A New Characteristic of the Identity Function

Journal of Number Theory  NT2108 journal of number theory 63, 325338 (1997) article no. NT972108

A New Characteristic of the Identity Function Jean-Marie De Koninck* Departement de Mathematiques et de Statistique, Universite Laval, Quebec G1K 7P4, Canada

and Imre Katai - and Bui Minh Phong Computer Algebra Department, Eotvos Lorand University, Muzeum krt. 68, 1088 Budapest, Hungary Received March 4, 1996; revised November 5, 1996

In 1992, C. Spiro [7] showed that if f is a multiplicative function such that f (1)=1 and such that f ( p+q)= f ( p)+ f (q) for all primes p and q, then f(n)=n for all integers n1. Here we prove the following: Theorem. Let f be a multiplicative function such that f (1)=1 and such that f( p+m 2 )= f ( p)+ f (m 2 )

for all primes p and integers m1,

(1)

then f (n)=n for all integers n1. Proof.

First we show that f ( p 2 )= f ( p) 2.

(2)

Indeed, using (1) and the fact that f is multiplicative, we have f ( p)+ f ( p 2 )= f ( p+ p 2 )= f ( p(1+ p))= f ( p) f ( p+1 2 ) = f ( p)( f ( p)+1)= f ( p) 2 + f ( p), from which (2) follows immediately. We now show that f (n)=n

for all positive integers n12.

(3)

* Research partially supported by Grants from NSERC of Canada and FCAR of Quebec. Research partially supported by the Hungarian Research Foundation (OTKA).

325 0022-314X97 25.00 Copyright  1997 by Academic Press All rights of reproduction in any form reserved.

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Repeated use of (1) gives f (3)= f (2+1)= f (2)+ f (1)= f (2)+1 and thus f (4)= f (3+1)= f (3)+1= f (2)+2. Then f (6)= f (4)+ f (2)= f (2)+2+ f (2)=2+2f (2), f (7)= f (4)+ f (3)=3+2f (2), f (8)=1+f (7)= 4+2f (2), f (9)= f (4)+ f (5)= f (2)+2+ f (5). Moreover f (11)= f (4)+ f(7)=5+3f (2), while also f (11)= f (9) + f (2) = f (4) + f (5) + f (2) = 2+2f (2)+f (5). Finally f (12)= f (4) f (3)= f (11)+1=6+3f (2), which implies that (2+ f (2)) f (3)=6+3f(2), that is 2f (3)+ f (6)=6+3f (2) and therefore (2+2f (2))+(2+2f (2))=6+3f (2), from which we deduce that f(2)=2. It easily follows from this that f(n)=n successively for n=3, 4, 6, 7, 8, 11, 5, 9, 10, 12. This proves (3). It is clear that the Theorem will follow if we can prove the following: If T is an integer such that f (n)=n for all n
(4)

Because of (3), we can assume that T>12. We proceed by contradiction. Hence assume that (4) is false for a certain T>12, that is that f (n)=n for each positive integer n
(5)

Clearly T+1 is composite. Letting P*(n) denote the largest prime power which divides n, then either T+1 is a prime power or else T+1=AB with 1
with :3 and some prime q.

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IDENTITY FUNCTION CHARACTERISTIC

327

We also note that : must be an odd number. Indeed, if : is even, then f (q : +q)= f (q(q :&1 +1))= f (q) f (q :&1 +1)=q(q :&1 +1)=q : +q,

(6)

while on the other hand f (q : +q)= f (q : )+ f (q)= f (q : )+q{q : +q, which contradicts (6). With the help of a computer, we found all those prime powers r k 10 6, with k3, which cannot be written as r k = p+m 2 ( p is a prime, m an integer), namely 2 6, 5 4, 2 10, 3 8, 5 6, 2 14, 3 10, 2 16, 7 6, 19 4, 2 18, 23 4, 3 12, 29 4 and 31 4. Using the results established above including the induction hypothesis and the fact that each of these 15 numbers r k have an even exponent k, it follows that T>10 6. Our next step is to prove three important lemmas. Lemma 1. f( p)= p.

Assume that T>10 6. Then for all primes p
Proof of Lemma 1. Let p be the smallest prime, if any, for which f( p){p and assume that T
(&=1, 3, 5, ..., [- p], & odd).

