Erhan Gürel 1, Received October 14, 2015; accepted November 24, 2015

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1 MATHEMATICAL COMMUNICATIONS 109 Math Commun 21(2016), A note on the products ((m+1) 2 +1)((m+2) 2 +1)(n 2 +1) and ((m+1) +1)((m+2) +1)(n +1) Erhan Gürel 1, 1 Middle East Technical University, Northern Cyprus Campus, TZ-2 Kalkanlı, Güzelyurt, Mersin 10, Turkey Received October 14, 2015; accepted November 24, 2015 Abstract We prove that for any positive integer m there exists a positive real number N m such that whenever the integer n N m, neither the product P n m = ((m+1) 2 +1)((m+ 2) 2 + 1)(n 2 + 1) nor the product Q n m = ((m + 1) + 1)((m + 2) + 1)(n + 1) is a square AMS subject classifications: 11A41, 11A51 Key words: polynomial products, diophantine equations 1 Introduction In 2008, J Cilleruelo proved that n k=1 (k2 +1) is a square only for n = [] His technique was applied to the products of consecutive values of other polynomials like 4x 2 +1 and 2x 2 2x+1 by Fang [5], and to x +1 by Gürel and Kişisel [7] Later, an idea due to W Zudilin was applied to x p + 1 by Zhang and Wang [8] and to x pk +1 by Chen et al [2] for odd prime p In the very recent article [6], the author proved that while the product n k=1 (4k4 +1) becomes a square infinitely often as the integer n changes, the product n k=1 (k4 +4) becomes a square only for n = 2 using techniques different from the articles mentioned above The problem is still open for polynomials like x 2k +1 In this paper, our purpose is to extend results of [] and [7] for products of sufficiently many consecutive values of the polynomial x 2 +1 and x +1 starting from any positive integer x = m+1 up to x = n by using techniques similar to the ones used in the above mentioned articles The main result of this paper is the following theorem Theorem 1 For any positive integer m, there exists a positive real number N m such that whenever the integer n N m, neither the product nor the product is a square P n m = ((m+1)2 +1)((m+2) 2 +1)(n 2 +1) Q n m = ((m+1) +1)((m+2) +1)(n +1) Corresponding author address: egurel@metuedutr (EGürel) c 2016 Department of Mathematics, University of Osijek

2 110 E Gürel In order to prove the main result, we need following lemmas Lemma 1 If p is a prime such that p 2 P n m or p 2 Q n m, then p < 2n Proof See the proof of Theorem 1 in [] and the proof of Lemma 1 in [7] Now, we can write P n m = p αp, Q n m = and n! pᾱp m! = p βp Comparing the products term by term, we can easily deduce that ( n! m! )2 < Pm n and ( n! m! ) < Q n m After taking the natural logarithms of both sides, we obtain the following inequalities under the condition that both Pm n and Q n m are square-full Lemma 2 β p < 1 α p (1) 2 β p < 1 ᾱ p (2) If p 1 (mod 4) and p n, then αp 2 β p ln(n2 +1) If p 1 (mod ) and p n, then ᾱp β p ln(n +1) If p 2 (mod ) and p n, then ᾱ p β p ln(n +1) Proof When p 1 (mod 4), the equation x (mod ) has two solutions in an interval of length That can be used to bound α p as follows: α p = #{m < k n, (k 2 +1)} 2 2 j ln(n2 +1) j ln(n +1) j ln(n2 +1) j ln(n +1) j ln(m2 +1) When p 1 (mod ), the equation x +1 0 (mod ) has at most three solutions in an interval of length That can be used to bound ᾱ p as follows: ᾱ p = #{m < k n, (k +1)} j ln(m +1) When p 2 (mod ), the equation x (mod ) has exactly one solution in an interval of length (see Lemma 2 in [7]) That can be used to bound ᾱ p as follows: ᾱ p = j ln(n +1) #{m < k n, (k +1)} j ln(n +1) j ln(m +1)

