On a problem of Nicol and Zhang

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1 Journal of Number Theory On a problem of Nicol and Zhang Florian Luca a, József Sándor b, a Instituto de Matemáticas, Universidad Nacional Autonoma de México, CP 58089, Morelia, Michoacán, Mexico b Department of Mathematics, Babes-Bolyai University, Str. Kogalniceanu, Cluj-Napoca, Romania Received 3 October 2006; revised 9 March 2007 Available online 24 May 2007 Communicated by David Goss Abstract In this note, we look at the composite integers n which divide φn+ σn Elsevier Inc. All rights reserved. For a positive integer n, letφn and σn be the Euler function of n and the sum of divisors function of n, respectively. Recall that φn= p α p and σn= p α n p α n p α+ p. We also put ωn = p n and Ωn = p α n α for the number of prime divisors and prime power divisors of n, respectively. Note that if n is prime then n φn+ σn. Here, we put A = { n composite: n φn+ σn }, and study the elements of A. It is known that n is prime if and only if 2n = φn+ σn. Hence, n A if and only if ln = φn + σn with some l 3. In particular, A contains no perfect number since if n is perfect, then σn= 2n and φn < n, therefore 2n<φn+ σn<3n. C.A. Nicol [] was the first to study the positive integers n A when l is fixed. He showed that they cannot be square-free and conjectured that they are all even. He also showed that if α is * Corresponding author. addresses: fluca@matmor.unam.mx F. Luca, jsandor@math.ubbcluj.ro J. Sándor X/$ see front matter 2007 Elsevier Inc. All rights reserved. doi:0.06/j.jnt

2 F. Luca, J. Sándor / Journal of Number Theory such that p = 2 α 2 7 is prime, then n = 2 α 3 p is in A. Since it is likely that there are infinitely many primes of the form 2 α 2 7 for some positive integer α, it is unreasonable to expect that one may succeed in proving that there are only finitely many positive integers n A with ωn = 3. In [5], Zhang showed that A contains no elements of the form p α q, where p and q are distinct primes and α is a positive integer. See [2, p. 203] for more information. Here, we look at the positive integers n A with a fixed number of prime factors. We have the following results. Proposition. If K 2 is any fixed positive integer then A contains only finitely many positive integers n with Ωn K. Our next results answer an open question from [5]. Proposition 2. There is no positive integer n A with ωn = 2. While we cannot prove Nicol s conjecture, we have the following partial result. Proposition 3. For any fixed positive integer K 2 there are only finitely many odd positive integers n A with ωn = K. Using the ideas from the proof of the above results, we can give a complete characterization of those n A such that ωn = 3. Proposition 4. If n A is such that ωn = 3, then either n = 2 α 3 p with p = 2 α 2 7 prime, or n {560, 588, 400}. While we cannot prove that there are infinitely many composite integers in A, we can at least prove that there are not too many of them. For a positive integer x put Ax = A [,x]. Proposition 5. The estimate holds as x. #Ax x exp 2 /2 + o log x log log x 2 By partial summation, it follows easily that n A n <. We point out that A contains only 7 elements 0 7 all of which are even.. Proof of Proposition Let μn be the Möbius function of n which equals zero if p 2 n for some prime p and ωn otherwise. Assume that Ωn K. Letm be the number of divisors d of n with

3 046 F. Luca, J. Sándor / Journal of Number Theory μd. Clearly, m τn 2 Ωn 2 K. Here, τn is the total number of divisors of n. Write ln = φn+ σn. Then l = φn n + σn n = d n μd d + d n d n = d n μn/d n/d + d n n/d = d n μd +. d Noting that μd + {0,, 2} and it is non-zero precisely for m divisors d of n, we get an equation of the form m a j n j = l, where a j {, 2} and n < <n m = n are positive integers. We show by induction that n i 2m 2i holds for all i =,...,m. Indeed, 2m n m a j n j = l, therefore n 2m, which proves the above assertion at i =. Assuming now that i>, we have 2m 2m i + > n i n i m j=i a j i a j = l. n j n j The right-hand side of the above equation is a positive rational number, therefore it is at least as large as /n n i. Thus, 2m >, n i n n i leading to n i < 2mn n i <2m i 2 = 2m 2i, which completes the proof of the induction step. Hence, n = n m 2m 2m 2 K+22K.

