Diophantine equations with products of consecutive values of a quadratic polynomial

Yang, Shichun; Togbé, Alain; He, Bing

doi:10.1016/j.jnt.2011.03.006

Cited by 6 publications

(5 citation statements)

References 20 publications

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“…In subsequent work by Fang [8], his technique was applied to products A f (m) corresponding to the polynomials 4x 2 + 1 and 2x 2 − 2x + 1. Yang, Togbé and He [12] proved that for any irreducible quadratic polynomial f (x) ∈ Z[x], there exists a prime p with ord p (A f (m)) = 1 if m ≥ C, where C is a computable constant depending on the coefficients of f (x). Furthermore, the above problem has been investigated by many authors for polynomials of the form f (x) = x k + 1, where k is an odd positive integer.…”

Section: Introduction and Main Resultsmentioning

confidence: 99%

Prime powers dividing products of consecutive integer values of $$x^{2^n}+1$$

Baier

Dey

2019

Res. number theory

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Let n be a positive integer and f (x) := x 2 n + 1. In this paper, we study orders of primes dividing products of the form P m,n := f (1)f (2) · · · f (m). We prove that if m > max{10 12 , 4 n+1 }, then there exists a prime divisor p of P m,n such that ord p (P m,n ) ≤ n · 2 n−1 . For n = 2, we establish that for every positive integer m, there exists a prime divisor p of P m,2 such that ord p (P m,2 ) ≤ 4. Consequently, P m,2 is never a fifth or higher power. This extends work of Cilleruelo [6] who studied the case n = 1. Introduction and main resultFor a prime p and a nonzero integer s, define ord p (s) to be the unique non-negative integer i for whichbe a polynomial of degree k ≥ 2 with positive leading coefficient which does not vanish at any positive integer. Setand note that this is nonzero for all m ∈ N by the above assumption.A major unsolved problem in analytic number theory concerns the question whether f represents infinitely many primes if f is irreducible and there exists no prime p dividing f (m) for all integers m. If f represents infinitely many primes, then trivially, for infinitely many integers m, there exists a prime such that ord p (A f (m)) = 1. For particular polynomials f , several authors investigated the related question whether for all sufficiently large integers m there exists a prime p with ord p (A f (m)) = 1. If this is the case, then, in particular, A f (m) is a perfect power for at most finitely many m ∈ N.Below we summarize a number of results from the literature. For the polynomial f (x) = x 2 + 1, J. Cilleruelo [6] proved the following result which we shall generalize in this paper. Theorem 1 (Cilleruelo). Let f (x) = x 2 + 1 and m > 3. Then there exists a prime divisor p of A f (m) with ord p (A f (m)) = 1. Consequently, A f (m) is a perfect power only for m = 3, in which case we have A f (3) = 10 2 . Cilleruelo's work [6] used only elementary tools such as Chebyshev's upper bound inequality for the primes counting function. In subsequent work by Fang [8], his

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Section: Introduction and Main Resultsmentioning

confidence: 99%

Prime powers dividing products of consecutive integer values of $$x^{2^n}+1$$

Baier

Dey

2019

Res. number theory

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“…Clearly, (8,19) is a solution of (3). For each solution ( , ) of (3), the map sends ( , ) to (9 + 4 + 4, 20 + 9 + 10), which gives another solution of (3).…”

Section: Resultsmentioning

confidence: 99%

“…Fang [4] confirmed another similar conjecture posed by Amdeberhan et al [1] to the products of quadratic polynomials ∏︀ =1 (4 2 +1) and ∏︀ =1 (2 2 −2 +1) are not perfect squares. Yang et al [8] studied the Diophantine equation…”

Section: Introductionmentioning

confidence: 99%

On products of quartic polynomials over consecutive indices which are perfect squares

Kuhapatanakul¹,

Meeboomak²,

Thongsing³

2018

NNTDM

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“…Shortly, Cilleruelo [3] proved the conjecture for P 1,1 (n), where he showed that P 1,1 (n) is square only for n = 3, and Fang [4] established the conjecture for P 4,1 (n). Yang, Togbé, and He [9] found all positive integer solutions to the equation P a,c (n) = y l for coprime integers a, c ∈ {1, . .…”

Section: Introductionmentioning

confidence: 99%