On equations $σ(n)=σ(n+k)$ and $\varphi(n)=\varphi(n+k)$

0.12 MB·

by  Tomohiro Yamada

On equations σ(n) = σ(n + k) and
ϕ(n) = ϕ(n + k) ∗†
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2 Tomohiro Yamada

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Abstract

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We study the distribution of solutions of equations σ(n)= σ(n+
.
h k) and ϕ(n) = ϕ(n+k). We give new upper bounds for these solu-
t tions.
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1 1 Introduction
v
1
1
5 In this paper, we study equations σ(n) = σ(n + k) and ϕ(n) = ϕ(n + k).
2
As far as the author knows, an equation of these types was first referred by
.
1
Ratat [13], who asked for which values of n the equation ϕ(n) = ϕ(n+1)
0
0 holds and gave n = 1,3,15,104 for examples. In 1918, answering to Ratat’s
1
question, Goormaghtigh [7] gave n = 164,194,255,495.
:
v
Xi After then, several authors such as Klee [9], Moser [11], Lal and Gillard
r [10], Ballew, Case and Higgins [3], Baillie [1] [2] and Graham, Holt and
a
Pomerance [8] searched for solutions to ϕ(n) = ϕ(n+k).
Klee [9] and Moser [11] noted that if p,2p − 1 are both odd primes
and n = 2(2p − 1), then ϕ(n) = 2p − 2 = ϕ(4p) = ϕ(n + 2). Under the
quantitative prime k-tuplet conjecture, the number of such primes ≤ x is
≫ x/(logx)2. Similarly to Moser’s result, we can see that σ(n) = σ(n+22)
if 3l −1,14l−1 are both primes and n = 28(3l−1).
On the other hand, Erdos, Pomerance and Sarkozy [6] shows that the
number of solutions n ≤ x to ϕ(n) = ϕ(n+1) is at most xexp(−(logx)1/3)
for sufficiently large x and a similar result holds for σ(n) = σ(n+1). They
∗MSC subject classification: 11A05, 11A25.
†Key words: Arithmetic functions.
1
1 INTRODUCTION 2
also conjectured that the number of such solutions below sufficiently large
x is at least x1−ǫ for every ǫ > 0.
Graham, Holt andPomerance [8]generalized these results. They showed
that, if j and j+k have the same prime factors with g = (j,j+k), both of
jr/g +1,(j +k)r/g +1 are primes which do not divide j and
j +k
n = j r +1 , (1)
(cid:18) g (cid:19)
then ϕ(n) = ϕ(n+k). Moreover, they gave the corresponding result of [6].
According to them, we denote by N(k,x) the number of integers n ≤ x with
ϕ(n) = ϕ(n + k) and N (k,x) the number of integers n ≤ x with ϕ(n) =
1
ϕ(n + k) which are in the form (1). Then they showed that N (k,x) ≤
1
xexp(−(logx)1/3) for sufficiently large x and N(k,x) = N (k,x) if k is odd,
1
so that their result implies the result of [6].
Our purpose is to prove corresponding results on the equation σ(a n+
1
b ) = σ(a n + b ) and ϕ(a n + b ) = ϕ(a n + b ) and improve the upper
1 2 2 1 1 2 2
bound xexp(−(logx)1/3) of [8]. The main results are the following two
theorems.
Theorem 1.1. Let a ,b ,a ,b be integers such that a a (a b −a b ) 6= 0.
1 1 2 2 1 2 1 2 2 1
Assume that m ,m ,k ,k satisfy the relations
1 2 1 2
k (a b −a b ) k (a b −a b )
2 1 2 2 1 1 1 2 2 1
m = ,m = ,k σ(m ) = k σ(m ). (2)
1 2 1 1 2 2
a (k −k ) a (k −k )
2 2 1 1 2 1
and q = k l−1(i = 1,2) are both prime. If
i i
m q −b
1 1 1
n = , (3)
a
1
then we have a n+b = m q ,a n+b = m q and σ(a n+b ) = σ(a n+b ).