4 F. Luca, J. Sándor / Journal of Number Theory Proof of Proposition 2 We write n = p α pα 2 2, where p <p 2.Ifp 3, then σn n n φn = p p p 2 p < 2, therefore σn<2n. Since φn < n, we get that φn+ σn<3n, which contradicts the fact that l 3. Thus, p = 2. In particular, If p 2 5, then φn n = p 2 2p 2 < 2. σn n n φn = 2p 2 p 2 5 2, therefore φn + σn < n/2 + 5n/2 = 3n, which is again a contradiction. Thus, p 2 = 3 and therefore n = 2 α 3 α 2 for some positive integers α and α 2.Now which gives φn+ σn= 2 α 3 α2 + 2 α+ 3 α2+, 2 2 α + 3 α 2 l = 2 α + 3 α α + 3 α 2+ 2 α + 3 α Putting u = 2 α + and v = 3 α 2, we get so 3uvl = uv + 9uv u 9v +, uv3l 0 = u 9v +. Since l 3 but the right-hand side of the above equation is negative, we get that l = 3, so uv = u + 9v. Hence, u = 9v v = v note that v, therefore v 8. This leads to the possibilities v = 3 or 9, which lead to u = 3 and 0, respectively, which are not convenient because none of them is a power of 2.

5 048 F. Luca, J. Sándor / Journal of Number Theory Proof of Proposition 3 We start with the following lemma. Let S be a fixed finite set of primes. A positive integer n is called an S-unit if all its prime factors are in S. Lemma. Given any finite set of primes S, there are only finitely many S-units in A. Proof. Let n = p α pα k k.wefixp < <p k in S. LetX = max{α i : i =,...,k}. Wefix i {,...,k} such that X = α i note that both the k primes p < <p k in S and the index i can be fixed in only finitely many ways. Since pi X n and pi X φn, the relation leads to p X i σn. However, σn= n φn+ σn k α p j + j p j Hence, pi X k p α j + j. Letν p m be the p-adic order of the positive integer m; i.e., the order at which p appears in the factorization of m. The above relation implies that X k. ν pi p α j + j. 3 For each j i, letf j be the multiplicative order of p j modulo p i if p i > 2; i.e., the smallest positive integer k such that p i p k j. When p i = 2, we take f j = 2 for all j i. By Fermat s little theorem, f j p i ifp i > 2. Further, it is known see e.g. [3] that the inequality ν pi p t j ν pi p f j j + ν pi t ν pi p f j j + log t log p i 4 holds for all positive integers t. Hence, using inequality 4 for each one of the terms appearing in the right-hand side of inequality 3 as well as the fact that α j X for all j =,...,k,we arrive at: X k j i ν pi p f j j + k logx +. 5 log p j The above inequality certainly shows that X is bounded by some number which can be computed in terms of the elements of S only, which completes the proof of this lemma. Recall that a unitary divisor of a positive integer n is a divisor u of n such that u and n/u are coprime. j i

6 F. Luca, J. Sándor / Journal of Number Theory Lemma 2. If n A is odd then there do not exist coprime divisors u and v> of n with u unitary such that l = φuv uv + σuv uφv. 6 Proof. If u =, then writing x = φv/v, we get that x is a rational solution of the equation x 2 lx + = 0. Thus, it is an algebraic integer, hence an integer, which is impossible since x 0,. From now on, we assume that relation 6 holds with some u>. Let again n = p α pα k k, where 2 <p < <p k are primes. Assume that 6 holds for some divisors u and v of n. Clearly, we may assume that v is square-free. So, let I and J are non-empty disjoint subsets of {,...,k} such that relation 6 holds with u = i I p α i i and v = j J p j. Putting x = φv/v, we get therefore The two roots of the above quadratic are l φu u x σu ux = 0, φu u x2 lx + σu u = 0. x,2 = l ± l 2 4φuσ u/u 2. 2φu/u Since φuσ u/u 2 <, φv/v = x< and l 3, we get that the solution x with the + sign exceeds 3 + 5/2 > >x. Hence, Further, note that x = l l 2 4φuσ u/u 2 2φu/u = lu lu 2 4σ uφu 2φu = l = φuv uv 2σu lu + lu 2 4σ uφu. 7 + σuv uv < + uv φuv,

