On the Diophantine equation .
On the diophantine equation .
On the diophantine equation
On the diophantine equation
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)...
On the Diophantine equation
On the diophantine equation
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.
On the diophantine equation
On the Diophantine equation (x² ± C)(y² ± D) = z⁴
On the Diophantine Equation x2 + D = 4yq.
On the Diophantine equation x² - dy⁴ = 1 with prime discriminant II
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...
On the diophantine equation x²+x+1 = yz
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.
On the diophantine equation xl + yl = czln.
On the diophantine equation xp - x = yq - y.
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.
On the diophantine equation x(x-1)...(x-(m-1)) = λy(y-1 )...(y-(n-1)) + l
On the diophantine equation
On the diophantine equation
On the diophantine equations
On the diophantine equations 2y2 = 7k + 1 and x2 + 11 = 3n.
On the Diophantine equations of type .
On the diophantine Frobenius problem.