1 1 1 1 2 2 2 2 1 1 2 2
Similarly, assume that m ,m ,k ,k satisfy the relations
1 2 1 2
k (a b −a b ) k (a b −a b )
2 1 2 2 1 1 1 2 2 1
m = ,m = ,k ϕ(m ) = k ϕ(m ). (4)
1 2 1 1 2 2
a (k −k ) a (k −k )
2 1 2 1 1 2
and q = k l+1(i = 1,2) are both prime. If
i i
m q −b
1 1 1
n = , (5)
a
1
then we have a n+b = m q ,a n+b = m q and ϕ(a n+b ) = ϕ(a n+b ).
1 1 1 1 2 2 2 2 1 1 2 2
Furthermore, if a = a and the condition (4) holds, then m ,m =
1 2 1 2
m +b −b must have the same prime factors.
1 2 1
2 PRELIMINARY LEMMAS 3
Theorem 1.2. Let a ,b ,a ,b be integers with a > 0,a > 0,a b −
1 1 2 2 1 2 1 2
a b 6= 0. Let N(a ,b ,a ,b ;x) denote the number of integers n ≤ x with
2 1 1 1 2 2
ϕ(a n + b ) = ϕ(a n + b ) that are not in the form (3) given in Theo-
1 1 2 2
rem 1.1. Similarly, let M(a ,b ,a ,b ;x) denote the number of integers
1 1 2 2
n ≤ x with σ(a n + b ) = σ(a n + b ) that are not in the form (5) given
1 1 2 2
in Theorem 1.1. Then N(a ,b ,a ,b ;x) and M(a ,b ,a ,b ;x) are both
1 1 2 2 1 1 2 2
≪ xexp(−2−1/2(logxlogloglogx)1/2).
Appliedinsomeparticularcases, thesetheoremsgivethefollowingcorol-
laries.
Corollary 1.3. If k is odd, then the number of solutions n ≤ x to ϕ(n) =
ϕ(n+k) is ≪ xexp(−2−1/2(logxlogloglogx)1/2).
Corollary 1.4. If there exists no integer m such that m divides σ(m) and
m+1 divides σ(m+1), then the numberof solutions n ≤ x to σ(n) = σ(n+1)
is ≪ xexp(−2−1/2(logxlogloglogx)1/2).
The proof of Theorem 1.1 is straightforward. The proof of 1.2 depends
on one of many results of Banks, Friedlander, Pomerance, Shparlinski [4]
concerning to multiplicative structures of values of Euler’s totient function.
It is unlikely that there exists an integer m such that m divides σ(m)
and m+1 divides σ(m+1). However, the proof of the nonexistence of such
an integer will be difficult. We note that the nonexistence of such an integer
would follow from the conjecture that there exists no odd integer m > 1 for
which m divides σ(m).
2 Preliminary Lemmas
In this section, we shall introduce some basic lemmas on distributions of
integers with special multiplicative structures.
We denote by P(n),p(n) the largest and smallest prime factor of n re-
spectively. For the positive real number x, let us denote x = x,x =
0 i+1
max{1,logx }. We denote by c some positive constant not necessarily same
i
at every occurence. Furthermore, we denote by x,y,z real numbers and we
put u = logx/logy and v = logy/logz. These notations are used in later
sections.
Lemma 2.1. Denote by Ψ(x,y) the number of integers n ≤ x divisible by
no prime > y. If y > x2, then we have Ψ(x,y) < xexp(−ulogu+o(u)) as
1
x,u tend to infinity.
2 PRELIMINARY LEMMAS 4
Proof. This follows from a well-known theorem of de Bruijn [5]. A simpler
proof is given by Pomerance [12].
Lemma 2.2. Let
S = {n | p2 | n for some p,a with pa > y,a ≥ 2}. (6)
Then we have the number of elements in S below x is ≪ xy−1/2.