Our plan is to show that there exists an odd integer &[- p] satisfying both f(l & )=l & and f (& 2 )=& 2. For such a &, it will follow that f (l & )= f( p)+ f (& 2 )= f ( p)+& 2 {p+& 2 since f ( p){p, thereby contradicting the fact that f (l & )=l & = p+& 2, thus establishing the proof that f ( p)= p. By definition, l & <2p,

(7)

the inequality being strict because - p is not an integer. Now write l & =A & } B & ,

where A & is the largest prime power dividing l & .

(8)

We first look for &'s such that f (l & )=l & . First consider the case where A & is a prime. It is clear that we cannot have A & p; indeed, since B & is even, it would follow from (7) that 2B & =l & A & <2pp=2, a non sense. Hence A &
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using (7), we have Q 2 <2p
\

+

(x59).

On the other hand, one can verify that x 3 x 1+ <(1+') log x 2 log x log x

\

+

with

'=

1 10

(x>3.3_10 6 )

and also, using a computer, that ?(x)<(1+')

x log x

(10 6
It follows that x x
'=

1 , 10

(x>10 6 )

(9)

and since the largest integer ; such that Q ; < p, for ;3, satisfies ;< log plog Q, we may thus conclude that *H<4 : 1+2 2 ;< p ;3

:

1

Q;< p Q>2, ;3

log p log p &2 +2 ?( p 13 )+?(x 14 ) log 2 log 3

\ + \ + 4p log p 3p <4 \ log 2 &2+ +2(1+') \log p+log 3+.

<4

13

14

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(10)

IDENTITY FUNCTION CHARACTERISTIC

329

Since we have already shown that, if &  H, then f (l & )=l & , we now look for odd &'s not exceeding - p with the property f (& 2 )=& 2. We call such a & a good &; the other ones being called bad &'s. Certainly those &'s for which each prime power ? $ dividing exactly & is such that $=1 or ? $ <- T are good. Possible bad &'s must therefore have a prime power ? $ - T (with $2). Hence the number N bad of bad &'s up to - p is small, indeed it is N bad <

: - T? $
-p 1 <- p : . $ ? ?$ - T? $
_ &

(11)

$ 2

Now, for each integer $2, real R2, using Stieltjes integral, then integration by parts and finally (9), we obtain 1 1 $= : ? ?$ ? $ >R ?>R 1$ :

=

=

|

 R 1$

?(t) t$

<&

<&

d?(t) t$

}



+$ R 1$

|

 R 1$

?(t) dt t $+1

?(R 1$ ) +$(1+') R $ R 1&1$ log R

+

|

 R 1$

dt t $ log t

$ 2(1+') . ($&1) R 1&1$ log R

(12)

Note that in the case $=2, we have 4(1+') 2+4' 2 1 + 12 = 12 . < & 12 2 ? R log R R log R R log R ?2 >R :

Observe also that, for $3, we have

&$+

$ 2(1+') 3 < +2($+1) '. $&1 2

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(13)

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DE KONINCK, KATAI, AND PHONG

By treating separately the two cases $=2 and $3, we have : $2

1 1 = : +: $ ? ? 2 $3 R
1 ?$ R
<

2+4' (32)+2($+1) ' + : R 12 log R 3$2 log Rlog 2 R 1&1$ log R

<

2+4' 2((32)+8') + R 12 log R R 23 log 2

Letting R=- T, we obtain : $2

2(2+4') 2((32)+8') 1 + . < 14 $ ? T log T T 13 log 2 - T
(14)

Hence, using (14), inequality (11) can be written as N bad <- p

\

3+16' 8+16' + 16 . 18 p log p p log 2

+

(15)

Hence, combining (10) and (15), and if p>10 6, it follows that [- p]>N bad +*H, which proves that there exists at least one odd integer &[- p] such that f(& 2 )=& 2 and f (l & )=l & = p+& 2 while f (l & )= f ( p)+ f (& 2 )= f ( p)+& 2. Hence if f ( p){p, it would follow that f (l & )= f ( p)+& 2 {p+& 2 = f (l & ), a non sense. The proof of Lemma 1 is thus completed. Lemma 2. then

Let ? be a prime and 2 a positive integer such that ? 2
(16)

Proof of Lemma 2. First assume that ? is odd. Then, for each prime 2< p
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(17)

IDENTITY FUNCTION CHARACTERISTIC

331

If E p
\

T2 ; &? 22, 2 ; , 2

+

where ?(x ; k, l)=*[rx, r prime: r#k (mod l)]. Using the sieve result (see Halberstam and Richert [1], formula (4.10), p. 110) ?(x ; k, l)<

3x ,(k) log(xk)

