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Math 117-01
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Math 117
Primes
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J.M.Basilla
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Institute of Mathematics
University of the Philippines-Diliman
jbasilla@math.upd.edu.ph
http://www.math.upd.edu.ph/faculty/jbasilla/
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Julius Magalona Basilla
22 November 2013
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Outline
Math 117-01
J.M.Basilla
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1. Prime Numbers
2. Prime Factorization
3. Distribution of Primes
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Prime Numbers
Math 117-01
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1. Remark: The integer 1 has only one positive divisor,
namely 1. All integers n 6= 1 has at least two positive
divisors, namely 1 and itself.
2. Definition: A positive integer p with exactly two positive
divisors, namely one and itself, is called a prime
number. An integer with more than two divisor is called
a composite integer.
3. Lemma: Every number n has a prime divisor p.
Proof.
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4. Remark If p is a prime number and p - a, then a and p
are relatively prime. By an exercise, it can be shown
that if p|ab and p - a then p|b.
5. Proposition. Let p be a prime number and let a1 , a2 ,
. . ., an be integers such that p|a1 a2 · · · an . Then p|ai for
some i, 1 ≤ i ≤ n.
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Let p be the least divisor of n greater than 1. We claim that
p is prime. If q > 0 divides p then q|n. Also since q|p then
0 < q ≤ p. By minimality of p then q = 1 or q = p. therefore,
p is indeed a prime number
Proof.
The proof is by induction on n. If n = 1, then we the
proposition is true trivially. Let k ≥ 1 and suppose that p|a1
for some 1 ≤ i ≤ k whenever p|a1 a2 . . . ak . If
p|a1 a2 . . . ak ak+1 = a1 a2 . . . ak (ak+1 ). If p|ak+1 then we are
done. If p - ak+1 then p|a1 a2 . . . ak ak+1 . By the hypothesis of
induction, p|ai for some 1 ≤ i ≤ k.
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Fundamental theorem of arithmetic
Math 117-01
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Observe that 2 is a product of 2 and 2 is prime. Let
n = k ≥ 2. Assume that every positive integer 2 ≤ t ≤ k
can be written as a product of prime. Consider n = k + 1. If
n is prime, then n is a product of itself and we are done. If n
is not prime, then n has a prime divisor, say p. Then np ≤ k
so np can be written as a product of primes. Hence, n can
be written as a product of primes.
To prove uniqueness, suppose that
n = p1 p2 · · · pr = q1 q2 · · · qs , where p1 , p2 , . . . , pr , q1 , q2 , . . . ,
qt are primes, not necessarily distinct.
Observe that p1 |q1 q2 · · · qs so p1 |qj for some 1 ≤ j ≤ s.
Since qj is prime we have p1 = qj and
p2 p3 · · · pr = q1 q2 · · · qj−1 qj+1 · · · qs . Repeating the process
would prove that r = s and pi = qσ(i) for 1 ≤ i ≤ r and
some bijection σ : {1, 2, . . . , r} → {1, 2, . . . , r}.
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Proof.
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6. Proposition. Fundamental Theorem of Arithmetic.
Every positive integer n > 1 can be written uniquely as
a product of primes.
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Distribution of primes
Math 117-01
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Suppose that there are only finitely many primes. Let p1 , p2 ,
. . ., pn be the list of all primes. Consider the number
N = p1 p2 · · · pn + 1. The number N should have at least
prime divisor (by the lemma above) but it is easy to show
that all the primes p1 , p2 , . . ., pn does not divide N . Thus,
there must be a prime not in the list. And therefore, the
prime numbers are infinitely many.
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Proof.
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7. Proposition. Euclid. There are infinitely many prime
numbers
8. Problem. How do we generate an infinite set of prime?
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Mersenne Prime
Math 117-01
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9. Proposition Let m be a composite positive integer.
Then am − 1 is composite for any positive number a.
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Recall that for all n > 1,
xn − y n = (x − y)(xn−1 + xn−2 y + · · · + xy n−2 + y n−1 ). If
m = m1 m2 , then
am1 m2 − 1 = (am1 )m2 − 1m2
= (am1 − 1)((am1 )m2 −1 + (am1 )m2 −2 + · · · + am1 + 1).
The factorization given above will be nontrivial; hence
am − 1 is composite.
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10. Remark. The preceding proposition implies that if
am − 1 is prime then m has to be prime. The number
am − 1 is much larger than m, thus if it would be true
that m and am − 1 are both prime then this will allows
us to jump from a small prime m to a much bigger
prime m.
11. Example. Mersenne Prime
m
2
3
5
7
11
2m − 1 3
7
31
127
2047
prime prime prime prime 23 ∗ 89
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Perfect number
Math 117-01
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12. Definition Perfect number. A positive number n is
perfect if and only if the sum of its positive divisors is
2n.
13. Example
perfect number positive divisors
Mersenne Prime
6
1, 2, 3, 6
3
28
1, 2, 4, 7,14, 28
7
496
1, 2, 4, 8, 16,
31
31, 62, 124, 248, 496
14. Proposition. An even number n is a perfect number if
and only if n = 2p−1 (2p − 1), where p and 2p − 1 are
primes.
Proof.
The positive divisors of 2p−1 (2p − 1) are 1, 2, . . . , 2p−1 ,
2p − 1, 2(2p − 1), . . . , 2p−1 (2p − 1). This sum is given by
(1 + 2 + 22 + 2p−1 )(1 + 2p − 1) = (2p − 1)(2p ) = 2n.
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Sieve of Eratosthenes
Math 117-01
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16. Remark. This proposition gives us a simple way of
enumerating all primes less than a certain number n.
For example we would like to enumerate all the primes
less than or equal to 100. The proposition tells us that
we need only check divisibility by the primes 2, 3, 5,
and 7 since every number n √
≤ 100 must have a prime
divisor less than or equal to 100 = 10.
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Let n be a composite number. The integer n has a divisor
other than 1 and n, say q. We have 1 < q < n. Let n = qr.
√
√
Observe that if q ≥ n, then r ≤ n. Without loss of
√
generality, assume q ≤ n. Then applying the lemma
above to q, q has a prime divisor say p and we have
√
p ≤ q ≤ n.
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Proof.
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15. Proposition. Eratosthenes. If n is a composite
number, then n has a prime divisor less than or equal
√
to n.
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