Multiplicative properties of sets of residues

Transcription

1 Multiplicative properties of sets of resiues C. Pomerance Hanover an A. Schinzel Warszawa Abstract: We conjecture that for each natural number n, every set of resiues mo n of carinality at least n/2 contains elements a,b,c with ab c. It is prove that the set of numbers n failing to have this property has upper ensity smaller than Introuction In a recent paper [2] it has been shown that every set of positive integers with lower asymptotic ensity greater than /2 contains three integers whose prouct is a square. It follows that for every positive integer n, every set of resiues mo n of carinality larger than n/2 contains resiues a,b,c, with abc 2. We conjecture that more is true an every set of resiues mo n of carinality at least n/2 contains resiues a,b,c with ab c. That is, say a set S is prouct free if has no solution with a,b,c S. We say a moulus n has property P if the largest prouct-free subset S of Z n has carinality strictly smaller than n/2. We enote the ring of integers mo n by Z n. Conjecture. Every natural number n has property P. If true, Conjecture is best possible, since for n an o prime, the set of quaratic nonresiues mo n is prouct free an has carinality n /2. Perhaps the following slightly stronger form of Conjecture hols: The largest prouct-free subset of Z n has carinality equal to the maximum value of q /2 n/q, where q runs over the prime powers iviing n. It is easy to see that Z n has a prouct-free subset of this carinality: For q 2 k with k 2 take those resiues mo n of the form 2 i 4j +3 with i < k as a prouct-free set if k, we take the empty set, an for q p k with p o take resiues of the form p i j, where i < k an j is a quaratic nonresiue mo p. In every case if n q, this construction prouces a prouct-free subset of Z q of size q /2 an so, more generally if q an n/q are coprime, the Chinese remainer theorem gives a prouct-free subset of Z n Zq Z n/q of size q /2 n/q. In this paper we make some progress towars Conjecture as follows. Let sn enote the largest square-full ivisor of n an let ωn enote the number of istinct prime factors of n. Theorem. A natural number n has property P if ωsn 5. The first author was supporte in part by NSF grants DMS , DMS-0080.

2 2 C. Pomerance an A. Schinzel Theorem 2. The asymptotic ensity of the set of integers n with ωsn 6 is smaller than In particular, the set of integers failing to have property P has upper ensity at most In the final section we present a secon conjecture an prove that it implies Conjecture. The secon conjecture, which is state in the language of linear programming, may be the preferre vehicle for further progress. We remark that there has been some consieration in the literature of large prouct-free subsets of finite groups. For a recent survey, with pointers to other papers, see [3]. Acknowlegments. We thank the referee for a careful reaing an the eitor for suggesting references [] an [4]. In aition we thank Robin Pemantle for help in recasting our secon conjecture as a linear program. 2. Preliminary results For a natural number n an a prime p, we let v p n be the number of factors p in the prime factorization of n. We introuce some special notation that we will use throughout the paper. Suppose that n,m are coprime natural numbers. We shall later take n square-full an m squarefree, but this is not necessary to assume in this section. We consier the multiplicative monoi Z n Z m, where Z m is the unit group mo m. By the Chinese remainer theorem, Z n Z m may be thought of as {a Z nm : a,m }. For n, let Further, if S Z n Z m, let T n,m T {a Z n Z m : a,n }. S n,m S T S, R n,m R T \S. Lemma. Let n be a natural number. Suppose for each squarefree number m coprime to n, if S Z n Z m is prouct free, then S 2ϕmn, with strict inequality holing in the case m. Then for every squarefree number m coprime to n, we have that mn has property P. Proof. Let m be squarefree an coprime to n. For each j m, let A j {a Z mn : a,m j}. Then A j is a multiplicative monoi with ientity a, where a mo mn/j an a 0 mo j that is isomorphic to Z n Z m/j. If S Z mn is prouct free, then so is S A j for each j m. By hypothesis then, S A j 2ϕm/jn for each j m, with strict inequality holing in the case j m. Thus, S S A j < m 2 n j j m j mϕ 2 mn. We conclue that mn has property P, completing the proof.