7 050 F. Luca, J. Sándor / Journal of Number Theory therefore l < uv φuv. 8 Let l 0 = ν 2 l and let λ = ν 2 2φuv. We first show that λ l Writing s for the cardinality of I J, it follows that ωuv = s, and all prime factors of uv are odd. Hence, λ + ωuv = + s. So, it suffices to show that s l 0 +. Inequality 8 shows that 2 l 0 l < uv φuv p p p s p s s s + = s + 2, therefore s 2 l 0 2. Since the inequality 2 l 0 l holds for all l 0 3, it follows that it suffices to assume that l 0 2, and it suffices to show that s 3. If l 0, we only need that s 2, which is certainly true because u> and v> are coprime. Finally, if l 0 = 2 and s = 2 we then have 3 l < uv φuv = 5 8 < 2, which is false. Hence, ν 2 2φuv ν 2 l + 2. Now rewrite Eq. 7 as 2φvu= v lu lu 2 4φuσ u. Since v is odd, we get that ν 2 lu lu 2 4φuσ u = ν 2 2φuv 2 + ν2 l = 2 + ν 2 lu. On the other hand, the relation ν 2 a b = min{ν 2 a, ν 2 b} holds provided that a and b have different 2-adic valuations. Thus, we conclude that it must be the case that ν 2 lu = ν 2 lu 2 4φuσ u note that lu 2 4φuσ u is an integer, so it makes sense to speak about its 2-adic valuation. Further, it also follows that ν 2 lu + lu 2 4φuσ u = ν 2 lu +. In the above, we used the fact that if A = 2 β X and B = 2 β Y, where X and Y are odd integers and 2 β+2 A B, then 4 X Y, therefore 2 X + Y ; hence, β + = ν 2 A + B. We now use again Eq. 7 under the form φv v = 2σu lu + lu 2 4φuσ u, to get φv lu + lu 2 4φuσ u = 2vσ u.

8 F. Luca, J. Sándor / Journal of Number Theory By computing 2-adic valuations and using again that v is odd we get that + ν 2 σu = ν2 2vσu = ν2 φv + ν2 lu + lu 2 4φuσ u giving 2 + ν 2 lu, ν 2 σu ν2 lu +. Since v is odd, we get that ν 2 lu = ν 2 n. Further, since u is a unitary divisor of n, wehave σu σn, therefore ν 2 σn ν2 σu ν2 ln +. 9 Finally, v> and odd, and since uv n, we get that φuv φn, therefore ν 2 φn ν2 2φuv ν2 l + = ν 2 ln +. 0 Comparing divisibilities 9 and 0, we see that in the formula ln = φn + σn, the righthand side of it is divisible by a larger power of 2 then the left-hand side of it, which is the final contradiction. Proof of Proposition 3. We may assume that k 3. Further, by Lemma, it suffices to show that the largest prime factor p k of n is bounded. We now start looking at the small primes. Clearly, φn<n and σn n n φn = k i= + < + k k < exp. p i p p The inequality e x < + 2x holds for all x 0, /2. Thus, if p > 2k, then therefore σn n < + 2k p, 3n ln = φn+ σn<2n + 2kn p < 3n, which is a contradiction. Hence, p 2k +. We now prove recursively on i =,...,k, that there is a function f i k, which we will explicitly give, such that p i f i k. Clearly, we take f k = 2k +. Assume that i k, and that f j k have been constructed such that p j f j k for all j =,...,i. Write n i = j i pα j j, m i = n/n i and rewrite the equation ln = φn+ σn,

