(x+y^2)(x^2+y)=(xy)^3
expanding and moving to one side gives
2y^3+xy3xy^2+3x^2y+x^2y^2=0
since we are interested in nonzero solutions we can divide out the common factor of y
2y^2+x3xy+3x^2+x^2y=0
grouping around y we get
2y^2+(x^23x)y+3x^2+x=0
using the quadratic equation to solve for y we get
y=[3xx^2+sqrt((x^23x)^28(3x^2+x))]/4
y=[3xx^2+sqrt(x(x8)(x+1)^2)]/4
y=[3xx^2+(x+1)sqrt(x(x8))]/4
so we need x(x8) to be a perfect square
x(x8)=k^2
x^28x=k^2
x^28x+16=k^2+16
(x4)^2=k^2+16
(x4)^2k^2=16
(x4k)(x4+k)=16
let a*b=16 be an integer factorization of 16, then we can get an integer solution fo x,k with
x4k=a
x4+k=b
adding these together gives us
2x8=a+b
solving for x we get
x=(a+b+8)/2
due to symetry we can restrict ourselves to when a<=b
so now we can just test using the factorizations of 16 namely
(1,16),(2,8),(4,4),(4,4),(8,2),(16,1)
testing each of these the only nonzero integer value for x is with (2,8) (4,4) and (2,8)
giving the possible x values 9, 8, and 1
if x=9 y is 21 or 6, if x=8 y is 10, if x=1 y=1
so the only solutions are the onese Charlie found, namely
(x,y): (1,1),(8,10),(9,6),(9,21)

Posted by Daniel
on 20180620 14:40:55 