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Register a new userPrime powers dividing products of consecutive integer values of $x^{2^n}+1$
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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 $P_{m,n}:=f(1)f(2)cdots f(m)$. 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$_{p}(P_{m,n} )leq ncdot 2^{n-1}$. For $n=2$, we establish that for every positive integer $m$, there exists a prime divisor $p$ of $P_{m,2}$ such that ord$_{p} (P_{m,2}) leq 4$. Consequently, $P_{m,2}$ is never a fifth or higher power. This extends work of Cilleruelo who studied the case $n=1$.
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A recent construction by Amarra, Devillers and Praeger of block designs with specific parameters depends on certain quadratic polynomials, with integer coefficients, taking prime power values. The Bunyakovsky Conjecture, if true, would imply that each of them takes infinitely many prime values, giving an infinite family of block designs with the required parameters. We have found large numbers of prime values of these polynomials, and the numbers found agree very closely with the estimates for them provided by Lis recent modification of the Bateman-Horn Conjecture. While this does not prove that these polynomials take infinitely many prime values, it provides strong evidence for this, and it also adds extra support for the validity of the Bunyakovsky and Bateman-Horn Conjectures.
We show that the largest prime factor of $n^2+1$ is infinitely often greater than $n^{1.279}$. This improves the result of de la Bret`eche and Drappeau (2019) who obtained this with $1.2182$ in place of $1.279.$ The main new ingredients in the proof are a new Type II estimate and using this estimate by applying Harmans sieve method. To prove the Type II estimate we use the bounds of Deshouillers and Iwaniec on linear forms of Kloosterman sums. We also show that conditionally on Selbergs eigenvalue conjecture the exponent $1.279$ may be increased to $1.312.$
In this paper, we consider the problem about finding out perfect powers in an alternating sum of consecutive cubes. More precisely, we completely solve the Diophantine equation $(x+1)^3 - (x+2)^3 + cdots - (x + 2d)^3 + (x + 2d + 1)^3 = z^p$, where $p$ is prime and $x,d,z$ are integers with $1 leq d leq 50$.
We show that the diophantine equation $n^ell+(n+1)^ell + ...+ (n+k)^ell=(n+k+1)^ell+ ...+ (n+2k)^ell$ has no solutions in positive integers $k,n ge 1$ for all $ell ge 3$.
This work determine the entire family of positive integer solutions of the diophantine equation. The solution is described in terms of $frac{(m-1)(m+n-2)}{2} $ or $frac{(m-1)(m+n-1)}{2}$ positive parameters depending on $n$ even or odd. We find the solution of a diophantine system of equations by using the solution of the diophantine equation. We generalized all the results of the paper [5].
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