9 052 F. Luca, J. Sándor / Journal of Number Theory as l φn i σn i = φn i φmi + σn i σmi. n i n i n i m i n i m i Here, we distinguish two cases. Case. The left-hand side of relation is 0. Equation then leads to Note that and σm i /m i m i /φm i φn i σn i φn i n i m i φm i = < σm i m i k j=i+ k j=i+ k j=i+ k k j=i+ + k j=i+ p j + pj + p j + + pj I {i+,...,k} #I> + 23k+ p j pi+ 2. i + pj I {i+,...,k} I j I k j=i+ + 23k p 2 i+ f j k k j=i+ := g i k. p j, p j pj 2 2 p 2 j I j j/ I p j + 23k p 2 i+ + pj < k j=i+ + 2k + 2 3k p j pi+ 2 Hence, putting L = k j=i+ /p j, we get that σm i /m i m i /φm i L L + 2 3k+ /pi+ 2 = + 2 3k+ /pi+ 2 L.

10 F. Luca, J. Sándor / Journal of Number Theory Since p 2 i+ L p i+, we get that so so we may take f i+ k 2 3k+ / g i k. g i k σm i/m i m i /φm i + 2 3k+ /p i+, p i+ g ik2 3k+ g i k < 23k+ g i k, Case 2. The left-hand side of relation is positive. Here, we put h i k = i f j k. To simplify notations, we let A = h i k 2 and B = 2 2i.We put q j = p α j j for all j =,...,i.welets i be maximal such that there exist a set of s distinct indices {j,...,j s } all i such that the inequalities q jt 2AB 2t hold for all t =,...,s. By the maximality of s, we get that the inequality q j >2AB 2s+ holds whenever j i is not in {j,...,j s }. Note that both instances s = 0ifallq j >2AB 2 for j =,...,i and s = i can actually occur. We now look for a lower bound on the expression appearing in the left-hand side of relation. Note that l φn i n i σn i n i = l i = l φc C p j p j σd D i pi j i j/ {j,...,j s } where C = j i p j, and D = s t= p α jt j t. Hence, putting we may write E = j i j/ {j,...,j s } p j, l φn i n i σn i n i = F G, p i p i q i pj p j, q j p j

11 054 F. Luca, J. Sándor / Journal of Number Theory where F = l φc C σde DφE, and G = σd D j i j/ {j,...,j s } pj p j q j p j E. φe Let us take a closer look at the number F.Ifs = i, then G = 0 and so F 0 because we are in Case 2. If s<i, then C>, φc/c = φde/de, D is a unitary divisor of n, E> and D and E are coprime, and now the fact that F 0 follows from Lemma 2. Thus, F 0. Now the denominator of F is j i p j p j In particular, when s = i we have s q jt = A2AB t A2AB 2s+. t= l φn i σn i = F n i n i A2AB >. 2 2i+ 2AB 2i+ If s<i,letj 0 be such that q j0 = min{q j : j i, j / {j,...,j s }}. Then G 2i p j 22i. q j0 p j q j0 j i Since q j0 >2AB 2s+, we get that F A2AB 2s+ = 2B 22i+ = 2AB 2s+ q j0 Since F G is positive, we get that F is positive, and so > 2 G. l φn i σn i = F G F n i n i 2 2AB. 3 2s+ 2AB 2i+ Comparing estimates 2 and 3, we get that estimate 2 is always true. Returning to, we get that 2AB <l φn i σn i < n i σmi, 4 2i+ n i n i φn i m i

12 F. Luca, J. Sándor / Journal of Number Theory where we used the fact that the left-hand side of estimate is positive, estimate 2, as well as the facts that φm i /m i and σn i /n i <n i /φn i. Since σm i m i m i φm i k i k i + < exp, p i+ p i+ we get again from the inequality e x < 2x for x 0, /2, that if p i+ > 2k +, then σm i m i which combined with 4 leads to < exp k p i+ p i+ 2k2AB 2i+ +. 2k p i+, Hence, we may take f i+ k 2k2AB 2i+ +. Thus, we showed recursively that if we let f k = 2k + and f i+ k be defined for i as { 2 i f i+ k = max 2k 2 2i f j k i+2, 2 3k+ i }, f j k then the inequality p i+ f i+ k holds, which concludes the induction and the proof of Proposition Proof of Proposition 4 We write n = p α pα 2 2 pα 3 3. We first show that unless n = 2α 3 p 3, where p 3 = 2 α 2 7 is a prime, then p = 2, p 2 {3, 5} and p We next show that if X = max{α,α 2,α 3 }, then X 9 after which a quick computation finishes the proof. We start with p.ifp 5, then 3 l = φn n which is impossible, while if p = 3, then + σn n < + n φn < 2.6, 3 l = φn n + n φn < < 2.86, which is again impossible. Hence, p = 2. If p 2, then which is impossible, while if p 2 = 7, then 3 φn n + n φn < 2.39, 3 φn n + σn n < 2.996,

