General Mathematics Vol. 16, No. 1 (2008), 93-99
Concerning the Euler totient 1
Nicuşor Minculete and Petrică Dicu
Abstract
In this paper we intend to establish several properties for the
Euler totient denoted ϕ. This totient provides us with a series of
arithmetic inequalities in relation with the functions τ and σ respectively, where τ (m) is the number of natural divisors of m, and σ(m)
is the sum of natural divisors of m.
2000 Mathematics Subject Classification: 11A25
Key words: prime number, arithmetic inequality, Euler’s totient.
1
Introduction
Let m be a natural number, m ≥ 1. We note with ϕ(m) the number of
positive integers less than or equal to m that are coprime to m. Hence
ϕ(m) = card{k|(k, m) = 1, k ≤ m}.
The function ϕ so defined is the totient function. The totient is usually
called the Euler totient or Euler’s totient. It follows from the definition
1
Received 15 December 2007
Accepted for publication (in revised form) 22 January 2008
93
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Nicuşor Minculete and Petrică Dicu
that ϕ(1) = 1 and ϕ(pk ) = (p − 1)pk−1 when p is a prime number and
k ∈ N∗ , also, ϕ(2k t) = 2k−1 ϕ(t), when t is odd number.
Remark several properties of these functions, found in some books: [3],
[5], [11], [12], [15].
In paper [7], Gauss proved the equality
X
(1)
ϕ(d) = m.
d/m
ϕ is also a multiplicative function; if (m, n) = 1, then
(2)
ϕ(mn) = ϕ(m)ϕ(n).
For m > 1 and m = pα1 1 p2α2 . . . pαr r we have the relation
(3)
2
ϕ(m) = m(1 −
1
1
1
)(1 − ) . . . (1 − ).
p1
p2
pr
Properties of Euler’s totient
Proposition 1. The number ϕ(m) is odd if m = 1 or m = 2, and even for
every m ≥ 3.
Proof. For m = 1 or m = 2 we have ϕ(1) = ϕ(2) = 1. If m ≥ 3 and
m = pα1 1 pα2 2 . . . pαr r , we obtain
ϕ(m) = pα1 1 −1 pα2 2 −1 . . . prαr −1 (p1 − 1)(p2 − 1) . . . (pr − 1),
and because at least one of the terms p1 , p2 , . . . pr , p1 − 1, p2 − 1, . . . , pr − 1
is even, then ϕ(m) is even.
Proposition 2. a1 + a2 + . . . + aϕ(m) =
mϕ(m)
,
2
for every m ≥ 2.
Proof. Let m be a natural number, m ≥ 2, and a1 < a2 < . . . < aϕ(m) ,
are the natural numbers less than or equal to m that are coprime to m, so
(ai , m) = 1,for all i ∈ {1, 2, . . . , ϕ(m)}.
Concerning the Euler totient
95
As (m − ai , m) = 1, we obtain {a1 , a2 , . . . , aϕ(m) } = {m − a1 , m − a2 , . . . ,
m − aϕ(m) }. Hence, because both sets have the same elements, the sum of
either set is the same, so
a1 + a2 + . . . + aϕ(m) = m − a1 + m − a2 , . . . , m − aϕ(m) ,
which is equivalent to a1 + a2 + . . . + aϕ(m) =
mϕ(m)
.
2
Proposition 3. For every m ≥ 3, show that a1 + a2 + . . . + aϕ(m) ≡
0( mod m).
Proof. Because a1 + a2 + . . . + aϕ(m) = mϕ(m)
and for every m ≥ 3, ϕ(m)
2
is even, then a1 + a2 + . . . + aϕ(m) ≡ 0( mod m).
Proposition 4. ai = m − aϕ(m)−i+1 , for all i ∈ {1, 2, . . . , ϕ(m)}
Proof. From a1 < a2 < . . . < aϕ(m) , we obtain m − aϕ(m) < . . . < m − a2 <
m − a1 , hence, we deduce that ai = m − aϕ(m)−i+1 , for all
i ∈ {1, 2, . . . , ϕ(m)}.
Remark. Because (ai , m) = 1 and ai < m, for all i ∈ {1, 2, . . . , ϕ(m)}, we
deduce that a1 = 1 and aϕ(m) = m − 1.
