On the Diophantine equation
We attack the equation of the title using a Frey curve, Ribet’s level-lowering theorem and a method due to Darmon and Merel. We are able to determine all the solutions in pairwise coprime integers if is prime and . From this we deduce some results about special cases of this equation that have been studied in the literature. In particular, we are able to combine our result with previous results of Arif and Abu Muriefah, and those of Cohn to obtain a complete solution for the equation for...
1. Introduction. Let ℤ, ℕ, ℚ be the sets of integers, positive integers and rational numbers respectively. In [7], Ribenboim proved that the equation (1) , x,y,m,n ∈ ℕ, x > 1, n > m ≥ 1, has no solution (x,y,m,n) with 2|x and (1) has only finitely many solutions (x,y,m,n) with 2∤x. Moreover, all solutions of (1) with 2∤x satisfy max(x,m,n) < C, where C is an effectively computable constant. In this paper we completely determine all solutions of (1) as follows. Theorem. Equation (1)...
Applying results on linear forms in p-adic logarithms, we prove that if (x,y,z) is a positive integer solution to the equation with gcd(x,y) = 1 then (x,y,z) = (2,1,k), (3,2,k), k ≥ 1 if c = 1, and either , k ≥ 1 or if c ≥ 2.
Let p denote a prime number. P. Samuel recently solved the problem of determining all squares in the linear recurrence sequence {Tₙ}, where Tₙ and Uₙ satisfy Tₙ² - pUₙ² = 1. Samuel left open the problem of determining all squares in the sequence {Uₙ}. This problem was recently solved by the authors. In the present paper, we extend our previous joint work by completely solving the equation Uₙ = bx², where b is a fixed positive squarefree integer. This result also extends previous work of the second...
All solutions of the equation x²+x+1 = yz in non-negative integers x,y,z are given in terms of an arithmetic continued fraction.
We consider the diophantine equation(*) xp - x = yq - y in integers (x, p, y, q). We prove that for given p and q with 2 ≤ p < q, (*) has only finitely many solutions. Assuming the abc-conjecture we can prove that p and q are bounded. In the special case p = 2 and y a prime power we are able to solve (*) completely.