Article
Keywords:
exponential diophantine equation; Lucas number; positive divisor
Summary:
Let $a$, $b$, $c$, $r$ be positive integers such that $a^{2}+b^{2}=c^{r}$, $\min (a,b,c,r)>1$, $\gcd (a,b)=1, a$ is even and $r$ is odd. In this paper we prove that if $b\equiv 3\hspace{4.44443pt}(\@mod \; 4)$ and either $b$ or $c$ is an odd prime power, then the equation $x^{2}+b^{y}=c^{z}$ has only the positive integer solution $(x,y,z)=(a,2,r)$ with $\min (y,z)>1$.
References:
[1] Y. Bilu, G. Hanrot and P. Voutier (with an appendix by M. Mignotte):
Existence of primitive divisors of Lucas and Lehmer numbers. J. Reine Angew. Math. 539 (2001), 75–122.
MR 1863855
[3] Z.-F. Cao and X.-L. Dong:
The diophantine equation $a^{2}+b^{y}=c^{z}$. Proc. Japan Acad. 77A (2001), 1–4.
MR 1934716
[4] Z.-F. Cao, X.-L. Dong and Z. Li:
A new conjecture concerning the diophantine equation $x^{2}+b^{y}=c^{z}$. Proc. Japan Acad. 78A (2002), 199–202.
MR 1950170
[5] L. Jeśmanowicz:
Several remarks on Pythagorean number. Wiadom. Mat. 1 (1955/1956), 196–202. (Polish)
MR 0110662
[6] C. Ko:
On the diophantine equation $x^{2}=y^{n}+1, xy\ne 0$. Sci.Sin. 14 (1964), 457–460.
MR 0183684
[9] T. Nagell: Sur I’impossibilité de quelques equation á deux indéterminées. Norsk Matem. Forenings Skrifter 13 (1921), 65–82.