Fun with the Fundamental Theorem of Arithmetic
1
Divisibility
1.1
Useful notation
Definition. Given an integer with n 6= 0 and a prime p, the valuation of n at p, denoted vp (n), is the power
to which p is raised in the prime factorization of n. If p does not appear in the prime factorization of n, then
vp (n) = 0.
To be more precise, from FTA it follows that any integer n 6= 0 can be written uniquely as
Y
p np ,
n = (±1)
p prime
where np ≥ 0. This is an infinite product, but note that (again by FTA) all but finitely many of these
exponents will be 0. Then for a prime p, we define vp (n) = np , the exponent of p appearing in this
factorization.
Y
Example. Consider n = ±1. Then n = (±1)
pnp where all the np are 0. Thus for any prime p we
p prime
have vp (±1) = np = 0.
Example. Consider n = −1500. Let’s find vp (n) for each prime p. Write −1500 = (−1)22 · 3 · 53 . Then
v2 (n) = 2, v3 (n) = 1, v5 (n) = 3, and vp (n) = 0 for all primes p 6= 2, 3, 5.
Some easy properties of vp follow from FTA.
Lemma. Let m and n be positive integers. Let p be a prime. Then
(i) vp (m · n) = vp (m) + vp (n) for all primes p;
(ii) m = n iff vp (m) = vp (n) for all primes p;
(iii) m | n iff vp (m) ≤ vp (n) for all primes p.
Proof of (iii). If m | n, then there is an integer c such that n = m · c. Then for all primes p we have
vp (n) = vp (m · c) = vp (m) + vp (c), by part (i). But since c is an integer, vp (c) ≥ 0. This implies that
vp (n) ≥ vp (m), or equivalently, vp (m) ≤ vp (n).
Qr
Proof of (iii) contd. Going the other way, suppose vp (m) ≤ vp (n) for all primes p. Write n = i=1Q
pni i , with
r
i
ni ≥ 1 for all i. Then since the powers
appearing in m are at most ni , we can write m = i=1 pm
i ,
Qr of primes
ni −mi
where mi ≤ ni for all i. Set c = i=1 pi
. Then c is an integer (all the exponents of the primes are
nonnegative), and m · c = n. Thus m | n.
1.2
Number of divisors
Qr
i
Let n = i=1 pniQ
be a prime factorization of n. According to our lemma, any positive divisor m of n can be
r
i
written as m = i=1 pm
i , where 0 ≤ mi ≤ ni .
In fact by the uniqueness claim in FTA our divisors m are in 1-1 correspondence with sequences
(m
ni . How many such sequences are there? There are
1
Qr , m2 , . . . , mr ) of exponents with 0 ≤ miQ≤
r
i=1 (ni + 1) such sequences. Thus there are
i=1 (ni + 1) positive divisors of n.
Example. Let n = 300 = 22 · 3 · 52 . Then n has (2 + 1)(1 + 1)(2 + 1) = 18 positive divisors. Note that the
divisor m = 1 corresponds to the sequence (m1 , m2 , m3 ) = (0, 0, 0) and the divisor m = 300 corresponds to
the sequence (m1 , m2 , m3 ) = (2, 1, 2).
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1.3
GCD and LCM
Given prime factorizations of m and n, we can read off their greatest common divisor (gcd) and least common
multiple (lcm).
Qr
Qr
i
and n = i=1 pni i be prime factorizations.
Theorem. Let m = i=1 pm
i
Qr
(i) Let di = min(mi , ni ). Then gcd(m, n) = i=1 pdi i .
Qr
(ii) Let li = max(mi , ni ). Then lcm(m, n) = i=1 plii .
Example. Let m = 2450 = 2 · 52 · 72 , and let n = 2100 = 22 · 3 · 52 · 7. Then gcd(m, n) = 2 · 52 · 7 = 350 and
lcm(m, n) = 22 · 3 · 52 · 72 = 14, 700.
2
2.1
Rationality and integral closure
Rationality and irrationality
We can also use the FTA to create simple proofs concerning the rationality of a number.
√
Theorem. The square-root of 2 is irrational; i.e., 2 ∈
/ Q.
√
√
Proof. By contradiction. Suppose 2 ∈ Q. Then we can write 2 = m
n with m, n ∈ Z. Now square both
sides of this equation and rearrange to conclude that 2n2 = m2 . Call r = 2n2 and s = m2 . Then v2 (r) =
v2 (2n2 ) = v2 (2) + v2 (n2 ) = 1 + v2 (n) + v2 (n) = 1 + 2v2 (n), which is odd. Similarly
v2 (s) = v2 (m2 ) = 2v2 (m)
√
is even. But since r = s, we must have v2 (r) = v2 (s) Contradiction! Thus 2 ∈
/ Q.
The theorem above and its proof can be easily generalized. An integer n is a perfect square if n = a2
for some integer a.
Lemma. An integer n > 0 is a perfect square iff vp (n) is even for all primes p; i.e., iff all powers in the
prime factorization of n are even.
√
Theorem. Let n > 0 be an integer. Then n is rational iff n is a perfect square.
2.2
Integral closure
In fact we can prove an even more general fact. An integer polynomial is a polymonial f (x) = an xn +
an−1 xn−1 + · · · a1 x + a0 with integer coefficients; i.e., ai ∈ Z for all i. A monic integer polynomial is a
polynomial f (x) = xn + an−1 xn−1 + · · · a1 x + a0 with leading coefficient an = 1.
Theorem (Z is integrally closed). Let f (x) = xn + an−1 xn−1 + · · · a1 x + a0 be a monic integer polynomial,
and let α be a root of f ; i.e., we have f (α) = 0. If α is rational, then in fact α is an integer!
Proof. Look it up!
√
Example. Suppose n0 is not a perfect square, and consider the polynomial f (x) = x2 − n0 . Then α = n0
√
is a root of f (x). According to the theorem, if n0 were rational, then it would in fact be an integer; i.e.,
√
we would have n0 = a for some a ∈ Z. But this is impossible as then we would have a2 = n0 and yet n0 is
√
not a perfect square. Contradiction! Thus n0 cannot be rational!
2