13 056 F. Luca, J. Sándor / Journal of Number Theory which is again impossible. Thus, p 2 {3, 5}. When p 2 = 5, we have 3 φn n + n φn p 3 p 3, which leads to p From now on, we assume that p 2 = 3. If p 3 7, then l φn n + n φn < Hence, either l = 3, or l = 4 and p 3 = 5. From now on, we assume that l = 3. Then ln = 2 α 3 α 2+ p α 3 3, φn= 2 α 3 α 2 p α 3 3 p 3, σn= σ 2 α 3 α 2 p α pα Thus, the equation ln = φn+ σn can also be written as where p α 3 3 Ap 3 + B = C p α , 5 A = 2 α 3 α 2+ 2 α 3 α 2 σ 2 α 3 α 2 = 3α 2+ 2 B = 2 α 3 α 2 σ 2 α 3 α 2, C = σ 2 α 3 α 2 = 2 α + 3 α α 3 α 2, It is clear that the right-hand side of Eq. 5 is 0 and is zero if and only if α 3 =. Further, note that B<0 because 2 α 3 α 2 is a divisor of 2 α 3 α 2, therefore 2 α 3 α 2 <σ2 α 3 α 2. Thus, A>0. We now distinguish the following instances. Assume that α 2 =. Then A = 4, B = 2 α If α 3 =, we then get that Ap 3 + B = 0, therefore p 3 = 2 α 2 7. Assume now that α 3 >. Then C = σ2 α 3 α 2 = 42 α + and p 3 C. Hence, p 3 2 α +. If α =, we then get p 3 3, which is impossible. If α = 2, we then get that p 3 7. Assume now that p 3 > 7. Then α 3. Since Ap 3 + B>0, we get that p 3 > 2 α 2 7. Hence, p 3 2 α because α 3 and p 3 is odd, therefore p 3 2 α 2 7. However, we also have that α p 3 2 Ap 3 + B = C p α 3 3 < 42α α 3 < 32 7 < 5, C p 3 pα3 3 p α 3 3

14 F. Luca, J. Sándor / Journal of Number Theory which gives that p 3 < 2 α , contradicting the fact that p 3 2 α Assume now that α 2 >. Then the condition A>0 is equivalent to 2 α + < 3α 2+ 3 α 2. 6 The maximum of the function appearing in the right-hand side of the above inequality 6 for α 2 2 is 3 and is achieved when α 2 = 2. Hence, 2 α + 0, leading to α =, 2. Assume that α =. Then A = 5 3 α 2 + 3/2, B = 23 3 α 2 3/2, and C = σ2 3 α 2 = 33 α 2+ /2. Furthermore, either α 3 = and p 3 = B/A,orα 3 > and If α 3 =, then putting x = 3 α 2, we get that 0 Ap 3 + B C p 3 pα3 3 p α 3 3 < C p 3. p 3 = 23x 3/2 5x + 3/2 = 23x 3 5x + 3. Hence, 5x x 3 235x , leading to 5x Since x = 3 α 2 is a power of 3, we get that 5x/3 + is a divisor of 28, and x/3 = 3 α 2 2 is also a power of 3. However, 28 has no such divisors. Thus, α 3 > and since p 3 4, we get p 3 < B A + C 4A = 23x 3/2 5x + 3/2 + 39x /2 4 5x + 3/2 < = 5.95, so p 3 5. Finally, assume that α = 2. Then A = 3 α 2 + 7/2, B = 55 3 α 2 7/2 and C = σ4 3 α 2 = 73 α 2+ /2. We again get that either α 3 = and p 3 = B/A,orα 3 > and 0 <Ap 3 + B< C p 3. If α 3 =, we then get, again with 3 α 2 = x, that p 3 = 55x 7/x +7. Hence, x +7 55x 7. Since 55x 7 = 55x , we get that x , but there is no divisor of 392 of the form 3 α for some integer α 2 2. Hence, therefore p p 3 < B A + C 55x 7 79x /2 + < A x + 7 2x = 70.75,

