MATHEMATICAL COMMUNICATIONS
Math. Commun. 21(2016), 109–114
109
A note on the products ((m + 1)2 + 1)((m + 2)2 + 1) . . . (n2 + 1)
and ((m + 1)3 + 1)((m + 2)3 + 1) . . . (n3 + 1)
Erhan Gürel1,∗
1
Middle East Technical University, Northern Cyprus Campus, TZ-32 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
Nm such that whenever the integer n ≥ Nm , neither the product Pm
= ((m + 1)2 + 1)((m +
3
3
2)2 + 1) . . . (n2 + 1) nor the product Qn
=
((m
+
1)
+
1)((m
+
2)
+ 1) . . . (n3 + 1) is a
m
square.
AMS subject classifications: 11A41, 11A51
Key words: polynomial products, diophantine equations
1. Introduction
Qn
In 2008, J. Cilleruelo proved that k=1 (k 2 + 1) is a square only for n = 3 [3]. His
technique was applied to the products of consecutive values of other polynomials like
4x2 + 1 and 2x2 − 2x + 1 by Fang [5], and to x3 + 1 by Gürel and Kişisel [7]. Later,
an idea due to W. Zudilin was applied to xp + 1 by Zhang and Wang [8] and to
k
xp + 1 by Chen et al. [2] for odd
Qnprime p. In the very recent article [6], the author
proved that while the product k=1Q(4k 4 + 1) becomes a square infinitely often as
the integer n changes, the product nk=1 (k 4 + 4) becomes a square only for n = 2
using techniques different from the articles mentioned above. The problem is still
k
open for polynomials like x2 + 1. In this paper, our purpose is to extend results
of [3] and [7] for products of sufficiently many consecutive values of the polynomial
x2 + 1 and x3 + 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 Nm
such that whenever the integer n ≥ Nm , neither the product
n
= ((m + 1)2 + 1)((m + 2)2 + 1) . . . (n2 + 1)
Pm
nor the product
Qnm = ((m + 1)3 + 1)((m + 2)3 + 1) . . . (n3 + 1)
is a square.
∗ Corresponding
author. Email address: [email protected] (E. Gürel)
http://www.mathos.hr/mc
c
2016
Department of Mathematics, University of Osijek
110
E. Gürel
In order to prove the main result, we need following lemmas.
n
Lemma 1. If p is a prime such that p2 |Pm
or p2 |Qnm , then p < 2n.
Proof. See the proof of Theorem 1 in [3] and the proof of Lemma 1 in [7].
Now, we can write
n
Pm
=
Y
pαp , Qnm =
p<2n
Y
pᾱp and
p<2n
Y
n!
=
pβ p .
m!
p≤n
n! 2
n
Comparing the products term by term, we can easily deduce that ( m!
) < Pm
and
n! 3
n
( m! ) < Qm . After taking the natural logarithms of both sides, we obtain the
n
following inequalities under the condition that both Pm
and Qnm are square-full.
X
βp ln p <
1 X
αp ln p.
2 p<2n
(1)
X
βp ln p <
1 X
ᾱp ln p.
3 p<2n
(2)
p≤n
p≤n
Lemma 2.
If p ≡ 1 (mod 4) and p ≤ n, then
αp
2
− βp ≤
ln(n2 +1)
.
ln p
If p ≡ 1 (mod 3) and p ≤ n, then
ᾱp
3
− βp ≤
ln(n3 +1)
.
ln p
If p ≡ 2 (mod 3) and p ≤ n, then ᾱp − βp ≤
ln(n3 +1)
.
ln p
Proof. When p ≡ 1 (mod 4), the equation x2 + 1 ≡ 0 (mod pj ) has two solutions
in an interval of length pj . That can be used to bound αp as follows:
X
X
X
n
m
αp =
#{m < k ≤ n, pj |(k 2 + 1)} ≤
2 j −
2 j .
p
p
2
2
2
+1)
j≤ ln(n
ln p
+1)
j≤ ln(n
ln p
+1)
j≤ ln(m
ln p
When p ≡ 1 (mod 3), the equation x3 + 1 ≡ 0 (mod pj ) has at most three solutions
in an interval of length pj . That can be used to bound ᾱp as follows:
X
X
X
n
m
j
3
#{m < k ≤ n, p |(k + 1)} ≤
3 j −
3 j .
ᾱp =
p
p
ln(m3 +1)
ln(n3 +1)
ln(n3 +1)
j≤
ln p
j≤
ln p
j≤
ln p
When p ≡ 2 (mod 3), the equation x3 + 1 ≡ 0 (mod pj ) has exactly one solution
in an interval of length pj (see Lemma 2 in [7]). That can be used to bound ᾱp as
follows:
X
X
X n
m
ᾱp =
−
.
#{m < k ≤ n, pj |(k 3 + 1)} ≤
j
p
pj
3
3
3
+1)
j≤ ln(n
ln p
+1)
j≤ ln(n
ln p
+1)
j≤ ln(m
ln p
111
A note on the products
We also know that
βp =
X m
X n
−
.