3 Multiplicative properties of sets of resiues 3 Lemma 2. Suppose n,m are coprime natural numbers, S Z n Z m is prouct free, an D is a nonempty set of ivisors of n with S for each D. Let σ D /. If ϕn n > 2σ, then S < 2 ϕmn. Proof. We have ϕmn S n R D T Dϕ mn ϕmnσ > 2 ϕmn. Thus, S < 2ϕmn, completing the proof. Lemma 3. Suppose n,m are coprime natural numbers, S Z n Z m is prouct free, an S. Then S 2ϕmn. Further, in the case m, the inequality is strict. Proof. Let s S an let n. Note that multiplication by s is a bijection of T an the image of S uner this map is isjoint from S, that is, it is containe in R. Thus, S 2 T so that S n S T 2 2 ϕmn. n Now assume that m. We nee only show that at least one of the inequalities S 2 T is strict, an inee this is the case for n, since T n. This completes the proof of the lemma. Remark. The multiplication-by-s argument in the proof is use in various guises throughout the paper. Corollary. Suppose n,m are coprime natural numbers, ϕn > 2n, an m is squarefree. Then mn has property P. In particular, every squarefree number has property P. Proof. By Lemma it suffices to consier prouct-freesubsets S of Z n Z m. Lemma2hanlesthe case S an Lemma3hanles thecase S. For any natural number n, let ran enote the largest squarefree ivisor of n an let σn enote the sum of the ivisors of n. Another way of stating Corollary is that if u n/ran an u/ϕu < 2, then n has property P. It is possible to strengthen Corollary to the assertion: σu/u < 2 implies n has property P. However we will not nee this stronger assertion. We o not know how to replace 2 in either of these assertions with any larger number.

4 4 C. Pomerance an A. Schinzel 3. Propositions The heart of our metho is containe in the three propositions in this section. With some effort, they likely can be extene to more complicate cases an so allow an improvement in our main result. It is possible that an ultimate strengthening woul allow a proof of Conjecture. Proposition. Suppose n, m are coprime natural numbers an S is a prouct-free subset of Z n Z m. Suppose too that p is a prime factor of n an that S p. Let D be a nonempty set of ivisors of n not ivisible by p with S for each D, an let σ D/. If ϕn n > p 2pσ, 2 then S < 2 ϕmn. Proof. SupposenotanS isacounterexampleforn. For anyk, letn n 2 p k an let π k be the projection from Z n Z m to Z n Z m given by reucing the first coorinate moulo n. Since π k ab π k aπ k b for each pair a,b Z n Z m, it follows thats π k S is prouctfree. We claim that S is a counterexample for n. Inee, S np k S 2 ϕmn2 p k 2 ϕmn. Further, for D, S implies that S, an S p implies that S p. Since ϕn/n ϕn /n, we have an exact corresponence. In the sequel we will not use the ash an instea we will assume that 2 n for each D an that v p n is very large. In aition, we will enote v p n with the letter k. For a ivisor of n with p an D, consier the sets S p 2i,S p 2i+ for 0 i < k /2. Say s p S p. Multiplication by s p is a p : mapping of T p 2i onto T p 2i+. Since S is prouct free it follows that s p S p 2i is isjoint from S p 2i+. We conclue that mn p S p 2i + S p 2i+ T p 2i+ ϕ p 2i+ mn p 2i+ϕ. In aition, S p 2i T p 2i mn p 2iϕ, S p 2i+ T p 2i+ mn p 2i+ϕ. These inequalities imply that S p 2i + S p 2i+ mn p 2iϕ an so We conclue that k R p j j0 0 i<k /2 R p 2i + R p 2i+ mn p 2i+ϕ. mn p p 2i+ϕ p 2 +O p k mn ϕ,

5 Multiplicative properties of sets of resiues 5 where O-constants may epen on p. Thus, k p R p j p 2 +O p k ϕm n:p, D j0 p p 2 +O p k ϕmϕp k p+ +O p k ϕmn n:p, D n p k ϕ D n p k n ϕ p p 2 +O p k ϕmnσ. For D, we consier pairs S p 2i+,S p 2i+2 for 0 i < k 2/2 an we fin in the same way that Hence, D j0 ϕmn S n k R p j T + D D D T + By the hypothesis of the proposition, ϕmn 0 i<k 2/2 mn p 2i+2ϕ p 2 +O p k p 2 p 2 +O p k ϕmnσ. D ϕ mn R p+ ϕmn+ p p+ ϕmnσ +O p k ϕmnσ. p+ ϕmn+ p p+ ϕmnσ Thus, if k is sufficiently large, then ϕmn S > 2 ϕmn > p+ + p 2p+ 2. an the proposition follows. Proposition 2. Suppose n,m are coprime natural numbers, 4 n, S Z n Z m is prouct free, S 2, S 4, an D is a set of o ivisors of n containing with S for each D. Let σ D/. If then S < 2 ϕmn. ϕn n > 3 4+8σ, 3 Proof. As with the proof of Proposition, we may assume that 2 n for each D anwemay assumethatk v 2 n isverylarge. For n, o,

