(x+y^2)(x^2+y)=(x-y)^3
expanding and moving to one side gives
2y^3+xy-3xy^2+3x^2y+x^2y^2=0
since we are interested in non-zero solutions we can divide out the common factor of y
2y^2+x-3xy+3x^2+x^2y=0
grouping around y we get
2y^2+(x^2-3x)y+3x^2+x=0
using the quadratic equation to solve for y we get
y=[3x-x^2+-sqrt((x^2-3x)^2-8(3x^2+x))]/4
y=[3x-x^2+-sqrt(x(x-8)(x+1)^2)]/4
y=[3x-x^2+-(x+1)sqrt(x(x-8))]/4
so we need x(x-8) to be a perfect square
x(x-8)=k^2
x^2-8x=k^2
x^2-8x+16=k^2+16
(x-4)^2=k^2+16
(x-4)^2-k^2=16
(x-4-k)(x-4+k)=16
let a*b=16 be an integer factorization of 16, then we can get an integer solution fo x,k with
x-4-k=a
x-4+k=b
adding these together gives us
2x-8=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 non-zero 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)
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Posted by Daniel
on 2018-06-20 14:40:55 |
analytical solution
