Let $D$ be a positive nonsquare integer, $p$ a prime number with $p\nmid D$ and $0<\unicode[STIX]{x1D70E}<0.847$. We show that there exist effectively computable constants $C_{1}$ and $C_{2}$ such that if there is a solution to $x^{2}+D=p^{n}$ with $p^{n}>C_{1}$, then for every $x>C_{2}$ with $x^{2}+D=p^{n}m$ we have $m>x^{\unicode[STIX]{x1D70E}}$. As an application, we show that for $x\neq \{5,1015\}$, if the equation $x^{2}+76=101^{n}m$ holds, then $m>x^{0.14}$.

We improve some previously known deterministic algorithms for finding integer solutions $x,y$ to the exponential equation of the form $af^{x}+bg^{y}=c$ over finite fields.

In this paper, we consider the Diophantine equations

$$\begin{eqnarray}\displaystyle F_{n}^{q}\pm F_{m}^{q}=y^{p} & & \displaystyle \nonumber\end{eqnarray}$$

$q,p\geq 2$

$\gcd (F_{n},F_{m})=1$

$F_{k}$

$q=2$

$q$

$q\equiv 3\;(\text{mod}\;4),3<q<1087$

$q=3$

We shall show that, for any positive integer D > 0 and any primes p1, p2, the diophantine equation x2 + D = 2sp1kp2l has at most 63 integer solutions (x, k, l, s) with x, k, l ≥ 0 and s ∈ {0, 2}.

Let $c\geq 2$ be a positive integer. Terai [‘A note on the Diophantine equation $x^{2}+q^{m}=c^{n}$’, Bull. Aust. Math. Soc.90 (2014), 20–27] conjectured that the exponential Diophantine equation $x^{2}+(2c-1)^{m}=c^{n}$ has only the positive integer solution $(x,m,n)=(c-1,1,2)$. He proved his conjecture under various conditions on $c$ and $2c-1$. In this paper, we prove Terai’s conjecture under a wider range of conditions on $c$ and $2c-1$. In particular, we show that the conjecture is true if $c\equiv 3\hspace{0.6em}({\rm mod}\hspace{0.2em}4)$ and $3\leq c\leq 499$.

In 1956, Jeśmanowicz conjectured that, for any positive integers $m$ and $n$ with $m>n$, $\gcd (m,n)=1$ and $2\nmid m+n$, the Diophantine equation $(m^{2}-n^{2})^{x}+(2mn)^{y}=(m^{2}+n^{2})^{z}$ has only the positive integer solution $(x,y,z)=(2,2,2)$. In this paper, we prove the conjecture if $4\nmid mn$ and $y\geq 2$.

Let $a,b,c$ be a primitive Pythagorean triple and set $a=m^{2}-n^{2},b=2mn,c=m^{2}+n^{2}$ , where $m$ and $n$ are positive integers with $m>n$ , $\text{gcd}(m,n)=1$ and $m\not \equiv n~(\text{mod}~2)$ . In 1956, Jeśmanowicz conjectured that the only positive integer solution to the Diophantine equation $(m^{2}-n^{2})^{x}+(2mn)^{y}=(m^{2}+n^{2})^{z}$ is $(x,y,z)=(2,2,2)$ . We use biquadratic character theory to investigate the case with $(m,n)\equiv (2,3)~(\text{mod}~4)$ . We show that Jeśmanowicz’ conjecture is true in this case if $m+n\not \equiv 1~(\text{mod}~16)$ or $y>1$ . Finally, using these results together with Laurent’s refinement of Baker’s theorem, we show that Jeśmanowicz’ conjecture is true if $(m,n)\equiv (2,3)~(\text{mod}~4)$ and $n<100$ .

Let $a$ and $m$ be relatively prime positive integers with $a>1$ and $m>2$. Let ${\it\phi}(m)$ be Euler’s totient function. The quotient $E_{m}(a)=(a^{{\it\phi}(m)}-1)/m$ is called the Euler quotient of $m$ with base $a$. By Euler’s theorem, $E_{m}(a)$ is an integer. In this paper, we consider the Diophantine equation $E_{m}(a)=x^{l}$ in integers $x>1,l>1$. We conjecture that this equation has exactly five solutions $(a,m,x,l)$ except for $(l,m)=(2,3),(2,6)$, and show that if the equation has solutions, then $m=p^{s}$ or $m=2p^{s}$ with $p$ an odd prime and $s\geq 1$.

