General Mathematics Vol. 17, No. 1 (2009), 65–70
Several inequalities about the number of
positive divisors of a natural number m1
Nicuşor Minculete1 , Petrică Dicu2
Abstract
In this paper we intend to establish several properties of the number of positive divisors of a natural number m. Among these,we
remark the inequality τ 2 (mn) ≥ τ (m2 )τ (n2 ), for all m, n ∈ N⋆ .
2000 Mathematics Subject Classification: 11A25
Key words and phrases: prime number, arithmetic inequality, the
number of natural divisors of m.
1
Introduction
For a positive integer m number, we will note τ (m) the number of positive
divisors of m. We remark that: τ (1) = 1 and if p is a prime number, then
τ (p) = 2 , τ (pα ) = α + 1 .
In papers [1]–[5], [7] we find the following properties of τ (m):
For m = pα1 1 pα2 2 ...pαr r , m > 1 we have the relation:
(1)
τ (m) = (α1 + 1)(α2 + 1)...(αr + 1) .
1
Received 5 May, 2008
Accepted for publication (in revised form) 10 December, 2008
65
66
Nicuşor Minculete,Petrică Dicu
If (m, n) = 1, then
τ (mn) = τ (m)τ (n) , for all m, n ∈ N⋆ .
(2)
For m ≥ 2, we have the relation:
m h
X
m−1
mi
(3)
−
.
τ (m) =
k
k
k=1
In [6], for m ≥ 1, we have
(4)
√
τ (m) ≤ 2 m .
In [8] it is shown that
τ (m)τ (n) ≥ τ (mn) , for all m, n ∈ N⋆ .
(5)
In [9] are establish the following inequalities:
2
τ (m) < m 3 , f or any m > 12 ,
(6)
(7)
2
ln m
, f or any m ≥ 3 .
ln ln m
In this paper, we establish some new inequalities for the function τ .
lnτ (m) < 1, 066
Main results
We can remark several properties of these functions for two natural non-zero
numbers, m and n.
Theorem 2.1.
(8)
(9)
a) τ (mn) ≤ τ (m)n , for all m, n ∈ N⋆ ,
b) n|m , atunci
τ (n)
τ (m)
≤
, for all m, n ∈ N⋆ .
m
n
67
Several inequalities about ...
Proof. We will show that τ (m) ≤ m , for all m ∈ N⋆ . From the inequality
√
√
(4), τ (m) ≤ 2 m, but m ≥ 2 m for m ≥ 4, therefore τ (m) ≤ m , m ≥ 4 .
For m ∈ {1, 2, 3} it is easy to see that the inequality is true.
From the inequality (5), τ (m)τ (n) ≥ τ (mn) , for all m, n ∈ N⋆ , but
τ (n) ≤ n , so τ (mn) ≤ τ (m)n , for all m, n ∈ N⋆ .
Because n|m, we have m = nd, and from the inequality (8) we obtain
τ (nd) ≤ τ (n)d , which is equivalent with nτ (m) ≤ ndτ (n) = mτ (n) .
Corollary 2.1. We have
(10)
(11)
a)
τ (m) + τ (n)
τ (mn)
≤
, f or all m, n ∈ N⋆ ,
mn
m+n
b)τ (mn) ≤
m2 τ (n) + n2 τ (m)
, f or all m, n ∈ N⋆ .
m+n
Proof. We apply the inequality (8) and we deduce (m + n)τ (mn) =
mτ (mn) + nτ (mn) ≤ mnτ (m) + mnτ (n) = mn(τ (m) + τ (n)) , which means
that the proof is complete.
Similarly, we prove the inequality (m + n)τ (mn) = mτ (mn) + nτ (mn) ≤
m2 τ (n) + n2 τ (m) , consequently the inequality (11).
Theorem 2.2.
(12)
τ ((m, n))τ ([m, n]) = τ (m)τ (n) , for all m, n ∈ N⋆ ,
where (m, n) is the greatest common divisor of m and n and [m, n] is the
least common multiple of m and n.
Proof. Let m and n be two natural non-zero numbers. We will factorize
the numbers m and n in prime factors, thus:
m = pα1 1 pα2 2 ...pαk k ·q1β1 q2β2 ...qsβs , n = pγ11 pγ22 ...pγkk ·r1δ1 r2δ2 ...rtδt , qj 6= rl , for all
k
s
Q
Q
j ∈ {1, ..., s}and f or all l ∈ {1, ..., t} , therefore τ (m) = (αi + 1) (βj +
i=1
1) , τ (n) =
k
Q
i=1
(γi +1)
t
Q
l=1
(δl +1) , we obtain τ ((m, n)) =
k
Q
i=1
j=1
(min{αi , γi }+1)
68
Nicuşor Minculete,Petrică Dicu
k
Q
and τ ([m, n]) =
i=1
(max{αi , γi } + 1)
s
Q
(βj + 1)
j=1
t
Q
(δl + 1) , which means
l=1
that τ ((m, n))τ ([m, n]) = τ (m)τ (n) , f or all m, n ∈ N⋆ .
