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 $(**)$ .

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 $(a, b, c)$ 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 $(a, b, c)= (3, 4, 5)$, $(15, 8, 17)$, $(255, 32, 257)$, $(65535, 512, 65537)$.