6 6 C. Pomerance an A. Schinzel D, we consier the pairs S 4 2i,S 4 2i+ an also the pairs S 2 4 2i,S 2 4 2i+ an we fin that k R 2 j 4 2i+ + mn 2 4 2i+ ϕ j0 0 i<k 4/ O 2 k mn ϕ. Thus, n: o, D j0 k R 2 j O 2 k 5 +O 2 k ϕ2 k ϕm 5 +O 2 k ϕmn n: o, D ϕ n 2 k ϕ D mn n 2 k 2 5 +O 2 k ϕmnσ. For D \ {} we consier the pairs S 4 2i+,S 4 2i+2 an the pairs S 2 4 2i,S 2 4 2i+, an we fin that R 2 j T + 5 +O 2 k ϕmnσ D\{} D\{} 6 5 +O 2 k ϕmnσ. Finally, we consier the pairs S 4 2i+,S 4 2i+2 an the pairs S 2 4 2i+,S 2 4 2i+2 an we fin that k R 2 j T + T O 2 k ϕmn j O 2 k ϕmn. We conclue that ϕmn S R n 5 +O 2 k 4 ϕmn+ 5 σ O 2 k ϕmn, where the O-constant may epen on σ. By the hypothesis, 4 ϕmn 5 ϕmn+ 5 σ + 2 ϕmn σ + 2 ϕn 5 n > 4σ σ +4 2,

7 Multiplicative properties of sets of resiues 7 so for k sufficiently large, we have nϕm S > 2ϕmn. This proves the proposition. Proposition 3. Suppose n,m are coprime positive integers, S Z n Z m is prouct free, p,q n are ifferent primes, S p,s q, an D is a nonempty set of ivisors of n coprime to pq with S for each D. Let σ D /. If ϕn n > ϕpq 2pqσ, 4 then S < 2 ϕmn. Proof. Similarly as with the two previous propositions, we may assume that 2 n for each D an k v p n,l v q n are both large. Suppose that n,,pq, an D. Let 0 i < k 2/2, 0 j < l 2/2, an let u p 2i q 2j. We consier 4-tuples S u,s pu,s qu,s pqu. Using that S is prouct free an S p,s q, we show that R vu T pu + T qu v pq p 2i+ q 2j + p 2i q 2j+ ϕ mn. 5 To see this, let s p S p, s q S q. We have s p S u isjoint from S pu an s p S qu isjoint from S pqu. Similarly, s q S u is isjoint from S qu an s q S pu is isjoint from S pqu. Now multiplication by s p is a p : mapping of T u onto T pu an also of T qu onto T pqu, an similarly multiplication by s q is a q : mapping of T u onto T qu an of T pu onto T pqu. For v pq, let α v S vu / T vu, so that each α v [0,] an α +α p, α +α q, α q +α pq, α p +α pq. The maximal value of α + p α p + q α q + pq α pq subject to these constraints occurs when α α pq an α p α q 0. This proves 5. In the sequel, O-constants possibly epen on p,q, an σ,

8 8 C. Pomerance an A. Schinzel We have n:,pq, D p + q p + q +O k i0 j0 l R p i q j p 2 q 2 p 2 q 2 +O p k +q l p 2 q 2 p 2 q 2 ϕmϕpk q l p k +q l ϕmn n p k q l D p+q p+q + ϕmn pqp+q p 2 q 2 ϕmnσ +O p k +q l ϕmn. v pq mn ϕ n ϕ p k q l n:,pq, D Next suppose that D. With u p 2i+ q 2j, we consier the 4-tuple S u,s pu,s qu,s pqu as before, so that R vu T pu + T qu p 2i+2 q 2j + mn p 2i+ q 2j+ ϕ. We also consier pairs S q 2j+,S q 2j+2 an we have R q 2j+ + R q 2j+2 T q 2j+2 mn q 2j+2ϕ. Thus, k D i0 j0 D + l R p i q j T + q 2 +pq p 2 q 2 + q 2 +O p k +q l q 2 +pq p 2 q 2 + q 2 +O p k +q l ϕmnσ. We conclue that ϕmn S n R p+q p+q + ϕmn+ +O p k +q l ϕmn D ϕ mn + q2 +pq +p 2 pqp+q p 2 q 2 ϕmnσ p+q p+q + ϕmn+ pqσ p+q + ϕmn+o p k +q l ϕmn.

