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 $\left( x,\,y,\,z \right)\,=\,\left( 2,\,2,\,2 \right)$.

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 (a,b,c) be a primitive Pythagorean triple such that b is even. In 1956, Jeśmanowicz conjectured that the equation ax+by=cz has the unique solution (x,y,z)=(2,2,2) in the positive integers. This is one of the most famous unsolved problems on Pythagorean triples. In this paper we propose a similar problem (which we call the shuffle variant of Jeśmanowicz’ problem). Our problem states that the equation cx+by=az with x,y and z positive integers has the unique solution (x,y,z)=(1,1,2) if c=b+1 and has no solutions if c>b+1 . We prove that the shuffle variant of the Jeśmanowicz problem is true if c≡1 mod b.

Let a,b,c be relatively prime positive integers such that a2+b2=c2 with b even. In 1956 Jeśmanowicz conjectured that the equation ax+by=cz has no solution other than (x,y,z)=(2,2,2) in positive integers. Most of the known results of this conjecture were proved under the assumption that 4 exactly divides b. The main results of this paper include the case where 8 divides b. One of our results treats the case where a has no prime factor congruent to 1 modulo 4, which can be regarded as a relevant analogue of results due to Deng and Cohen concerning the prime factors of b. Furthermore, we examine parities of the three variables x,y,z, and give new triples a,b,c such that the conjecture holds for the case where b is divisible by 8. In particular, to prove our results, we shall show an important result which asserts that if x,y,z are all even, then x/2,y/2,z/2 are all odd. Our methods are based on elementary congruence and several strong results on generalized Fermat equations given by Darmon and Merel.