Math 3221 Number Theory - Homework Until Test 2 Worksheet With Answers - Philipp Braun

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MATH3221: Number Theory
Philipp BRAUN
Homework until Test #2
Section 3.1
2
page 43, 1. It has been conjectured that there are infinitely many primes of the form n
2.
Exhibit five such primes.
2
2
Solution. Five such primes are e.g. 2 = 4
2 = 2
2, 7 = 9
2 = 3
2, 23 = 25
2 =
2
2
2
5
2, 47 = 49
2 = 7
2 and 79 = 81
2 = 9
2.
2. Give an example to show that the following conjecture is not true: Every positive integer can
2
be written in the form p + a
, where p is either a prime or 1, and a
0.
2
2
2
Counterexample: 25. Since none of the numbers 25, 25
1
= 24, 25
2
= 21, 25
3
=
2
2
16, 25
4
= 9 are primes and a has to be less than 5 since 5
= 25 which forced p to be
2
zero, there is neither a prime p such that there is a such that p + a
= 25, nor there is a
2
such that 1 + a
= 25.
3. Prove each of the assertions below:
(a) Any prime of the form 3n + 1 is of the form 6m + 1
(b) Each integer of the form 3n + 2 has a prime factor of this form.
3
(c) The only prime of the form n
1 is 7.
(d) The only prime p for which 3p + 1 is a perfect square is p = 5.
2
(e) The only prime of the form n
4 is 5.
Proof.
(a) Let p = 3n + 1 be any prime. Since p is a prime by assumption, 3n has to be even
(since otherwise 3n + 1 would be even and therefore divisible by 2, which contradicts
the assumption that p is prime), i.e. 2 3n. Since 3 2, 2 n by Theorem 3.1 (since
2 is a prime). That means that there is an integer m such that n = 2m. Therefore
3n + 1 = 3(2m) + 1 = 6m + 1, as we wanted to show.
(b) Let m = 3n + 2 be any integer. If m is prime, we are done. Assume m is not a prime.
Therefore there are integers a, b such that a , b > 1 and ab = 3n + 2. Consider the
following cases: Case I : a = 3k, b = 3l for integers k, l. Then ab = 3k 3l = 3(3kl),
which is of the form 3z and therefore cannot equal 3n + 2. Case II : a = 3k, b = 3l + 1
(or vice versa) for integers k, l. Then ab = 3k(3l + 1) = 3(3kl + k), which is also of the
form 3z. Case III : a = 3k + 1, b = 3l + 1 for integers k, l. Then ab = (3k + 1)(3l + 1) =
3(3kl + k + l) + 1, which is of the form 3z + 1.
1

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