9 By 4, Multiplicative properties of sets of resiues 9 ϕmn > p+q p+q + ϕmn+ Thus, if k,l are sufficiently large, pqσ p+q + ϕmn p+q p+q + + pqσ ϕn p+q + n p+q p q + p+q + 2p+q + 2. ϕmn S > 2 ϕmn, which proves the proposition. 4. Proof of Theorem Let n be a square-full natural number with ωn 5. Via Lemma, to prove that mn has property P for every squarefree number m coprime to n it suffices to show that for each such m, the largest prouct-free subset of Z n Z m has carinality at most 2ϕmn, with strict inequality in the case m. So, we fix some integer m coprime to n an we take a prouct-free set S Z n Z m. By Lemma 3, we may assume that S. We consier the 4 cases epening on the 4 possibilities for 6,n. First, assume that 6,n. Then ϕn n > 2, so that Corollary hanles this case. Next assume that 6,n 3. Then ϕn/n 384/00. If S 3, Lemma 2 with D {,3} completes the proof, so we may assume S 3. Then Proposition with p 3 an D {} completes the argument. Now assume that 6,n 2. Then ϕn/n 288/00. If S 2, Proposition with p 2, D {} shows that S < 2ϕmn. Thus, we may assume that S 2. If 5 n, then ϕn/n > /3, an then Lemma 2 with D {, 2} completes the proof, so we may assume that 5 n. If S 5, Proposition with p 5, D {,2} implies that we are one with this case. So, assume that S 5. If S 4, the result follows from Proposition 2 with D {,5}. So assume that S 4. Then Lemma 2 with D {,2,4,5} completes the argument. Theharestcase is when 6,n 6. Inthis case we have ϕn/n 6/77. If S 2,S 3, the result follows from Proposition 3 with p 2, q 3, an D {}. Next assumethat S 2 ans 3. Thentheresultfollows from Proposition with p 2, D {,3}. Now assume that S 2 an S 3. If 5 n then ϕn/n 240/00 an the result follows from Proposition with p 3, D {,2}. So assume that 5 n. If S 5, the result follows

10 0 C. Pomerance an A. Schinzel from Proposition 3 with p 3, q 5, D {,2}, so we may take S 5. Then the result follows from Proposition with p 3, D {,2,5}. We are left with the case that 6 n an S S 2 S 3. Proposition 2 with D {,3} hanles the case S 4, so we may assume that S 4. We consier the four possibilities for 35,n. If 35,n, then ϕn/n 640/243, so that Lemma 2 with D {,2,3,4} hanles this case. Suppose that 35,n 7, so that ϕn/n 240/00. Proposition with p 7 an D {,2,3,4} hanles the case S 7, while Lemma 2 with D {,2,3,4,7} hanles the case S 7. Suppose that 35,n 5, so that ϕn/n 32/43. Proposition with p 5 an D {,2,3,4} hanles the case S 5, while Lemma 2 with D {,2,3,4,5} hanles the case S 5. Finally supposethat 35 n. If either S 5 or S 7, Proposition with D {,2,3,4} completes the proof. So assume that S 5 S 7. Then Lemma 2 with D {,2,3,4,5,7} completes the proof. We remark that with our existing tools it is possible to begin hanlingthe case ωsn 6 an perhaps it is even possible to complete this case. Even a partial result woul give a better ensity estimate in the next section. It may be that such efforts woul allow one to fin a general argument an so complete a proof of the Conjecture. 5. Density In this section we prove Theorem 2. For a natural number n, recall that ran is the largest squarefree ivisor of n. Let m be a squarefree integer an let m be the ensity of those integers n with rasn m. For rasn m it is necessary an sufficient that m 2 n an v p n for each prime p m. Thus, m m 2 p m Let fm p m /p2, so that p 2 6 p 2. π 2 m 2 p m m 6 π2fm. 6 It is our task in this section to compute the asymptotic ensity of the set of those integers n with ωsn 6. Namely, we wish to compute : ωm 6 µ 2 m m 6 π 2 ωm 6 µ 2 mfm 6 π 2 ωm 5 µ 2 mfm. Let δ j ωmj µ2 mfm. We thus must compute δ j for j 0,,...,5. We eviently have δ 0.