Proposition 5. For every m ≥ 3, show that a1 < a2 < . . . < a ϕ(m) ≤ m2 ≤
2
a ϕ(m) +1 < a ϕ(m) +2 < . . . < aϕ(m) (without equality in both pairs) or m−a1 >
2
2
m − a2 > . . . > m − a ϕ(m) ≥ m − a ϕ(m) +1 > m − a ϕ(m) +2 > . . . > m − aϕ(m)
2
2
2
(without equality in both parts).
Proof. We may suppose by absurd that a ϕ(m) ≤ m2 and a ϕ(m) +1 < m2 , then
2
2
a ϕ(m) + a ϕ(m) +1 < m, but a ϕ(m) + a ϕ(m) +1 = m, so we find a contradiction;
2
2
2
2
hence, for every m ≥ 3, we obtain a1 < a2 < . . . < a ϕ(m) ≤ m2 ≤ a ϕ(m) +1 <
2
2
a ϕ(m) +2 < . . . < aϕ(m) (without equality in both parts ).
2
Proposition 6. For every m ≥ 3, we obtain
a1 a2 . . . a ϕ(m) ≡ (−1)
2
ϕ(m)
2
a ϕ(m) +1 a ϕ(m) +2 . . . aϕ(m) ( mod m).
2
2
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Nicuşor Minculete and Petrică Dicu
Proof. From Proposition 4 we have ai = m − aϕ(m)−i+1 , for all i ∈
{1, 2, . . . , ϕ(m)}, hence ai ≡ −aϕ(m)−i+1 ( mod m), for all i ∈ {1, 2, . . . , ϕ(m)
},
2
ϕ(m)
consequently a1 a2 . . . a ϕ(m) ≡ (−1) 2 a ϕ(m) +1 a ϕ(m) +2 . . . aϕ(m) ( mod m).
Remark. If
and if
ϕ(m)
2
ϕ(m)
2
2
2
2
is even, then a1 a2 . . . a ϕ(m) ≡ a ϕ(m) +1 a ϕ(m) +2 . . . aϕ(m) ( mod m),
2
2
2
is odd, then a1 a2 . . . a ϕ(m) ≡ −a ϕ(m) +1 a ϕ(m) +2 . . . aϕ(m) ( mod m).
2
2
2
Proposition 7. For every m ≥ 3, we have
a21 +a22 +...+a2ϕ(m) +. . .+2(a1 aϕ(m) +a1 aϕ(m)−1 +. . .+a ϕ(m) ·a ϕ(m) +1 ) =
2
m2 ϕ(m)
.
2
2
Proof.From Proposition 4 we have ai = m−aϕ(m)−i+1 , for all i ∈ {1, 2, . . . , ϕ(m)},
so aϕ(m)−i+1 = m − ai , for all i ∈ {1, 2, . . . , ϕ(m)}, hence 2(a1 aϕ(m) +
ϕ(m)
ϕ(m)
X
X
ai aϕ(m)−i+1 =
ai (m − ai ) =
a1 aϕ(m)−1 + . . . + a ϕ(m) · a ϕ(m) +1 ) =
2
X
ai −
i=1
a2i , but a1 + a2 + . . . + aϕ(m) =
mϕ(m)
,
2
therefore a21 + a22 +
i=1
i=1
... +
X
i=1
ϕ(m)
ϕ(m)
m
2
a2ϕ(m)
+ . . . + 2(a1 aϕ(m) + a1 aϕ(m)−1 + . . . + a ϕ(m) · a ϕ(m) +1 ) =
2
2
m2 ϕ(m)
.
2
Proposition 8. For every i ∈ {1, 2, . . . , ϕ(m)}, we have the inequality
2
ai aϕ(m)−i+1 ≤ m4 .
Proof. It is easy to see that ( m2 −ai )( m2 −aϕ(m)−i+1 ) ≤ 0, which is equivalent
2
to m4 − m2 (ai + aϕ(m)−i+1 ) + ai aϕ(m)−i+1 ≤ 0, but ai + aϕ(m)−i+1 = m,
2
consequently ai aϕ(m)−i+1 ≤ m4 .
Proposition 9. a21 + a22 + ... + a2ϕ(m) ≥
m2 ϕ(m)
.