(1k
we conclude that, since ,(2 ; )=2 ;&1 >T2, M T <3

1 1 3T 3T 4T T2 < = < . 2 ; ; 2 2 log(T 2 } 2 ) ,(2 ) log(T 2 } (32) T) log T3 log T (18)

On the hand, assume that E p =Q ; for some prime Q>2 and integer ;2 and satisfying TT and ;3. This number N T satisfies 3 1 : , NT < T 2 ; 2 Q ; 2 T
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which, in view of (12) and (13) and observing that ; runs in the range 3;<3 log T, gives 3 33 N T < T 2 } 23 . 2 T

(19)

Using (18) and (19), and if T is large enough, we certainly have that ?

1 T 22 T2 >M T +N T , > 2 2 log(T 22)

\ +

which implies that there exists at least one prime pT and #3, satisfies (19). On the other hand, if E p =3 # T for some #3, we then have p+2 22 =3 #F p . Let # 0 be the smallest integer satisfying 3 #0 >T. But the number M T of primes p satisfying 3 #0 | p+2 22 is M T
\

T2 ; &2 22, 3 # 0 , 2

+

and as previously we obtain that MT<

4T . log T

Now, since (see McCurley [4]) %(x ; 2, 3) := px, p#2 (mod 3) log p 0, 49042x holds for x3761, it follows that (log x) ?(x ; 2, 3)>0, 49x in the same range. Hence it is enough to prove that 4T 0, 49T 22 99 2&23 > T + log(T 22) 2 log T

(T10 6 ).

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IDENTITY FUNCTION CHARACTERISTIC

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This clearly holds if 0, 245>

4 log(T 22) T2 50 log + T 23 2 T log T

\ +

(T10 6 ).

(20)

But the right hand side of (20) is certainly non increasing for T10 6 and, on the other hand in that range, log 10 12 6 T2 50 log 10<0, 14 log <50 = 23 T 2 10 4 100

\ +

while 1 4 log(T 22) 8 < < 5, T log T T 10 thereby proving (20). This ends the proof of Lemma 2. As will be seen below, a crucial element in the proof of the Theorem rests on the fact that, given an odd prime q, there exists a prime number p
}

:

}

/(n) .

unv

It is known (see Polya [5]) that W- q log q.

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(21)

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DE KONINCK, KATAI, AND PHONG

Further set f(n) := ` (1+/(?)+/(? 2 )+ } } } +/(? : ))= : /(d ). ?: & n

d|n

Observe that f is a multiplicative function and that, if f (n){0, then for each ?
x x x = : /(d ) & : /(d ) =E&J, d d dx d dx

_&

{=

say. Furthermore write E as 

/(d ) /(d ) /(d ) &x : =xL(1, /)&x : d d d d>x d>x d=1

E=x :

(22)

and define Sx (v) := :

/(n).

x
We have /(d ) = d d>x :

|



x

S x (u) 1 d S x (u)= u u

=0+

|



x

S x (u) du< u2

|

}



+

x

x



W x

|



S x (u) du u2

du W < . u2 x

So far, we have thus established, in view of (22), that |E&xL(1, /)|
(23)

To estimate J, we let y
x

{ d= +: L , m

m

where in L m we sum over those d>y such that [xd]=m. From this it follows that |J | y+

:

|L m |.

m(xy)

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(24)

335

IDENTITY FUNCTION CHARACTERISTIC

Now let I m =[d>y : [xd]=m], and let d 0 be the smallest d # I m and d 1 be the largest d # I m . We write L m = : /(d ) d # Im

\

x x x & + d d1 d1

+ { =

: /(d)=A+B, d # Im

say. First A=

|

d0 d1

\

x x & d S d 1(u) u d1

=S d 1(u)

+

\

x x & u d1

+}

d0

&x d1

|

d0 d1

1 S d 1(u) 1 duW+Wx & 2W. 2 u d0 d1

\

+

From this estimate and the fact that BW, it follows that L m 3W. Using this estimate, (24) becomes x |J |
(25)

We now set x=q 3, in which case and in view of the remark made above on f, we may write S as S=*[n : n 2 q 3 ]+*[n : n 2qq 3 ]=[q 32 ]+q.

