In this paper, we find all integer solutions $(x,y,n,a,b,c)$ of the equation in the title for non-negative integers $a,b$ and $c$ under the condition that the integers $x$ and $y$ are relatively prime and $n\ge 3$. The proof depends on the famous primitive divisor theorem due to Bilu, Hanrot and Voutier and the computational techniques on some elliptic curves.

We find all the solutions of the Diophantine equation $x²+2^{\alpha }{13}^{\beta }=yⁿ$ in positive integers x,y,α,β,n ≥ 3 with x and y coprime.

In this paper, we find all solutions of the Diophantine equation $x^2+2^{\alpha }{5}^{\beta }{17}^{\gamma }=y^n$ in positive integers $x,y\ge 1$, $\alpha ,\beta ,\gamma ,n\ge 3$ with $gcd(x,y)=1$.

Consider the equation in the title in unknown integers $(x,y,k,l,n)$ with $x\ge 1$, $y>1$, $n\ge 3$, $k\ge 0$, $l\ge 0$ and $gcd(x,y)=1$. Under the above conditions we give all solutions of the title equation (see Theorem 1).

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 $x,y,z$ if $p\ge 7$ is prime and $k\ge 2$. 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 $x^2+2^k=y^n$ 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) $(x^m+1)(x^n+1)=y²$, 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 $x^y-y^x=c^z$ with gcd(x,y) = 1 then (x,y,z) = (2,1,k), (3,2,k), k ≥ 1 if c = 1, and either $(x,y,z)=(c^k+1,1,k)$, k ≥ 1 or $2\le x<y\le max1.5\times {10}^{10},c$ if c ≥ 2.

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 &lt; 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.