On the Diophantine equation $x(x + 1)\dotsm (x + n) + 1 = y^2$ $(17 \le n = \mbox {odd} \le 27)$

” THE DIOPHANTINE EQUATION x n = Dy 2 + 1 ”

On the Diophantine equation q n − 1 q − 1 = y

There exist many results about the Diophantine equation (qn − 1)/(q − 1) = ym, where m ≥ 2 and n ≥ 3. In this paper, we suppose that m = 1, n is an odd integer and q a power of a prime number. Also let y be an integer such that the number of prime divisors of y − 1 is less than or equal to 3. Then we solve completely the Diophantine equation (qn − 1)/(q − 1) = y for infinitely many values of y....

On the nature of solutions of the difference equation $mathbf{x_{n+1}=x_{n}x_{n-3}-1}$

We investigate the long-term behavior of solutions of the difference equation[ x_{n+1}=x_{n}x_{n-3}-1 ,, n=0 ,, 1 ,, ldots ,, ]noindent where the initial conditions $x_{-3} ,, x_{-2} ,, x_{-1} ,, x_{0}$ are real numbers.  In particular, we look at the periodicity and asymptotic periodicity of solutions, as well as the existence of unbounded solutions.

ON THE DIOPHANTINE EQUATION xn − 1 x −

We prove that if (x, y, n, q) 6= (18, 7, 3, 3) is a solution of the Diophantine equation (xn−1)/(x−1) = y with q prime, then there exists a prime number p such that p divides x and q divides p − 1. This allows us to solve completely this Diophantine equation for infinitely many values of x. The proofs require several different methods in diophantine approximation together with some heavy comput...

ON THE DIOPHANTINE EQUATION G n ( x ) = G m ( P ( x ) ) : HIGHER - ORDER RECURRENCES CLEMENS

Let K be a field of characteristic 0 and let (Gn(x))n=0 be a linear recurring sequence of degree d in K[x] defined by the initial terms G0, . . . , Gd−1 ∈ K[x] and by the difference equation Gn+d(x) = Ad−1(x)Gn+d−1(x) + · · ·+A0(x)Gn(x), for n ≥ 0, with A0, . . . , Ad−1 ∈ K[x]. Finally, let P (x) be an element of K[x]. In this paper we are giving fairly general conditions depending only on G0, ...