Proof. Let Π(t) be the number of perfect powers below t. It is clear that
Π(t) < t1/2 + t1/3 + ... + t1/k < t1/2 + kt1/3 = t1/2(1 + o(1)), where k =
⌊(logt)/(log2)⌋.
Let us denote by γ the smallest integer γ for which pγ > y and γ > 1.
p
Clearly we have #S(x) ≤ x p−γp. Since pγp > y, we have by partial
p≤x
summation P
1 Π(x) Π(y) x Π(t)
≤ − + dt ≪ y−1/2. (7)
pγp x y Z t2
Xp≤x y
This prove the lemma.
We use an upper bound for Φ(x,y) the number of integers n ≤ x such
that P(σ(n)) ≤ y or P(ϕ(n)) ≤ y.
Lemma 2.3. Let Φ(x,y) denote the number of integers n ≤ x such that
P(ϕ(n)) ≤ y. and Σ(x,y) denote the number of integers n ≤ x such that
P(σ(n)) ≤ y. For any fixed ǫ > 0 and (loglogx)1+ǫ < y ≤ x, we have
Φ(x,y) < x(exp(−u(1+o(1))loglogu)), (8)
and
Σ(x,y) < x(exp(−u(1+o(1))loglogu)), (9)
when u = (logx)/(logy) → ∞.
Proof. ThefirstresultisTheorem3.1in[4]. Thesecondresultcanbeproved
similarly, noting that the number of integers n ≤ x such that pa | n for some
primepwithσ(pa) > y,a ≥ 2is≪ x/y−1/2 ≪ x(exp(−u(1+o(1))loglogu))
by Lemma 2.2.
3 PROOF OF THEOREM ?? 5
3 Proof of Theorem 1.1
If (2) holdsandq = k l−1(i = 1,2)arebothprime, thenwe have σ(m q ) =
i i i i
σ(m )k l(i = 1,2) must be equal since k σ(m ) = k σ(m ). Moreover, we
i i 1 1 2 2
have a m q −a m q = l(a m k −a m k )−(a m −a m ) = a b −a b
1 2 2 2 1 1 1 2 2 2 1 1 1 2 2 1 1 2 2 1
since a m k = a m k and a m −a m = −a b +a b .
1 2 2 2 1 1 1 2 2 1 1 2 2 1
The corresponding statement with σ replaced by ϕ can be similarly
proved.
Finally, we note that if ϕ(m )/m = ϕ(m )/m , then m ,m must have
1 1 2 2 1 2
the same prime factors. This prove the last statement. This completes the
proof of Theorem 1.1.
4 Proof of Theorem 1.2
We prove Theorem 1.2 for σ. The corresponding statement for ϕ can be
proved in a similar, but slightly simpler way since we need not to be careful
about square factors.
Let B(x) be the set of integers n ≤ x not in the form (3) given in
Theorem 1.1 for which the equation σ(a n+b ) = σ(a n+b ) hold.
1 1 2 2
We put y = exp(21/2(logxlogloglogx)1/2),z = y1/2,z = z/log(x/z)
1
andz = Y log(x/z). sothatu = (logx)/(logy) = 2−1/2(logx)1/2(logloglogx)−1/2.
2
Theorem 1.2 for σ canbe formulated that #B(x) < xz−1+o(1). We note that
we can take x to be sufficiently large so that y is also sufficiently large.
Let us consider the following sets of integers:
B (x) ={n | n ∈ B(x),a n+b ∈ S or a n+b ∈ S},
1 1 1 2 2
B (x) ={n | n ∈ B(x),P(σ(a n+b )) < y},
2 1 1
B (x) =B(x)\(B (x)∪B (x)).
0 1 2
We have #B (x) ≪ xy−1/2 = x/z by Lemma 2.2. Moreover, we have
1
#B (x) ≪ xz−1+o(1) by Lemma 2.3.
