Math 3221 Number Theory - Homework Until Test 2 Worksheet With Answers - Philipp Braun Page 2
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16So none of these cases can occur, so at least one of the factors a, b has to be 3z + 2 for
an integer z. Either this factor is a prime, or we can use the same argument to show
that 3z + 2 = (3x + 2)y with integers x, y. Since a , b > 1, 3z + 2 < 3n + 2. Thus
in every step the absolute value of the factor of the form 3a + 2 will be smaller, so the
method has to terminate with a prime of that form.
3
2
3
(c) Since n
1 = (n
+ n + 1)(n
1), (n
1) (n
1). So n must be less than or equal
3
to 2. For n = 1 we have 1
1 = 0, which is not the prime. So the only prime of the
3
3
form n
1 is 2
1 = 7.
2
1
2
1
(d) Assume (3p+1) = n
for a prime p and an integer n. Then p =
(n
1) =
(n+1)(n
3
3
1
1
1). Since p is a prime,
(n
1) = 1, which implies n = 4 and therefore p =
5 3 = 5.
3
3
2
2
(e) Since n
4 = (n + 2)(n 2), n 2 has to equal 1 (because otherwise n
4 is no prime
2
2
since (n
2) (n
4)), i.e. n = 3, so the only prime of that form is 3
4 = 5.
2
4. If p
5 is a prime number, show that p
+ 2 is composite.
2
Proof. p
+ 2 = (p + 1)(p
1) + 3. Since p is a prime bigger than 3, p is not divisible by
3. But since one of three consecutive numbers is divisible by three, so is either (p
1) or
2
(p + 1). Since 3 3, by Theorem 2.2 (vii) 3 (p + 1)(p
1) + 3, so p
+ 2 is divisible by 3
and therefore composite.
6. Establish each of the following statements:
4
(a) Every integer of the form n
+ 4, with n > 1, is composite.
(b) If n > 4 is composite, then n divides (n
1)!.
(c) Any integer of the form 8 + 1, where n
1, is composite.
(d) Each integer n > 11 can be written as the sum of two composite numbers.
Proof.
4
2
2
(a) n
+ 4 = (n
+ 2n + 2)(n
2n + 2), and for n > 1, each of these factors is bigger than
4
1, hence n
+ 4 is composite.
(b) If n > 4 is composite, then it is a square number (then 2 n < n since n > 4, and
therefore n 1 2 . . .
n . . . 2 n . . . (n 1) = (n 1)!) or it can be written as a product of
a prime 1 < p <
n and an integer
n < m < n. Since (n 1)! = 1 2 p . . . m . . . (n 1),
pm (n
1)!, i.e. n (n
1)!.
n
n
3
n
n
2
n
(c) 8
+ 1=(2
)
+ 1 = (2
+ 1)((2
)
(2
) + 1). Since both factors are bigger than 1 for
n
n
1, 8
+ 1 is not prime and therefore composite.
(d) In case n > 11 is even, there is k > 5 such that n = 2k. Therefore n = 6 + 2(k
3),
where 6 = 2 3 is a composite number and 2 (k
3) is a composite number (since
k
3 > 1 since k > 5). In case n is odd, there is k > 5 such that n = 2k + 1. Therefore
n = 9 + 2(k
4), where 9 = 3 3 and 2(k
4) are composite numbers (since k
4 > 1
because k > 5).
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