2
Proof. Form Proposition 7 and Proposition 8 we have clearly Proposition
9, because we have the equality 2(a1 aϕ(m) +a1 aϕ(m)−1 +. . .+a ϕ(m) ·a ϕ(m) +1 ) =
X
2
2
ϕ(m)
ai aϕ(m)−i+1 .
i=1
Remark. Let d1 , d2 . . . , dk be all the divisors of m(d1 = 1, dk = m).
As (ai , m) = 1, the numbers d2 , d3 . . . , dk belong to the set {1, 2, . . . , m}\
{a1 , a2 , . . . , aϕ(m) }, but d1 = 1, dk = m, and 1 < d2 , d3 . . . , dk−1 ≤ m2 , hence
max{a ϕ(m) , dk−1 } ≤ m2 .
2
Concerning the Euler totient
97
Proposition 10. a1 + a2 + . . . + a ϕ(m) + σ(m) ≤
2
Proof. We remark that max{a ϕ(m) , dk−1 } ≤
2
m2 +10m+8
.
8
m
,therefore
2
a1 + a2 + . . . + a ϕ(m) + d2 + d3 + . . . + dk−1 ≤ 1 + 2 + . . . + [
2
m
].
2
so
1 m m
1 m m
m2 + 2m
a1 +a2 +. . .+a ϕ(m) +σ(m)−m−1 ≤ [ ]([ ]+1) ≤ ( )( +1) =
,
2
2 2 2
2 2 2
8
which infers
m2 + 10m + 8
a1 + a2 + . . . + a ϕ(m) + σ(m) ≤
.
2
8
Proposition 11. m2 + m + 2 ≥ mϕ(m) + 2σ(m)
Proof. Because a1 +a2 +. . .+a ϕ(m) +d2 +d3 +. . .+dk ≤ 1+2+. . .+m, and
2
and d2 +d3 +. . .+dk = σ(m)−
form the relations a1 +a2 +. . .+aϕ(m) = mϕ(m)
2
mϕ(m)
m(m+1)
1, we obtain 2 + σ(m) − 1 ≤
,so m2 + m + 2 ≥ mϕ(m) + 2σ(m)
2
Proposition 12. a1 a2 . . . aϕ(m) ≤ m−
τ (m)
2
· m!.
Proof. We remark the inequality
a1 a2 . . . aϕ(m) d1 d2 . . . dk ≤ 1 · 2 · . . . · m = m!,
τ (m)
τ (m)
2
, which infers a1 a2 . . . aϕ(m) ≤ m! · m− 2 .
τ (m)−2
m
Proposition 13. a1 a2 . . . a ϕ(m) ≤ m− 2 · [ ]!.
2
2
Proof. it is easy to see that
m
m
a1 a2 . . . a ϕ(m) d2 d3 . . . dk−1 ≤ 1 · 2 · . . . · [ ] = [ ]!,
2
2
2
but d1 d2 . . . dk = m
τ (m)
τ (m)−2
but d1 d2 . . . dk = m 2 , so d2 d3 . . . dk−1 = m 2 , hence the inequality
τ (m)−2
m
becomes a1 a2 . . . a ϕ(m) ≤ m− 2 · [ ]!.
2
2
Proposition 14. m + τ (m) ≤ σ(m) + 1, for all m ≥ 2.
X
Proof. From Gauss’s Theorem,
ϕ(d) = m, and, as ϕ(m) ≤ m − 1, for
all m ≥ 2, we get m =
X
d/m
ϕ(d) = ϕ(d1 ) + ϕ(d2 ) + ϕ(d3 ) + . . . + ϕ(dk ) ≤ 1 +
d/m
d2 −1+d3 −1+. . .+dk −1 = σ(m)−τ (m)+1, therefore m ≤ σ(m)−τ (m)+1.
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Nicuşor Minculete and Petrică Dicu
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Nicuşor Minculete
”Dimitrie Cantemir” University of Braşov,
Str.Bisericii Române, nr 107,
Braşov - Romania
E-mail: minculeten@yahoo.com
Petrică Dicu
”Lucian Blaga” University of Sibiu
Faculty of Sciences
Department of Mathematics
Str. Dr. I. Raţiu, no. 5–7
550012 Sibiu - Romania
E-mail: petrica.dicu@ulbsibiu.ro