11 Multiplicative properties of sets of resiues For δ, we accelerate the convergence of the series as follows: δ π 2 p 2 log + 6 p 2 +log p 2, p p an so we fin that δ roune to 5 ecimal places. The computation for δ j for j > is simplifie by applying the Newton Girar formula for symmetric functions. In particular, with η j p 2 j, p we have δ j j j i η i δ j i. 7 i Note that 7 allows one to compute each δ j recursively in terms of previous values of δ i an values of the very rapily converging series η i where η δ has alreay been compute. To 5 ecimal places, we have Thus, via 7, we have We conclue that η , η , η , η δ , δ , δ , δ π 2δ 0 +δ +δ 2 +δ 3 +δ 4 +δ 5. 6 π , which proves Theorem Further remarks One might consier large prouct-free subsets of N, the set of natural numbers. As shown in [2], the upper asymptotic ensity of a prouct-free subset of N can be arbitrarily close to. It is easy to see that there are prouct-free subsets of N with asymptotic ensity equal to /2. Here are some examples:

12 2 C. Pomerance an A. Schinzel the set of natural numbers n that are the prouct of an o number of primes; the set of natural numbers n that are the prouct of a number that is 3 mo 4 an a power of 2; the set of natural numbers n that are the prouct of a number that is 2 mo 3 an a power of 3; more generally, for any o prime p, the set of natural numbers n which are a prouct of a quaratic nonresiue mo p an a power of p. These examples, the first of which was note in [2], also show that the principal result of [2] is best possible. A further example is supplie in Fish [] where it is shown that there are normal subsets of N which are prouct free. A subset S of N is normal if the characteristic function of S, written as a sequence of 0 s an s, is normal. Necessarily a normal subset of N has ensity /2. Must the lower asymptotic ensity of a prouct free subset of N be at most /2? If so, this woul establish Conjecture for all o numbers. Schur [4] showe that if N is k-colore there must be a monochromatic solution to a+b c. A. Sárközy suggeste to us that one might consier the multiplicative analog: If N is k-colore, must there be a monochromatic solution to ab c? Since, the number shoul not be allowe in the set, so we are k-coloring N \{}. By consiering the powers of 2, one sees that the multiplicative analog immeiately follows from the original aitive version. So, it is reasonable to consier then the multiplicative problem for squarefree numbers larger than. Here s a proof in the case k 2: Let p,...,p 9 be any 9 primes, an so without loss of generality, we may assume that each of p,...,p 5 is re. We then may assume that each prouct of 2 of these is blue an so each prouct of 4 of these is re. Then the prouct of all 5 is blue, an since a prouct of 4 can be written as one of the primes times the other 3, each prouct of 3 primes is blue. But then p p 2 p 3 p 4 p 5 p p 2 p 3 p 4 p 5 is all blue. It is possible, maybe even likely, that these thoughts generalize to k colors, an perhaps this an relate topics woul be interesting to explore. Consier the following conjecture, which may be tractable. Let Ωn enote the number of prime factors of n counte with multiplicity. Conjecture 2. Let p,p 2,...,p k be istinct primes, let b be a positive integer, an let n p p 2...p k b. For u n let α u be a real variable in [0,] such that if uv n an α u > 0, then α v + α uv. Further suppose that α 0. Then u n α u u < u: rau n Ωu o Remark. Note that the secon sum in the conjecture is over an infinite set of numbers u. u.

13 Multiplicative properties of sets of resiues 3 Theorem 3. Conjecture 2 implies Conjecture. Proof. Assume Conjecture 2. Let p,p 2,...,p k be arbitrary istinct primes an let m p p 2...p k. It is sufficient to show that every n with ran m has property P. Since every every ivisor of a number with property P also has property P, it is thus sufficient to show that n m b+ has property P for every large integer b. Suppose that n m b+ an S Z/nZ is prouct free. For u n, let T u T u n,, S u S T u as in Section 2, an let α u S u / T u. By Lemma 3 with m being, we may assume that α 0. Assume uv m b an α u > 0. Then S u, say s u S u, an multiplication by s u is a u : mapping of T v onto T uv. Since S is prouct free, we have s u S v S uv, so that u S v + S uv T uv ; that is, ϕn α v uv +α ϕn uv uv ϕn uv, or α v +α uv. Thus, the numbers α u for u m b satisfy the hypotheses of Conjecture 2, an so u m b α u Note that u < rau m Ωu o S u n u 2 rau m S u n α u ϕ u u n u Ωu m u 2 ϕm m. σm ϕn u m b α u u + u n u m b ϕu. The first sum here is boune as above, an the secon sum is boune by m m 2 ϕm u ϕm p b+ < m 2σm u n u m b if b is sufficiently large b k+4 is sufficient. For such b, S < ϕn m 2 ϕm m + ϕn m σm 2 σm ϕnm 2ϕm 2 n. Thus, n has property P. It is possible to recast Conjecture 2 as a linear program. The feasible set for the variables is not a polytope, but rather a union of polytopes, since the conition α u > 0 implies α v +α uv may be viewe as the union of α u 0 an α v +α uv. For each polytope in the union, the maximum of the objective function occurs at a vertex, so the maximum over the union p m