15 058 F. Luca, J. Sándor / Journal of Number Theory The above analysis shows that it suffices to look for the numbers n A with p = 2, p 2 = 3, 5 and p To bound X, we use the inequality 5 together with the fact that ν 2 p ± 5 for all odd primes p 67, and if p is any odd prime 67, then ν p q p 5 for all primes q 67 as well. Hence, inequality 5 immediately shows that X logX +, log 2 giving X 9. A computation with Mathematica now shows that the only possibilities are the ones given in the statement of Proposition Proof of Proposition 5 Let x be large positive real number and y = yx be a function depending on x which tends to infinity in a way we will make more precise later. Put Pn for the largest prime factor of n and Ψx; y = #{n x: Pn y}. It is known see Section III.5.4 in [4], that the estimate Ψx,y x exp + o u log u 7 holds as x tends to infinity with u = log x/log y uniformly when exp log log x 5/3+ε y x. 8 From now on, we count only positive integers n x such that Pn>y.IfPn 2 n, it follows that n is such that p 2 n for some prime p y. Forafixedprimep, the number of such n x does not exceed x/p 2. Summing up over p x /2, we get that the totality of such n x does not exceed y p x /2 x p 2 x x /2 y dt t 2 x y. 9 Finally, let us count the number of n Ax with Pn > y and Pn 2 n. Since φn < n and σn/n log log n log log x, it follows that there exists a constant c such that if ln = φn + σn and n x, then l c log log x. We fix the number l. Further, we write n = mp, where p = Pn, and we note that m and p are coprime. We fix m. Note that m x/y. Then the equation ln = φn+ σn leads to lm φm σm p = σm φm. Since m> because n is not prime, we get that σm>φm. Hence, lm φm σm is non-zero and Pn= p = σm φm lm φm σm.

16 F. Luca, J. Sándor / Journal of Number Theory Thus, p = Pn is defined uniquely by m and l. Since m x/y and l c log log x, it follows that the totality of such n does not exceed Comparing the three bounds 7, 9 and 20, we see that c x log log x. 20 y #Ax x exp + o u log u + and the optimal bound that follows from this argument arises when This gives x exp + o u log u = x log log x, 2 y x log log x. y + o u log u = log y + Olog log x, therefore that + o log x log y log log x log y = log y + Olog log x. Solving for y versus x we get log y = 2 /2 + o log x log log x /2, as x which satisfies 8 with ε = forlargex, and now estimate 2 and the above choice of y lead to the bound claimed by Proposition 5. Acknowledgments We thank the referee for suggestions which improved the quality of this paper. We also thank Professor William D. Banks for his help with the computations. This work was done in May of 2006, while the first author was in residence at the Centre de Recherches Mathemátiques of the Université de Montréal for the thematic year Analysis and Number Theory. This author thanks the organizers for the opportunity of participating in this program. During the preparation of this paper, F.L. was also supported in part by grants PAPIIT IN04505, SEP-CONACyT and a Guggenheim Fellowship and J.S. was partially supported by grant Nr. 88/2005 of Sapientia Foundation from Cluj, Romania. References [] C.A. Nicol, Some Diophantine equations involving arithmetic functions, J. Math. Anal. Appl [2] J. Sándor, B. Crstici, Handbook of Number Theory II, Springer-Verlag, [3] J. Sándor, D.S. Mitrinovic, B. Crstici, Handbook of Number Theory I, Springer-Verlag, [4] G. Tenenbaum, Introduction to Analytic and Probabilistic Number Theory, Cambridge University Press, 995. [5] M. Zhang, On a divisibility problem, J. Sichuan Normal Univ. Nat. Sci. Ed