2
Now let n ∈ B (x). Since n 6∈ B (x), σ(a n+b ) must have some prime
0 2 1 1
factor p ≥ y. Therefore a n+b must have some prime power factor qa with
1 1
σ(qa) > y. However, we must then have a = 1 since n 6∈ B (x). So that
1
a n + b must have some prime factor of the form k p − 1, where k ≥ 1
1 1 1 1
is an integer. Similarly, a n+b must have some prime factor of the form
2 2
4 PROOF OF THEOREM ?? 6
k p−1, where k ≥ 1 is an integer. So that We can write
2 2
a n+b = m (k p−1)(i = 1,2), (10)
i i i i
where p is a prime ≥ y and m ,m ,k ,k are positive integers such that
1 2 1 2
k p−1 is a prime not dividing m for each i = 1,2. Now we have
i i
σ(m )k = σ(m )k (11)
1 1 2 2
since
σ(m )k p = σ(a n+b ) = σ(a n+b ) = σ(m )k p. (12)
1 1 1 1 2 2 2 2
Now we divide B (x) into two sets
0
B (x) = {n | n ∈ B (x),a n+b = m (k p−1),m m ≤ x/z}
3 0 i i i i 1 2
and
B (x) = {n | n ∈ B (x),a n+b = m (k p−1),m m > x/z}.
4 0 i i i i 1 2
By (10), we have p(a m k −a m k ) = a (b +m ) −a (b +m ). If
2 1 1 1 2 2 2 1 1 1 2 2
a m k −a m k = 0, then (11) gives a m σ(m ) = a m σ(m ), contrary
2 1 1 1 2 2 1 2 1 2 1 2
to the assumption. Thus we see that a (b +m )−a (b +m ) 6= 0 and p
2 1 1 1 2 2
divides a (b +m )−a (b +m ).
2 1 1 1 2 2
We show that #B (x) < xz−1+o(1). Multiplying (10) by a and sub-
3 3−i
tracting one from the other, we have
a m (k p−1)−a m (k p−1) = a b −a b . (13)
2 1 1 1 2 2 2 1 1 2
Let us denote c = (k ,k ) and k = k /c,k = k /c. By virtue of (11),
1 2 1 1 2 2
k ,k are uniquely determined by m ,m .
1 2 1 2
If a m k 6= a m k , then p can be expressed by
2 1 1 1 2 2
a (m +b )−a (m +b )
2 1 1 1 2 2
p = . (14)
a m k −a m k
2 1 1 1 2 2
Therefore we have
a (m +b )−a (m +b )
2 1 1 1 2 2
pc = . (15)
a m k −a m k
2 1 1 1 2 2
So that c can be uniquely determined by p,m ,m . Since p divides d =
1 2
a (m +b )−a (m +b ), the number of possibility of p is at most ω(d) ≪
2 1 1 1 2 2
logd ≪ log(x/z) ≪ logx. The number of possibility of a pair (m ,m ) is ≪
1 2
4 PROOF OF THEOREM ?? 7
(x/z)log(x/z) = x/z . Hence we obtain #B (x) ≪ xlogx/z = xz−1+o(1)
1 3 1
provided that a m k 6= a m k .
2 1 1 1 2 2
If a m k = a m k , then (13) gives a (b +m ) = a (b +m ). Hence
2 1 1 1 2 2 2 1 1 1 2 2
we have
k (a b −a b ) k (a b −a b )
2 1 2 2 1 1 1 2 2 1
m = ,m = . (16)
1 2
a (k −k ) a (k −k )
2 2 1 1 2 1
Thereforek ,k ,m ,m ,l = p,q = k p−1,q = k p−1satisfythecondition
1 2 1 2 1 1 2 2
in Theorem 1.1.