#{m < k ≤ n, p |k} =
pj
pj
ln(n)
ln(m)
ln(n)
X
j≤
j
j≤
ln p
Therefore, when p ≡ 1 (mod 4),
X n n αp
− j
+
− βp ≤
2
pj
p
ln(n)
j≤
ln p
X
−
ln(m2 +1)
ln(m)
ln p <j≤
ln p
m
pj
j≤
ln p
ln p
X
ln(n)
ln(n2 +1)
ln p <j≤
ln p
n
pj
(3)
ln(n2 + 1)
≤
.
ln p
When p ≡ 1 (mod 3),
X n n ᾱp
− j
+
− βp ≤
3
pj
p
ln(n)
j≤
ln p
X
−
ln(m)
ln(m3 +1)
ln p <j≤
ln p
m
pj
X
ln(n)
ln(n3 +1)
ln p <j≤
ln p
n
pj
(4)
ln(n3 + 1)
≤
.
ln p
When p ≡ 2 (mod 3),
X n n ᾱp − βp ≤
− j
+
pj
p
ln(n)
j≤
ln p
X
−
ln(m)
ln(m3 +1)
ln p <j≤
ln p
m
pj
X
ln(n)
ln(n3 +1)
ln p <j≤
ln p
n
pj
(5)
ln(n3 + 1)
≤
.
ln p
Using this Lemma 2, we can update inequalities (1) and (2) as follows:
X
βp ln p <(
p≡3(mod4)
p≤n
X
p≡2(mod3)
p≤n
α2
− β2 ) ln 2 +
2
X
ln(n2 + 1) +
p≡1(mod4)
p≤n
X
2βp ln p
ᾱ3
<(
− β3 ) ln 3 +
3
3
ln(n3 + 1) +
ᾱp ln p
.
+
3
n<p<2n
X
n−m
p−1
−
X
p≡2(mod3)
p≤n
p≡1(mod3)
p≤n
Lemma 3. For any prime p ≤ n, βp ≥
αp ln p
.
2
n<p<2n
X
ln(n+1)
ln p .
(6)
ln(n3 + 1)
3
(7)
112
E. Gürel
Proof. Using the lower and the upper bounds of the prime powers in a factorial as in
Lemma 1 [1] for n and m, respectively, we can easily obtain the desired inequality.
When p = 2, if k is even k 2 + 1 ≡ 1 (mod 4), otherwise k 2 + 1 ≡ 2 (mod 4), so,
lnm lmm n − m + 1
−
≤
α2 =
2
2
2
and using the preceding lemma,
β2 ≥ n − m −
ln(n + 1)
.
ln 2
When p = 3, k 3 + 1 ≡ 0 (mod 3t ) has only one solution for t = 1 and two solutions
for t > 1. So,
lnm jmk
lnm jmk
5(n − m)
3 ln n 3 ln m
−
) + 2(
−
) + ··· ≤
+3+
−
ᾱ3 ≤ (
3
3
9
9
6
ln 3
ln 3
and using the preceding lemma again,
n − m ln(n + 1)
−
.
2
ln 3
Lemma 4. If p > n, then αp ≤ 2 and ᾱp ≤ 3 .
β3 ≥
Proof. If p > n, and αp ≥ 3, 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|l2 + 1. 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 ᾱp , as in [7].
Using Lemma 3 and Lemma 4, inequalities (6) and (7) become
X (n − m) ln p
X
X
α2
ln(n + 1)
ln(n2 + 1) +
<(
− β2 ) ln 2 +
p−1
2
p≡3(4)
p≤n
p≡3(mod4)
p≤n
p≡1(mod4)
p≤n
+
X
(8)
ln p.
n<p<2n
X
p≡2(mod3)
p≤n
2(n − m) ln p
ᾱ3
<(
− β3 ) ln 3 +
3(p − 1)
3
+
X
p≡2(mod3)
p≤n
+
X
X
ln(n3 + 1)
p≡1(mod3)
p≤n
ln(n3 + 1)
+
3
X
p≡2(mod3)
p≤n
2 ln(n + 1)
3
(9)
ln p.
n<p<2n
Employing the estimates for α2 , β2 , replacing ln(n+1) by ln(n2 +1) and replacing
X
X
ln p +
ln p
p≡3(mod4)
p≤n
n<p<2n
A note on the products
113
by
X
ln p,
p<2n
the right-hand side of inequality (8) gets larger,
X (n − m) ln p
X
(1 − 3n + 3m) ln 2 X
<
+
ln(n2 + 1) +
ln p.
(p − 1)
4
p<2n
(10)
p≤n
p≡3(4)
p≤n
Dividing both sides by n − m we obtain
P
X
π(n) ln(n2 + 1)
3
ln p
1
p<2n ln p
ln 2 +
<
−
+
.
(p − 1)
4(n − m) 4
n−m
n−m
(11)
p≡3(4)
p≤n
Likewise, employing the estimates for ᾱ3 , β3 , replacing ln(n + 1) by ln(n3 + 1) and
replacing
X
2 ln p
3
p≡2(mod3)
p≤n
by
X
ln p,
p≤n
the right-hand side of inequality (9) gets larger,
X
p≡2(mod3)
p≤n
X
2(n − m)ln p
(9 − n + m) ln 3 X
<
+
ln(n3 + 1) +
ln p.
3(p − 1)
9
p<2n
p≤n
(12)
we obtain
Dividing both sides by 2(n−m)
3
P
X
3π(n) ln(n3 + 1) 3 p<2n ln p
ln p
1
3
ln 3 +
<
−
+
. (13)
(p − 1)
2(n − m) 6
2(n − m)
2(n − m)
p≡2(mod3)
p≤n
Since
π(n) ∼
X
n
and
ln p ∼ 2n,
ln n
p<2n
the limiting values of the right-hand sides in (12) and (13) are 16−34 ln 2 ∼
= 3.48013
3 ∼
and 45−ln
7.31689,
respectively.
However,
the
left
hand
sides
of
these
inequalities
=
6
are divergent series exceeding the right-hand sides for some values of Nm for each
inequality. That means any of these inequalities will no longer be satisfied and it
n
will contradict Pm
and/or Qnm is square-full, which proves our theorem.
Acknowledgement
The author would like to thank the referees for their helpful suggestions.
114
E. Gürel
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