Smarandache Notions Journal, Vol. 11, No. 1-2-3, Spring 2000, pp. 132-135.

SOLUTION OF TWO QUESTIONS CONCERNING THE DIVISOR FUNCTION AND THE PSEUDO-SMARANDACHE FUNCTION

Zhong Li

Abstract In this paper we completely solve two questions concerning the divisor function and the pseudo Smarandache function. Key words divisor function, pseudo Smarandache function, function-

al equation

1 Introduction

Let N be the set of all positive integers . For any n EN, let (1) d(n)= > 1,

(2) Z(n)=minlala€N,n|> j| Then d(n)and Z (n )are called the divisor function and the pseudo Smarandache function of 7 , respectively, Inf! , Ashbacher posed the follow- ing unsolved questions.

Question 1 How many solutions n are there to the functional equa- tion. (3) Z(n)=d(n),nEN?

Question 2 How many solutions n are there to the functional equa-

tion.

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(4) Z(n)+d(n)=n,n€.2? In this paper we completely solve the above questions as follows. Theorem 1 The equation (3) has only the solutions n =1,3 and 10. Theorem 2 The equation (4) has only the solution n =56.

2 Proof of Theorem 1

A computer search showed that (3) has only the solutions n = 1,3 and 10 with n<10000(see !1/) We now let n be a solution of (3) with 41,3 or 10 . Then we have n > 10000. Let (5) n = pin prz" pr? be the factorization of n .By [2, Theorem 273], we get from (1) and (5) that

(6) d(n)=(r1+1)(r2+1)=(r}+1).

On the other hand, since 2 j=a(at1)/2 for any a CN, we see

from (2) that 1|Z(n)(Z(n)+1)/2. It implies that Z(n)(Z(7) + 1)/2=n.So we have

| 1_1 (7) Z(n)>]2n+7 7 Hence, by (3),(5),(6) and a get (8) 1>v2 Hh 2 -e

itl 2išzir;+1 If p;>3,then from (8) we get p125 and

>AS- ha>,

a contradiction. Therefore, if (8) holds, then either p; =2 orp, =3 .By the same method, then n must satisfy one of the following condi- tions.

(i) p;=2 and r;<4 .

(ii) py=3 and r,=1. However, by (8), we can calculate that n < 10000,a contradiction.

Thus, the theorem is proved.

3 Proof of Theorem 2

A computer search showed that (4) has only the solution n = 56 with n <10000 (see l1). We now let n be a solution of (4) with n56. Then we have n >10000. We see from (4) that

(9) Z(n)=-d(n) (mod n) It implies that. - (10) Z(n)+1=1-d(n) (mod n)

By the proof of Theorem 1,we have n|Z(n)(Z(n)+1)⁄2, by (2). It can

be written as

(11) Z(n)(Z(n)+1)=0 (mod n).

Substituting (9) and (10) into (11), we get

(12) d(n)(d(n)—-1)=0 (mod n).

Notice that d(n)>1 if n >1. We see from (12)that

(13) (d(n))?>n

Let (5) be the factorization of n .By (5),(6) and (13) ,we obtain (14) les

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On the other hand, it is a well known fact that Z(p”) =p’ -1>(r+ 1)? for any prime power p” with p” >32. We find from (14) that k22.

If p,>3,then p; /(r; +1) 25/4>1 for 1=1,2,°--k, It implies that if (14) holds, then either =2 or =3 . By the same method, then n must satisfy one of the following conditions:

(i) py=2, po=3 and (74,72) = (1,1), (2,1), (3,1), (4,1), (5,1), (6,1),(1,2),(2,2), (3,2), (4,2)or (5,2).

(ii) p1=2,p2>3 and r,;SS.

(iii) p) =3 and r,=1. However, by (14), we can calculate that n < 10000,a contradiction. Thus,

the theorem is proved.

References

C. Ashbacher, The pseudo Smarandache function and the classical func- tions of number theory, Smarandache Notions J. ,9(1998) ,78 81.

G. H. Hardy and E. M. Wright, An Introduction to the Theory of Num- bers, Oxford, Oxford Univ. Press, 1937.

Department of Mathematics Maoming Educational College Maoming , Guangdong

P. R. China

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