Let p be a fixed odd prime. We prove the following results for positive integer solutions $(x,m,n)$ of the equation $(*)\ x^2=2^m+p^n$. (i) If $p \equiv 3 \pmod 8$, then $(*)$ has only the solution $(p,x,m,n)=(3,5,4,2)$. (ii) If $p \equiv 5 \pmod 8$, then $(*)$ has only the solution $(p,x,m,n)=(5,3,2,1)$. (iii) If $p \equiv 7 \pmod 8$, then $(*)$ has at most one solution $(x,m,n)$, except for $p=7$, $(x,m,n)=(3,1,1)$ and $(9,5,2)$. Moreover, if $p=2^q-1$ is a Mersenne prime with $p>7$, where q is an odd prime with $q>3$, then $(*)$ has exactly one solution $(x,m,n)=(2^q+1,q+2,2)$. If $p \equiv 7\pmod 8$, p is not a Mersenne prime and either $p<1.5\times 10^{12}$ or $p>C$, where C is an effectively computable absolute constant, then $(*)$ has only the solutions $p=a^2-2$, where a is an odd positive integer, $(x,m,n)=(a,1,1)$. (iv) If $p \equiv 1\pmod 8$ with $p \ne 17$, then $(*)$ has at most two solutions $(x,m,n)$.

It is conjectured that for any fixed relatively prime positive integers a,b and c all greater than 1 there is at most one solution to the equation $a^x+b^y=c^z$ in positive integers x, y and z, except for specific cases. We develop the methods in our previous work which rely on a variety from Baker’s theory and thoroughly study the conjecture for cases where c is small relative to a or b. Using restrictions derived from the hypothesis that there is more than one solution to the equation, we obtain a number of finiteness results on the conjecture. In particular, we find some, presumably infinitely many, new values of c with the property that for each such c the conjecture holds true except for only finitely many pairs of a and b. Most importantly we prove that if $c=13$ then the equation has at most one solution, except for $(a,b)=(3,10)$ or (10,3) each of which gives exactly two solutions. Further, our study with the help of the Schmidt Subspace Theorem among others more, brings strong contributions to the study of Pillai’s type Diophantine equations, notably a general and satisfactory result on a well-known conjecture of M. Bennett on the equation $a^x-b^y=c$ for any fixed positive integers a,b and c with both a and b greater than 1. Some conditional results are presented under the abc-conjecture as well.

For a wide class of integer linear recurrence sequences $(u(n))_{n=1}^\infty $, we give an upper bound on the number of s-tuples $\left (n_1, \ldots , n_s\right ) \in \left ({\mathbb Z}\cap [M+1,M+ N]\right )^s$ such that the corresponding elements $u(n_1), \ldots , u(n_s)$ in the sequence are multiplicatively dependent.

For an integer $k \geq 2$, let $P_{n}^{(k)}$ be the k-generalised Pell sequence, which starts with $0, \ldots ,0,1$ (k terms), and each term thereafter is given by the recurrence $P_{n}^{(k)} = 2 P_{n-1}^{(k)} +P_{n-2}^{(k)} +\cdots +P_{n-k}^{(k)}$. We search for perfect powers, which are sums or differences of two k-generalised Pell numbers.

We prove a nonabelian variant of the classical Mordell–Lang conjecture in the context of finite- dimensional central simple algebras. We obtain the following result as a particular case of a more general statement. Let K be an algebraically closed field of characteristic zero, let $B_1,\dots ,B_r\in \mathrm {GL}_m(K)$ be matrices with multiplicatively independent eigenvalues and let V be a closed subvariety of $\mathrm {GL}_m(K)$ not passing through zero. Then there exist only finitely many elements of $\mathrm {GL}_m(K)$ of the form $B_1^{n_1}\cdots B_r^{n_r}$ (as we vary $n_1,\dots ,n_r$ in $\mathbb {Z}$) lying on the subvariety V.

A generalisation of the well-known Pell sequence $\{P_n\}_{n\ge 0}$ given by $P_0=0$, $P_1=1$ and $P_{n+2}=2P_{n+1}+P_n$ for all $n\ge 0$ is the k-generalised Pell sequence $\{P^{(k)}_n\}_{n\ge -(k-2)}$ whose first k terms are $0,\ldots ,0,1$ and each term afterwards is given by the linear recurrence $P^{(k)}_n=2P^{(k)}_{n-1}+P^{(k)}_{n-2}+\cdots +P^{(k)}_{n-k}$. For the Pell sequence, the formula $P^2_n+P^2_{n+1}=P_{2n+1}$ holds for all $n\ge 0$. In this paper, we prove that the Diophantine equation

$$ \begin{align*} (P^{(k)}_n)^2+(P^{(k)}_{n+1})^2=P^{(k)}_m \end{align*} $$

has no solution in positive integers $k, m$ and n with $n>1$ and $k\ge 3$.