14 4 C. Pomerance an A. Schinzel of polytopes again occurs at one of the vertices of one of the polytopes in the union. Thus, we conclue that one may restrict to variables α u {0,}. With thevariables sorestricte, theconition α u > 0implies α v +α uv is equivalent to α u +α v +α uv 2. Thus, we may consier instea the linear program where each variable α u [0,], each triple α u,α v,α uv satisfies α u +α v +α uv 2, an α 0. Uner these conitions, we are to maximize αu /u. Perhaps this point of view can lea to a proof. We close this paper with a proofof Conjecture2whenk 2 usingtools close to those usein Section 3. Theorem 4. Conjecture 2 hols for k an k 2. Proof. For k with prime p an n p b, we have ivisors p i of n for i,...,b. If α p 0, then But u n i o α u u < i 2 p i pp. p i p p 2 p /p, which oes inee excee the prior estimate. Thus, we may assume that α p > 0. Then for i o an p i+ n, we have α p i +α p i+ so that α p i p i + α p i+ p i+ p i. Using also α p b, we have u n α u u i b i o p i < i o completing the case k. For k 2, we write n pq b where p,q are istinct primes. We wish to show that L < R, where L : u n α u u, R : rau pq Ωu o p i, u pq 2 ϕpq pq σpq pqp+q p 2 q 2, cf. the proof of Theorem 3. First assume that α p α q 0. Then L < p i q j pq p q p q pq +p2 +q 2 p q, pqp q i+j 2 so that if s p+q an m pq, we have L R < s2 m sp+q + sm 2 s2 m ss+m+ sm 2 s m s+ s m +.

15 Multiplicative properties of sets of resiues 5 As a function of m this expression is ecreasing. But m 2s 2, so we have L s R < 2s 2 s+ s 2s s ss 2. As a function of s this expression is ecreasing, an since s 5, we have L/R < 4/5 <. Now assume α p > 0 an α q 0 the case where α p 0, α q > 0 will follow in the same way. If p n, then α +α p, so that α + α p p. 8 We use 8 for p i with i o, for p i q with i o, an for p i q j with i even an j 2. But, if such a number n has v p b, we use α. We thus have L < an so + q i o p i + i even j 2 L R < pq +q2 +p 2 q + pq 2 p+q p i q j + p q p 2 + p 2 p 2 qq, < q2 +q +p+p/q q 2. +pq Since p /q 2 q >, it follows that pq > q +p/q, so L < R. Our last case is when α p > 0,α q > 0. If pq n an >, we have α +α p, α +α q, α p +α pq, α q +α pq, so that as in the proof of Proposition 3, we have α + α p p + α q q + α pq pq + pq. We apply this when p i q j when i is even an j is o. When i is o an j 0, we apply 8. But, if such n has either v p b or v q b, we merely use α. We thus have L < i even j o This conclues our proof. p i q j + p i+ q j+ + i o References p i rau pq Ωu o u R. [] A. Fish, Ranom Liouville functions an normal sets, Acta Arith., , [2] L. Haju, A. Schinzel, an M. Skalba, Multiplicative property of sets of positive integers, Arch. Math. Basel, , [3] K. Kelaya, Prouct-free subsets of groups, then an now, Communicating mathematics, 69 77, Contemp. Math., 479, Amer. Math. Soc., Provience, RI, 2009.

16 6 C. Pomerance an A. Schinzel [4] I. Schur, Über ie Kongruenz x m + y m z m mo p, Jahresb. Deutsche Math. Ver., 25 96, 4 7. C. Pomerance Department of Mathematics Dartmouth College Hanover, NH 03755, USA carl.pomerance@artmouth.eu A. Schinzel Institute of Mathematics Polish Acaemy of Sciences Sniaeckich 8, P.O. Box Warszawa, Polan schinzel@impan.pl