Download Quotients of Fibonacci Numbers

Quotients of Fibonacci Numbers
Stephan Ramon Garcia and Florian Luca
Abstract. There have been many articles in the M ONTHLY on quotient sets over the years.
We take a first step here into the p-adic setting, which we hope will spur further research. We
show that the set of quotients of nonzero Fibonacci numbers is dense in the p-adic numbers
for every prime p.
The nth Fibonacci number Fn is defined by the recurrence relation Fn+2 = Fn+1
+ Fn with initial conditions F0 = 0 and F1 = 1. Let F = {1, 2, 3, 5, 8, 13, 21, . . .}
denote the set of nonzero Fibonacci numbers and let R(F) = {Fm /Fn : m, n ∈ N} be
the set of all quotients of elements of F. Binet’s formula
1
Fn = √ (ϕ n − ϕ̃ n ),
5
(1)
in which
√
√
1+ 5
1− 5
ϕ=
= 1.618 . . . and ϕ̃ =
= −0.618 . . . ,
2
2
√
implies that |Fn − ϕ n / 5| tends to zero exponentially fast as n → ∞ [37, p.57]. So
as a subset of R, R(F) accumulates only at the points ϕ k for k ∈ Z [8, Ex.17].
On the other hand, R(F) is a subset of the rational number system Q, which can be
endowed with metrics other than the one inherited from R. For each prime p, there is
a p-adic metric on Q, with respect to which Q can be completed to form the set Q p of
p-adic numbers.
Our aim here is to prove the following theorem.
Theorem 2. R(F) is dense in Q p for every prime p.
By this we mean that the closure of R(F) in Q p with respect to the p-adic metric is
Q p . Although there have been many papers in the M ONTHLY on quotient sets over the
years [1, 4, 8, 9, 15, 17, 26, 33], we are unaware of similar work being undertaken in
the p-adic setting. We hope that this result will spur further investigations.
Definition 3. If p is a prime number, then each r ∈ Q\{0} has a unique representation
a
r = ± pk ,
b
(4)
in which k ∈ Z, a, b ∈ N, and a, b, p are pairwise relatively prime. The p-adic valuation ν p : Q → Z is defined by ν p (0) = +∞ and, for r as in (4), by ν p (r ) = k.
It satisfies ν p (x y) = ν p (x) + ν p (y) for x, y ∈ Q. The p-adic absolute value on Q is
∥r ∥ p = p −ν p (r ) and the p-adic metric on Q is d p (x, y) = ∥x − y∥ p .
http://dx.doi.org/10.4169/amer.math.monthly.123.10.1039
MSC: Primary 11B39, Secondary 11A41; 11S99
December 2016]
NOTES
1039
The p-adic number system Q p is the completion of Q with respect to the p-adic
metric. Ostrowski’s theorem asserts that Q p , for p prime, and R are essentially the
only completions of Q that arise from an absolute value [20]. Thus, it is natural to seek
results like Theorem 2, since similar questions in R have long since been explored.
In fact, Q p is a field that contains Q as a subfield. Its elements can be !
represented
n
as infinite series, convergent with respect to the p-adic metric, of the form ∞
n=k an p ,
in which k ∈ Z and an ∈ {0, 1, . . . , p − 1}. The p-adic valuation, norm, and metric
extend uniquely to Q p in a manner that respects these series representations.
Example 5. In Q2 we have 1 + 2 + 22 + 23 + 24 + · · · = −1 since
"
" N −1
"
"
"
"#
"
" 1 − 2N
"
"
" "
"
"
n
" = "1 − (1 − 2 N )" = "2 N " = 2−N
"
+
1
2
−
(−1)
=
"
"
"
"
2
2
"
"
1−2
2
n=0
2
tends to zero as N → ∞.
Manipulating p-adic numbers is analogous to handling decimal expansions of real
numbers. Instead of powers 10n with n running from some nonnegative N to −∞, we
have powers pn with n running from N (potentially negative) to +∞. Further details
on Q p can be found in [11, 20].
To prove our theorem, we require a few preliminary results. In what follows, we
write a|b to denote that the integer b is divisible by the integer a. Proofs of the following assertions can be found in [37, p.82, p.73, & p.81].
Lemma 6.
(a) If j|k, then F j |Fk .