Theorem 2.3.
τ 2 (mn) ≥ τ (m2 )τ (n2 ) , f or all m, n ∈ N⋆ .
(13)
Proof. We consider m =
k
Q
i=1
mn =
k
Q
i=1
k
Q
pαi i +γi
(γi +1)
t
Q
s
Q
j=1
β
qj j
t
Q
l=1
pαi i
s
Q
j=1
β
qj j , n =
rlδl , hence τ (m) =
(δl +1) , therefore τ (mn) =
k
Q
k
Q
i=1
k
Q
pγi i
t
Q
l=1
rlδl , which means that
(αi +1)
i=1
s
Q
(βj +1) and τ (n) =
j=1
s
Q
(αi +γi +1)
(βj +1)
t
Q
(δl +1) ,
k
k (α + 1)(γ + 1)
Q
Q
αi γi
i
i
= τ (mn)·
1+
≥
so τ (m)τ (n) = τ (mn)·
αi + γi + 1
αi + γi + 1
i=1
i=1
k
s
k
Q
Q
Q
τ (mn) . Because τ (m2 ) =
(2αi + 1) (2βj + 1) and τ (n2 ) =
(2γi +
i=1
i=1
l=1
i=1
1)
t
Q
j=1
j=1
l=1
i=1
(2δl + 1) , we obtain the equality:
l=1
τ (m2 )τ (n2 ) =
but τ 2 (mn) =
k
Q
i=1
k
Q
(2αi + 1)
s
Q
(2βj + 1)
j=1
(αi +γi +1)2
k
Q
(2γi + 1)
i=1
s
Q
(βj +1)2
t
Q
t
Q
(2δl + 1) ,
l=1
(δl +1)2 . It is easy to see the
i=1
j=1
l=1
s 2
2
k
β
Q
Q
(α
+
γ
)
i
i
j
·
·
1+
equality τ 2 (mn) = τ (m2 )τ (n2 )·
1+
(2αi + 1)(2γi + 1) j=1
2βj + 1
i=1
t
Q
δl2
1+
.
2δl + 1
l=1
βj2
(αi + γi )2
≥ 1 , for all i = 1, k , 1 +
≥
Since 1 +
(2αi + 1)(2γi + 1)
2βj + 1
δl2
1 , for all j = 1, s , 1 +
≥ 1 , for all l = 1, t , we obtain τ 2 (mn) ≥
2δl + 1
τ (m2 )τ (n2 ) .
Theorem 2.4. Let m and n be two natural non-zero numbers, then τ (mn) ≤
√
√
n m + m n.
Several inequalities about ...
69
√
Proof. We apply the inequality (4) for m and n, we have nτ (m) ≤ 2n m
√
and mτ (n) ≤ 2m n . By adding the inequalities, we obtain
(14)
√
√
nτ (m) + mτ (n) ≤ 2n m + 2m n ,
but using the inequality (8), we have τ (mn) ≤ τ (m)n and τ (mn) ≤
τ (n)m , f or all m, n ∈ N⋆ , we deduce
(15)
2τ (mn) ≤ τ (m)n + τ (n)m ,
so, from the inequalities (14) and (15), we obtain the inequality
√
√
τ (mn) ≤ n m + m n .
References
[1] Acu D., Aritmetică şi teoria numerelor, Editura Universităţii ”Lucian
Blaga” din Sibiu, 1999 (in romanian).
[2] Alexandru, V. and Goşonoiu, N., M., Elemente de teoria numerelor,
Editura Universităţii din Bucureşti, 1999 (in romanian).
[3] Creangă, I. and col., Introducere in teoria numerelor, Editura Didactică
şi Pedagogică, Bucureşti, 1965 (in romanian).
[4] Minculete N. şi Dicu P.,Concerning the Euler totient, General Mathematics, vol. 16, No. 1 (2008), 85-91.
[5] Panaitopol, L. and Gica, Al., Probleme celebre de teoria numerelor,
Editura Universităţii din Bucureşti, 1998 (in romanian).
[6] Panaitopol, L. and Gica, Al., O introducere in aritmetică şi teoria
numerelor, Editura Universităţii din Bucureşti, 2001 (in romanian).
70
Nicuşor Minculete,Petrică Dicu
[7] Panaitopol, L. and Gica, Al., Probleme de aritmetică şi teoria numerelor, Editura GIL, Zalău, 2006 (in romanian).
[8] Sandor, J., Geometric Theorems, Diophantine Equations and Arithmetic Functions, American Research Press, Rehoboth, 2002.
[9] -http://www.nationmaster.com/encyclopedia
1
University ”Dimitrie Cantemir” of Braşov
Department of REI Str. Bisericii Române, Nr. 107,
Braşov, Romania.
E-mail address: minculeten@yahoo.com
2
University ”Lucian Blaga” of Sibiu
Department of Mathematics
Str. Dr. I. Raţiu, Nr. 5–7,
550012 - Sibiu, Romania.
E-mail address: petrica.dicu@ulbsibiu.ro