Let p be a prime and let r, s be positive integers. In this paper, we prove that the Goormaghtigh equation $(x^m-1)/(x-1)=(y^n-1)/(y-1)$, $x,y,m,n \in {\mathbb {N}}$, $\min \{x,y\}>1$, $\min \{m,n\}>2$ with $(x,y)=(p^r,p^s+1)$ has only one solution $(x,y,m,n)=(2,5,5,3)$. This result is related to the existence of some partial difference sets in combinatorics.

Let a, b, c be fixed coprime positive integers with $\min \{ a,b,c \}>1$. Let $N(a,b,c)$ denote the number of positive integer solutions $(x,y,z)$ of the equation $a^x + b^y = c^z$. We show that if $(a,b,c)$ is a triple of distinct primes for which $N(a,b,c)>1$ and $(a,b,c)$ is not one of the six known such triples, then $c>10^{18}$, and there are exactly two solutions $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$ with $2 \mid x_1$, $2 \mid y_1$, $z_1=1$, $2 \nmid y_2$, $z_2>1$, and, taking $a<b$, we must have $a=2$, $b \equiv 1 \bmod 12$, $c \equiv 5\, \mod 12$, with $(a,b,c)$ satisfying further strong restrictions. These results support a conjecture put forward by Scott and Styer [‘Number of solutions to $a^x + b^y = c^z$’, Publ. Math. Debrecen 88 (2016), 131–138].

A multivariate, formal power series over a field K is a Bézivin series if all of its coefficients can be expressed as a sum of at most r elements from a finitely generated subgroup $G \le K^*$; it is a Pólya series if one can take $r=1$. We give explicit structural descriptions of D-finite Bézivin series and D-finite Pólya series over fields of characteristic $0$, thus extending classical results of Pólya and Bézivin to the multivariate setting.

Let a, b, c be fixed coprime positive integers with $\min \{a,b,c\}>1$. We discuss the conjecture that the equation $a^{x}+b^{y}=c^{z}$ has at most one positive integer solution $(x,y,z)$ with $\min \{x,y,z\}>1$, which is far from solved. For any odd positive integer r with $r>1$, let $f(r)=(-1)^{(r-1)/2}$ and $2^{g(r)}\,\|\, r-(-1)^{(r-1)/2}$. We prove that if one of the following conditions is satisfied, then the conjecture is true: (i) $c=2$; (ii) a, b and c are distinct primes; (iii) $a=2$ and either $f(b)\ne f(c)$, or $f(b)=f(c)$ and $g(b)\ne g(c)$.

We prove that if $s\ge 2$ is a fixed integer, then the equation $ns^n+1=(b^m-1)/(b-1)$ has only finitely many positive integer solutions $(n,b,m)$ with $b\ge 2$ and $m\ge 3$. When $s=2$, it has no solution.

We find all integer solutions to the equation $x^2+5^a\cdot 13^b\cdot 17^c=y^n$ with $a,\,b,\,c\geq 0$, $n\geq 3$, $x,\,y>0$ and $\gcd (x,\,y)=1$. Our proof uses a deep result about primitive divisors of Lucas sequences in combination with elementary number theory and computer search.

We show that in a parametric family of linear recurrence sequences $a_1(\alpha ) f_1(\alpha )^n + \cdots + a_k(\alpha ) f_k(\alpha )^n$ with the coefficients $a_i$ and characteristic roots $f_i$, $i=1, \ldots ,k$, given by rational functions over some number field, for all but a set of elements $\alpha $ of bounded height in the algebraic closure of ${\mathbb Q}$, the Skolem problem is solvable, and the existence of a zero in such a sequence can be effectively decided. We also discuss several related questions.

Jeśmanowicz conjectured that $(x,y,z)=(2,2,2)$ is the only positive integer solution of the equation $(*)\; ((\kern1.5pt f^2-g^2)n)^x+(2fgn)^y=((\kern1.5pt f^2+g^2)n)^x$, where n is a positive integer and f, g are positive integers such that $f>g$, $\gcd (\kern1.5pt f,g)=1$ and $f \not \equiv g\pmod 2$. Using Baker’s method, we prove that: (i) if $n>1$, $f \ge 98$ and $g=1$, then $(*)$ has no positive integer solutions $(x,y,z)$ with $x>z>y$; and (ii) if $n>1$, $f=2^rs^2$ and $g=1$, where r, s are positive integers satisfying$(**)\; 2 \nmid s$ and $s<2^{r/2}$, then $(*)$ has no positive integer solutions $(x,y,z)$ with $y>z>x$. Thus, Jeśmanowicz’ conjecture is true if $f=2^rs^2$ and $g=1$, where r, s are positive integers satisfying $(**)$.

We show that the set of natural numbers has an exponential diophantine definition in the rationals. It follows that the corresponding decision problem is undecidable.