(b) For each m ∈ N, there is a smallest index z(m) so that m|Fkz(m) for all positive
integers k.
(c) ν p (Fz( p) p j ) = ν p (Fz( p) ) + j for any odd prime p and j ∈ N.
We also need a convenient dense subset of Q p . We freely make use of the fact that
d p (x, y) ≤ 1/ pk if and only if ν p (x − y) ≥ k.
Lemma 7. For each prime p, the set { p−r n : n, r ∈ N} is dense in Q p .
Proof. Since Q p is the closure of Q with respect to the p-adic metric, it suffices to
show that each rational number can be arbitrarily well approximated, in
! the p-adic
j
metric, by rational numbers of the form p−r n with n, r ∈ N. If x = p k ∞
j=0 a j p ∈
! N −k−1
Q p , then let N ≥ k + 1 and n = j=0 a j p j so that
k
$
ν p (x − p n) = ν p p
k
∞
#
j=N −k
aj p
j
%
= k + νp
∞
$ #
j=N −k
%
a j p j ≥ k + (N − k) = N .
An algebraic
√ integer is a root of a monic polynomial with integer coefficients. For
instance, 2, 5, and ϕ are algebraic integers. They are roots of x − 2, x 2 − 5, and
x 2 − x − 1, respectively. The following is [24, Cor.1, p.15].
Lemma 8. A rational algebraic integer is an integer.
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√
We also require a few facts about arithmetic in the field K = Q( 5). Let OK denote
the set of algebraic integers in K. It √
is a ring; in fact, OK = {a + bϕ : a, b ∈ Z} [24,
Cor.2, p.15]. In particular, ϕ, ϕ̃, and 5 each belong to OK .
An ideal in OK is a additive subgroup i of OK so that αi ⊆ i for all α ∈ OK . A
prime in OK refers to a prime ideal in the ring OK . A prime ideal is an ideal p ⊆ OK
with the property that, for any α, β ∈ OK , the condition αβ ∈ p implies that α ∈ p or
β ∈ p. To avoid confusion, we refer to the primes 2, 3, 5, 7, . . . as rational primes.
The product of two ideals i, j in OK is defined by ij = {αβ : α ∈ i, β ∈ j}; it is
also an ideal in OK . Positive powers of ideals are defined inductively by i1 = i and
in = i(in−1 ) for n = 2, 3, . . .. We say that i divides j if i ⊆ j.
For an ideal i in OK , we write α ≡ β (mod i) to mean that β − α ∈ i. The familiar properties of congruences hold when working modulo an ideal. For instance,
α ≡ β (mod i) implies that α j ≡ β j (mod i) for j ∈ N. Similarly, if p and q are
distinct prime ideals, α ≡ β (mod p), and α ≡ β (mod q), then α ≡ β (mod pq).
Lemma 9. Let a, b ∈ Z and j ∈ N. If a ≡ b (mod p j ), then p j divides b − a in Z.
Proof. Suppose that a, b ∈ Z and a ≡ b (mod p j ). Then, b − a = p j α for some α ∈
OK . Since α = (b − a)/ p j is a rational algebraic integer, it is an integer by Lemma 8.
Consequently, p j divides b − a in Z.
Each rational prime p gives rise to an ideal p = pOK in OK This ideal
√ factors
uniquely as a product of prime ideals [24, Th.16, p.59]. In fact, since Q( 5) is a
quadratic extension of Q we have [24, p.74]:
√
Lemma 10. Let K = Q( 5) and let p be a rational prime. Then p = pOK is the
product of at most two prime ideals (not necessarily distinct).
We can be a little more specific. The only rational prime p√for which p is a square
of a prime ideal is p = 5; this reflects the factorization 5 = ( 5)2 [24, Th.24, p.72].
Proof of Theorem 2. There are two cases, according to whether p is odd or p = 2. Let
p be an odd rational
√ prime and let p = p OK . For each j ∈ N, Lemma 6 ensures that
p2 j |Fz( p) p2 j . Since 5 ∈ OK , (1) reveals that
2j
2j
ϕ z( p) p − ϕ̃ z( p) p =
√
5Fz( p) p2 j ≡ 0 (mod p2 j ),
and hence
2j
ϕ z( p) p ≡ ϕ̃ z( p) p
2j
(mod q2 j )
for each prime q in OK that divides p. Since ϕ̃ = −1/ϕ, we have
2j
ϕ 2z( p) p ≡ ϕ −2z( p) p
2j
(mod q2 j ),
so that
2j
ϕ 4z( p) p ≡ 1 (mod q2 j )
December 2016]
and
NOTES
2j
ϕ̃ 4z( p) p ≡ 1 (mod q2 j ).
