On the Diophantine equations
$$z^2=f(x)^2 \pm f(y)^2$$
z
2
=
f
(
x
)
2
±
f
(
y
)
2
involving quartic polynomials
2018 ◽
Vol 79
(1)
◽
pp. 25-31
◽
Keyword(s):
Keyword(s):
2011 ◽
Vol 105
(2)
◽
pp. 223-245
1990 ◽
Vol 34
(2)
◽
pp. 235-250
◽
1966 ◽
Vol s3-16
(1)
◽
pp. 153-166
◽
Keyword(s):