . , n] + n + 1
of all composite numbers.
One checks that both numbers 2 · 3 · 5 · 7 · 11 · 13 · 17 + 1 and 2 · 3 · 5 · 7 · 11 · 13 · 17 − 1
are composite. Indeed, 17 is the smallest prime for which both p] − 1 and p] + 1 are
composite. We thus know already that all the numbers in the string
510 492 = 17] − 18, . . . , 17] + 18 = 510 528
510491
510529
−
= 2.76
ln 510529 ln 510491
are composite, but my criterium (3.1) is not satisfied. . With a bid more calculation we
discover the even longer string
510 482, . . . , 510 528
510529
510481
−
= 3.37
ln 510529 ln 510481
of all composite numbers. This time even my criterium (3.1) is satisfied.
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