(11)
1041
2j
2j
Setting x = ϕ 4z( p) p and y = ϕ̃ 4z( p) p in the identity
x m − y m = (x − y)(x m−1 + x m−2 y + · · · + y m−1 ),
m ∈ N,
implies that
2j
2j
F4z( p) p2 j m
ϕ 4z( p) p m − ϕ̃ 4z( p) p m
=
F4z( p) p2 j
ϕ 4z( p) p2 j − ϕ̃ 4z( p) p2 j
2j
2j
2j
2j
= (ϕ 4z( p) p )m−1 + (ϕ 4z( p) p )m−2 (ϕ̃ 4z( p) p ) + · · · + (ϕ̃ 4z( p) p )m−1
≡ m (mod q2 j )
by (11). Now Lemma 10 ensures that
F4z( p) p2 j m
≡ m (mod p j )
F4z( p) p2 j
for any m ∈ N. However, F4z( p) p2 j m /F4z( p) p2 j is a rational integer by Lemma 6, so
F4z( p) p2 j m
≡ m (mod p j )
F4z( p) p2 j
(12)
by Lemma 9. Appealing to Lemma 6, we have
ν p (F4z( p) p2(k+2r ) ) = ν p (F4z( p) p2(k+r ) ) + 2r,
(13)
so
F4z( p) p2(k+2r )
= p2r ℓ,
F4z( p) p2(k+r )
ℓ ∈ Z,
gcd(ℓ, p) = 1.
(14)
Set m = pr nℓ and j = k + r in (12) and use (14) to conclude that
p2r ℓ ·
F4z( p) p2(k+r ) pr nℓ
F4z( p) p2(k+r ) pr nℓ
=
≡ pr nℓ (mod p k+r ).
F4z( p) p2(k+2r )
F4z( p) p2(k+r )
Since gcd(ℓ, p) = 1, the preceding yields
pr
F4z( p) p2k+3r nℓ
≡ n (mod p k ),
F4z( p) p2k+4r
and hence
νp
!
"
F4z( p) p2k+3r nℓ
− p−r n ≥ k.
F4z( p) p2k+4r
This holds for all n, r ∈ N and all k > r , so R(F) is dense in Q p by Lemma 7.
If p = 2 we must replace the exponents 4z( p) p j in (11) with 3 · 2 j+2 since z(2) = 3
and 4 = 22 . The proof can proceed as before if we replace (13) with the corresponding
statement for p = 2. It suffices to show that ν2 (F3·2 j ) = j + 2 for j ∈ N; see [22] for
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a more general result. We proceed by induction on j. The base case j = 1 is ν2 (F6 )
= ν2 (8) = 3. Now suppose that ν2 (F3·2 j ) = j + 2 for some j ≥ 2. Let L n denote the
nth Lucas number; these satisfy the recurrence L n+2 = L n+1 + L n with initial conditions L 0 = 2 and L 1 = 1. Since L 2n = 2(−1)n + 5Fn2 [37, p.177] and since 2 = F3
divides F3·2 j−1 , we have
!
"
j−1
2
ν2 (L 3·2 j ) = ν(L 2(3·2 j−1 ) ) = ν2 2(−1)3·2 + 5F3·2
j−1 = 1.
Because F2n = Fn L n [37, p.177], we conclude that
ν2 (F3·2 j+1 ) = ν2 (F2(3·2 j ) ) = ν2 (F3·2 j ) + ν2 (L 3·2 j ) = ( j + 2) + 1 = ( j + 1) + 2,
which completes the induction.
ACKNOWLEDGMENT. This work was partially supported by National Science Foundation Grant DMS1265973. We thank the anonymous referees for numerous helpful suggestions.
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Department of Mathematics, Pomona College, Claremont, CA 91711
[email protected]
School of Mathematics, University of the Witwatersrand, Private Bag X3, Wits 2050, Johannesburg, South
Africa
[email protected]
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