3 We also know that β p = A note on the products 111 #{m < k n, k} = n j ln(m) Therefore, when p 1 (mod 4), α p 2 β p When p 1 (mod ), ᾱ p β p When p 2 (mod ), ( ) n n + ln(m) <j ln(m2 +1) ᾱ p β p ( ) n n + ln(m) <j ln(m +1) ( ) n n + ln(m) ln p <j ln(m +1) ln(n) <j ln(n2 +1) ln(n2 +1) ln(n) <j ln(n +1) ln(n +1) ln(n) <j ln(n +1) ln(n +1) () (4) (5) Using this Lemma 2, we can update inequalities (1) and (2) as follows: p (mod4) β p <( α 2 2 β 2)ln2+ 2β p p 1(mod4) <(ᾱ β )ln+ p 1(mod) + ᾱ p n< ln(n 2 +1)+ n< ln(n +1)+ Lemma For any prime p n, β p n m p 1 ln(n+1) α p (6) 2 ln(n +1) (7)

4 112 E Gürel Proof Usingthe lowerandthe upperbounds ofthe primepowersin afactorialasin Lemma1[1]fornandm, respectively,wecaneasilyobtainthedesiredinequality When p = 2, if k is even k (mod 4), otherwise k (mod 4), so, n m α 2 = 2 n m and using the preceding lemma, β 2 n m ln(n+1) ln2 When p =, k +1 0 (mod t ) has only one solution for t = 1 and two solutions for t > 1 So, n m n m ᾱ ( )+2( )+ 9 5(n m) ++ lnn 9 6 ln lnm ln and using the preceding lemma again, β n m 2 Lemma 4 If p > n, then α p 2 and ᾱ p ln(n+1) ln Proof If p > n, and α p, then it is easy to see that there exist distinct j,k,l such that p j 2 + 1, p k 2 + 1, and p l Then p (j k)(j + k), so p j + k, and similarly, p j +l Then p k l, a contradiction A similar argument can be applied to ᾱ p, as in [7] Using Lemma and Lemma 4, inequalities (6) and (7) become p (4) (n m) p 1 2(n m) (p 1) <( α 2 2 β 2)ln2+ + n< p 1(mod4) ln(n 2 +1)+ <(ᾱ β )ln+ ln(n +1) p 1(mod) n< ln(n +1) p (mod4) 2ln(n+1) ln(n+1) (8) (9) Employingtheestimatesforα 2, β 2, replacingln(n+1)byln(n 2 +1)andreplacing + p (mod4) n<

5 by A note on the products 11, the right-hand side of inequality (8) gets larger, p (4) p (4) (n m) (p 1) < (1 n+m)ln2 4 Dividing both sides by n m we obtain ( (p 1) < 1 4(n m) 4 + ln(n 2 +1)+ (10) ) ln2+ π(n)ln(n2 +1) + n m n m (11) Likewise, employing the estimates for ᾱ, β, replacing ln(n+1) by ln(n +1) and replacing 2 by, the right-hand side of inequality (9) gets larger, 2(n m) (p 1) < (9 n+m)ln 9 Dividing both sides by 2(n m) we obtain ( (p 1) < 2(n m) 1 6 Since π(n) n lnn and + ln(n +1)+ ) ln+ π(n)ln(n +1) 2(n m) 2n, (12) + (1) 2(n m) the limiting values of the right-hand sides in (12) and (1) are 16 ln2 4 = 4801 and 45 ln 6 = 71689, respectively However, the left hand sides of these inequalities are divergent series exceeding the right-hand sides for some values of N m for each inequality That means any of these inequalities will no longer be satisfied and it will contradict Pm n and/or Q n m is square-full, which proves our theorem Acknowledgement The author would like to thank the referees for their helpful suggestions

6 114 E Gürel References [1] Y Bugeaud, M Laurent, Minoration effective de la distance p-adique entre puissances de nombres alge briques, J Number Theory 61(1996), [2] YChen, MGong, XRen, On the products (1 l +1)(2 l +1)(n l +1), J Number Theory 1(201), [] JCilleruelo, Squares in (1 2 +1)(n 2 +1), J Number Theory 128(2008), [4] PDusart, Sharper bounds for ψ,θ,π,p k, Rapport de recherche n , L Université de Limoges, 1998 [5] J-HFang, Neither n k=1 (4k2 + 1) nor n k=1 (2k(k 1) + 1) is a perfect square, Integers 9(2009), [6] EGürel, On the products n k=1 (4k4 +1) and n k=1 (k4 +4), to appear [7] EGürel, AUÖKisisel, A note on the products (1µ +1)(n µ +1), J Number Theory 10(2010), [8] WZhang, TWang, On the products (1 k +1)(2 k +1)(n k +1), J Number Theory 12(2012),