(In reply to
Puzzle Answer by K Sengupta)
By the problem we have:
x^3+x^2*y^2+xy+y^3=x^3-3x^2*y+3x*y^2-y^3
=> x^2*y^2 +xy+y^3 =-3x^2*y+3x*y^2-y^3
Since each of x and y is a non-zero integer, we cannot consider 0 as a valid value for each.
For y=/ 0, we obtain
2y^2+(x^2-3x)y+3x^2+x= 0 ......(€)
So the discriminate is given by:
D = n^2= (x^2-3x)^2-4*2(3x^2+x)
= x^4-6x^3-15x^2-8x
= x(x^3-6x^2-15x-8)
= x(x+1)^2(x-8)
Since D is a perfect square and (x+1)^2 is a perfect square, - ) some integer k such that:
k^2
= x(x-8)
= (x-4)^2 -4^2.....(#)
It's easy to do the necessary calculations to conclude that:
The only possible solutions to (#) are:
0^2 =4^2-4^2 and,
3^2= 5^2-4^2 ...,,(##)
This, comparing (#) with (##), we have:
(x-4) = +/-5 , +/- 4
=> x= 9, -1, 0, 8
From (€), we now observe that:
x=9 gives:
2y^2 + 54y +252=0
2(y+6)(y+21)=0
y=-6, -21
x=8 gives:
2y^2+40y+200=0
2(y+10)^2=0
y=-10
x=-1 gives:
2y^2+4y+2=0
2(y+1)^2=0
y=-1
We cannot consider x=0, since each of x any y are non-zero.
Consequently, (x,y) = (9,-6), (9,-21), (8, -10), (-1, -1) provides all possible solutions to the equation under reference.
Explanation to Puzzle Answer: Method 1