Let $K$ be an algebraic number field of degree $d\geqslant 3$ , $\unicode[STIX]{x1D70E}_{1},\unicode[STIX]{x1D70E}_{2},\ldots ,\unicode[STIX]{x1D70E}_{d}$ the embeddings of $K$ into $\mathbb{C}$ , $\unicode[STIX]{x1D6FC}$ a non-zero element in $K$ , $a_{0}\in \mathbb{Z}$ , $a_{0}>0$ and

$$\begin{eqnarray}F_{0}(X,Y)=a_{0}\mathop{\prod }_{i=1}^{d}(X-\unicode[STIX]{x1D70E}_{i}(\unicode[STIX]{x1D6FC})Y).\end{eqnarray}$$

$\unicode[STIX]{x1D710}$

$K$

$a\in \mathbb{Z}$

$F_{0}(X,Y)\in \mathbb{Z}[X,Y]$

$\unicode[STIX]{x1D710}^{a}$

$a\in \mathbb{Z}$

$\unicode[STIX]{x1D710}$

$$\begin{eqnarray}F_{a}(X,Y)=a_{0}\mathop{\prod }_{i=1}^{d}(X-\unicode[STIX]{x1D70E}_{i}(\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D710}^{a})Y).\end{eqnarray}$$

$m>0$

$(x,y,a)\in \mathbb{Z}^{3}$

$$\begin{eqnarray}0<|F_{a}(x,y)|\leqslant m\end{eqnarray}$$

$xy\not =0$

$\mathbb{Q}(\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D710}^{a})=K$

$m$

$K$

$F_{0}$

$\unicode[STIX]{x1D710}$

$d$

Let $a$ , $b$ , and $c$ be primitive Pythagorean numbers such that ${{a}^{2}}\,+\,{{b}^{2}}\,=\,{{c}^{2}}$ with $b$ even. In this paper, we show that if ${{b}_{0}}\,\equiv \,\in \,\,\,\left( \bmod \,a \right)$ with $\text{ }\!\!\varepsilon\!\!\text{ }\,\in \,\left\{ \pm 1 \right\}$ for certain positive divisors ${{b}_{0}}$ of $b$ , then the Diophantine equation ${{a}^{x}}\,+\,{{b}^{y}}\,=\,{{c}^{z}}$ has only the positive solution $\left( x,\,y,\,z \right)\,=\,\left( 2,\,2,\,2 \right)$ .

We solve the equation ${x}^{a} + {x}^{b} + 1= {y}^{q} $ in positive integers $x, y, a, b$ and $q$ with $a\gt b$ and $q\geq 2$ coprime to $\phi (x)$. This requires a combination of a variety of techniques from effective Diophantine approximation, including lower bounds for linear forms in complex and $p$-adic logarithms, the hypergeometric method of Thue and Siegel applied $p$-adically, local methods, and the algorithmic resolution of Thue equations.

Let d > 0 be a squarefree integer and denote by h = h(−d) the class number of the imaginary quadratic field . It is well known (see e.g. [25]) that for a given positive integer N there are only finitely many squarefree d's for which h(−d) = N. In [45], Saradha and Srinivasan and in [28] Le and Zhu considered the equation in the title and solved it completely under the assumption h(−d) = 1 apart from the case d ≡ 7 (mod 8) in which case y was supposed to be odd. We investigate the title equation in unknown integers (x, y, l, n) with x ≥ 1, y ≥ 1, n ≥ 3, l ≥ 0 and gcd(x, y) = 1. The purpose of this paper is to extend the above result of Saradha and Srinivasan to the case h(−d) ∈ {2, 3}.

We prove that for any positive integers x,d and k with gcd (x,d)=1 and 3<k<35, the product x(x+d)⋯(x+(k−1)d) cannot be a perfect power. This yields a considerable extension of previous results of Győry et al. and Bennett et al., which covered the cases where k≤11. We also establish more general theorems for the case where x can also be a negative integer and where the product yields an almost perfect power. As in the proofs of the earlier theorems, for fixed k we reduce the problem to systems of ternary equations. However, our results do not follow as a mere computational sharpening of the approach utilized previously; instead, they require the introduction of fundamentally new ideas. For k>11, a large number of new ternary equations arise, which we solve by combining the Frey curve and Galois representation approach with local and cyclotomic considerations. Furthermore, the number of systems of equations grows so rapidly with k that, in contrast with the previous proofs, it is practically impossible to handle the various cases in the usual manner. The main novelty of this paper lies in the development of an algorithm for our proofs, which enables us to use a computer. We apply an efficient, iterated combination of our procedure for solving the new ternary equations that arise with several sieves based on the ternary equations already solved. In this way, we are able to exclude the solvability of the enormous number of systems of equations under consideration. Our general algorithm seems to work for larger values of k as well, although there is, of course, a computational time constraint.