Let $m$, $a$, $c$ be positive integers with $a\equiv 3, 5~({\rm mod} \hspace{0.334em} 8)$. We show that when $1+ c= {a}^{2} $, the exponential Diophantine equation $\mathop{({m}^{2} + 1)}\nolimits ^{x} + \mathop{(c{m}^{2} - 1)}\nolimits ^{y} = \mathop{(am)}\nolimits ^{z} $ has only the positive integer solution $(x, y, z)= (1, 1, 2)$ under the condition $m\equiv \pm 1~({\rm mod} \hspace{0.334em} a)$, except for the case $(m, a, c)= (1, 3, 8)$, where there are only two solutions: $(x, y, z)= (1, 1, 2), ~(5, 2, 4). $ In particular, when $a= 3$, the equation $\mathop{({m}^{2} + 1)}\nolimits ^{x} + \mathop{(8{m}^{2} - 1)}\nolimits ^{y} = \mathop{(3m)}\nolimits ^{z} $ has only the positive integer solution $(x, y, z)= (1, 1, 2)$, except if $m= 1$. The proof is based on elementary methods and Baker’s method.

Let $q$ be an odd prime such that ${q}^{t} + 1= 2{c}^{s} $, where $c, t$ are positive integers and $s= 1, 2$. We show that the Diophantine equation ${x}^{2} + {q}^{m} = {c}^{n} $ has only the positive integer solution $(x, m, n)= ({c}^{s} - 1, t, 2s)$ under some conditions. The proof is based on elementary methods and a result concerning the Diophantine equation $({x}^{n} - 1)/ (x- 1)= {y}^{2} $ due to Ljunggren. We also verify that when $2\leq c\leq 30$ with $c\not = 12, 24$, the Diophantine equation ${x}^{2} + \mathop{(2c- 1)}\nolimits ^{m} = {c}^{n} $ has only the positive integer solution $(x, m, n)= (c- 1, 1, 2). $

Let $(a, b, c)$ be a primitive Pythagorean triple satisfying ${a}^{2} + {b}^{2} = {c}^{2} . $ In 1956, Jeśmanowicz conjectured that for any given positive integer $n$ the only solution of $\mathop{(an)}\nolimits ^{x} + \mathop{(bn)}\nolimits ^{y} = \mathop{(cn)}\nolimits ^{z} $ in positive integers is $x= y= z= 2. $ In this paper, for the primitive Pythagorean triple $(a, b, c)= (4{k}^{2} - 1, 4k, 4{k}^{2} + 1)$ with $k= {2}^{s} $ for some positive integer $s\geq 0$, we prove the conjecture when $n\gt 1$ and certain divisibility conditions are satisfied.

Let $a, b, c$ be relatively prime positive integers such that ${a}^{2} + {b}^{2} = {c}^{2} $. In 1956, Jeśmanowicz conjectured that for any positive integer $n$, the only solution of $\mathop{(an)}\nolimits ^{x} + \mathop{(bn)}\nolimits ^{y} = \mathop{(cn)}\nolimits ^{z} $ in positive integers is $(x, y, z)= (2, 2, 2)$. In this paper, we consider Jeśmanowicz’ conjecture for Pythagorean triples $(a, b, c)$ if $a= c- 2$ and $c$ is a Fermat prime. For example, we show that Jeśmanowicz’ conjecture is true for $(a, b, c)= (3, 4, 5)$, $(15, 8, 17)$, $(255, 32, 257)$, $(65535, 512, 65537)$.

Let p be an odd prime. In this paper, we consider the equation and we describe all its solutions. Moreover, we prove that this equation has no solution (x,y,m,n) when n>3 is an odd prime and y is not the sum of two consecutive squares. This extends the work of Tengely [On the diophantine equation x2+q2m=2yp, Acta Arith.127(1) (2007), 71–86].

Let a,b,c be relatively prime positive integers such that a2+b2=c2. Half a century ago, Jeśmanowicz [‘Several remarks on Pythagorean numbers’, Wiadom. Mat.1 (1955/56), 196–202] conjectured that for any given positive integer n the only solution of (an)x+(bn)y=(cn)z in positive integers is (x,y,z)=(2,2,2). In this paper, we show that (8n)x+(15n)y=(17n)z has no solution in positive integers other than (x,y,z)=(2,2,2).

Let n be a positive integer. In this paper, we consider the diophantine equation We prove that this equation has only the positive integer solutions (n, x, y, z) = (1, t, 1, 1), (1, t, 3, 2), (3, 2, 2, 2). Therefore we extend the work done by Leszczyński (Wiadom. Mat., vol. 3, 1959, pp. 37–39) and Makowski (Wiadom. Mat., vol. 9, 1967, pp. 221–224).

Let $n,d,k\geq2,b,y$ and $\ell\geq3$ be positive integers with the greatest prime factor of b not exceeding k. It is proved that the equation $n (n+d) \dotsb (n+(k-1)d)=b y^{\ell}$ has no solution if d exceeds d1, where d1 equals 30 if $\ell =3$; 950 if $\ell =4$; $5\times 10^4$ if $\ell=5$ or 6; 108 if $\ell=7$, 8, 9 or 10; 1015 if $\ell \geq 11$. This confirms a conjecture of Erdős on the above equation for a large number of values of d.