x^4+4y²=2017
4y² par, 2017 impar⇒x^4impar⇒ximpar
x=1, 4y²=2016, y²=504,y∉Q⇒y∉Z
x=3,, x^4=81, 4y²=1936, y²=484, y=+-√484 =+/- 22 este solutie
dar desigur si x=-3 va da x^4=81
x=5, x^4=625, 4y²=1392, y²=348, y∉Q⇒y∉Z
x=7, x^4=2401, 4y²<0, y∉R⇒y∉Z
solutii
x=3; y=22
x=-3;y=22
x=3; y=-22
x=-3; y=-22
4 solutii