Next we show that #B (x) is also at most xz−1+o(1). If k = k , then
4 1 2
k p − 1 divides a b − a b . Therefore a b − a b = 0 or y < a b −
1 1 2 2 1 1 2 2 1 1 2
a b , contrary to the assumption. Hence we see that k 6= k . Now (10)
2 1 1 2
implies that n ≡ ψ(a ,b ,a ,b ,k ,k ) (mod (k p − 1)(k p − 1)), where
1 1 2 2 1 2 1 2
ψ(a ,b ,a ,b ,k ,k )istheunique simultaneous solutionoftwo congruences
1 1 2 2 1 2
a ψ(a ,b ,a ,b ,k ,k ) ≡ −b (mod k p−1) and a ψ(a ,b ,a ,b ,k ,k ) ≡
1 1 1 2 2 1 2 1 1 2 1 1 2 2 1 2
−b (mod k p−1).
2 2
Since m m ≥ xY, we have (k p − 1)(k p − 1) ≤ (a x + b )(a x +
1 2 1 2 1 1 2
b )/(xY) ≪ x/Y. Now the number of elements of B (x) can be bounded
2 4
by
x
+1 , (17)
(cid:18)q q (cid:19)
pX,q1,q2 1 2
where p runs over the primes ≥ y and q ,q run over the primes such that
1 2
q ≡ q ≡ −1 (mod p) and q q ≤ (a x+b )(a x+b )/(m m ) ≪ x/z.
1 2 1 2 1 1 2 2 1 2
For each prime p, the sum can be estimated as
x 1 xY log(x/z) xZ
+1 ≪ x( )2+ ≪ 2, (18)
(cid:18)(k p−1)(k p−1) (cid:19) kp p2 p2
X 1 2 X
k1,k2 k
where k runs all positive integers up to cx for some suitable constant c.
Since p ≥ y, we have
xz xz
#B (x) ≪ 2 ≪ 2 ≪ xz−1+o(1). (19)
4 p2 y
Xp≥y
Clearly B(x) = B (x) and each #B (x) is at most xy−1+o(1).
1≤i≤4 i i
Therefore #B(x) < xSy−1+o(1). This proves Theorem 1.2 for σ. As we noted
inthebeginning ofthissection, Theorem 1.2forϕcanbeproved inasimilar
way. Now the proof is complete.
5 PROOF OF COROLLARIES 8
5 Proof of corollaries
Assume ϕ(n) = ϕ(n+k) with n satisfying the condition of Theorem 1.1 and
let m ,m be as appear in Theorem 1.1. then m and m = m +k must
1 2 1 2 1
have the same prime factors. Thus k must be even. This gives Corollary
1.3.
We derive Corollary 1.4 from Theorem 1.2. Assume that σ(n) = σ(n+
1),n = m q ,n + 1 = m q , m ,m ,k ,k satisfy the relations (2) and
1 1 2 2 1 2 1 2
q = k l − 1(i = 1,2) are both prime. We can take k ,k to be relatively
i i 1 2
prime by replacing k ,k ,l by k /(k ,k ),k /(k ,k ),l(k ,k ) respectively.
1 2 1 1 2 2 1 2 1 2
The relations (2) gives
m = k /(k −k ),m = k /(k −k ) (20)
1 2 2 1 2 1 2 1
and
k σ(m ) = k σ(m ). (21)
1 1 2 2
We show that m = k ,m = k and k = k +1. If d divides k −k , then
1 2 2 2 2 1 2 1
d must divide k and therefore d must divide (k ,k ). So that we must have
2 1 2
d = 1. Therefore k −k = 1. Now (21) gives k σ(k +1) = (k +1)σ(k ).
2 1 1 1 1 1
Since k ,k are clearly relatively prime, k must divide σ(k ) and k + 1
1 1 1 1 1
must divide σ(k +1). This proves Corollary 1.4.
1
References
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Tomohiro Yamada
Department of Mathematics
Faculty of Science
Kyoto University
Kyoto, 606-8502
Japan
e-mail: [email protected]