(26)

It follows from (23), (25) and (26) that x |L(1, /) q 3 | q+W+ y+3W +q 32. y

(27)

We now look for an optimal choice for y, namely one for which y= 3W (xy ), which means that y=- 3Wx. Hence, from (27), we get |L(1, /) q 3 | q+W+2 - 3 q 32 - W+q 32, which implies using (21) that |L(1, /)| <

2 - 3 - log q . q 54

(28)

We now look for a lower bound for L(1, /) which will contradict (28). First observe that, since q#1 (mod 4), it follows that (&1) (q&1)2 =1 and hence that /(n)=

n

q

\ q+ = \ n+ .

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This implies that, as is mentioned in Davenport [1], the discriminant q is positive. On the other hand, from the Dirichlet class number formula, we have that h(q)=

-q L(1, /), log =

where h(q) is the class number of the field Q(- q) with discriminant q ( >0) and where == 12 (t 0 +u 0 - q) is the smallest solution (with t 0 >0 and u 0 >0) of the Pell equation t 2 &qu 2 =4. Since => 12 (1+- q) and since h(q) is an integer 1, it follows that L(1, /)

log = -q

>

log[ 12 (1+- q)] -q

,

which contradicts (28), since we have assumed that q13. This ends the proof of Lemma 3. We may now complete the proof of the Theorem. For the moment, let us assume that q is odd, and let p be a prime smaller than q 3 such that (&pq)=1. Clearly, if q>3, one can show that such a prime exists by Lemma 3. It means in particular that there exists u 0 # [1, q2] such that &p#u 20

(mod q).

One can then show that, for each :2, there exists an integer v p # [1, q :2] such that (mod q : ).

&p#v 2p

(29)

If q=3, then (29) has a solution for :=2, and then consequently for each :2. Hence, in any case, there exists an integer m p such that v 2p + p=m p q :.

(30)

m p
(31)

First we note that

This is true because mp <

\

q 2: 1 q: +q : : = +1
+

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337

Then write (q : &v p ) 2 + p=q 2: &2q :v p +v 2p + p =q 2: &2q :v p +m p q : =q :(q : &2v p +m p )=M p q :,

(32)

say. Similarly it can be shown that M p
(33)

Similarly, if (M p , q)=1, then f (M p q : )= f ( p)+ f ((q : &v p ) 2 ).

(34)

By hypothesis, we have f ( p)= p, and, because of (31), we have that f(m p )=m p . Since v p and q : &v p are smaller than T, then by Lemma 2, we have f (v 2p )=v 2p , and similarly, if (33) holds, f ((q : &v p ) 2 )=(q : &v p ) 2. Assume that (33) holds, then, since we assumed that f (q : ){q :, we have f(m p q : )= f (m p ) f (q : )=m p f (q : ){m p q :, which contradicts the fact that f (m p q : )= f ( p)+ f (v 2p )= p+v 2p =m p q :. This implies that f (q : )=q :, as we wanted to establish. To complete the proof of the Theorem, it remains to consider the case q=2. We know that &7 is a quadratic residue modulo 2 : and therefore that for each :>3, there exists v : # [0, 2 :&1 ] such that 7+v 2: #0 (mod 2 : ),

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and consequently, 7+(v : +2 :&1 ) 2 #0 (mod 2 : ). Define m : and M : by 7+v 2: =m : 2 :, and 7+(v : +2 :&1 ) 2 =M : 2 :. We easily deduce from these two equations that M : &m : =v : +2 :&2. It follows from this relation and the fact that 2 does not divide v : that v : cannot both be even or odd at the same time, it follows that one of m : or M : is odd, that is that we either have (m : , 2)=1 or (M : , 2)=1, and the rest of the proof can thus be handled similarly as for the case ``q odd'' and we thus omit it.

REFERENCES 1. H. Davenport, ``Multiplicative Number Theory,'' Lectures in Advanced Mathematics, Markham, Chicago, 1967. 2. H. Halberstam and H. E. Richert, ``Sieve Methods,'' Academic Press, New York, 1974. 3. G. H. Hardy and E. M. Wright, ``An Introduction to the Theory of Numbers,'' 4th ed., Clarendon Press, Oxford, 1960. 4. K. S. McCurley, Explicit estimates for %(x ; 3, l) and (x ; 3, l), Math. Comp. 42 (1984), 287296. 5. G. Polya, Uber die Verteilung der quadratischen Reste und Nichtreste, Gottinger Nachr. (1918), 116141. 6. J. B. Rosser and L. Schoenfeld, Approximate formulas for some functions of prime numbers, Illinois J. Math. 6 (1962), 6494. 7. C. Spiro, Additive uniqueness sets for arithmetic functions, J. Number Theory 42 (1992), 232246.

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