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Full text of "Solution of Two Questions Concerning the Divisor Function And The Pseudo-Smarandache Function"

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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. 


132 


(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 


134 


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 pı =2 or pı =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 


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

[2]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 


135