(In reply to
Explanation to Puzzle Answer: Method 1 by K Sengupta)
x^3+x^2*y^2+xy+y^3=x^3-3x^2y+3xy^2-y^3
=> x^2*y^2 +xy +y^3 = -3x^2*y +3x*y^2-y^3
=> x^2*y +x+y^2 =-3x^2+3xy-y^2 ( as y is nonzero, we can divide by y)
=> 2y^2+x^2*y-3xy+3x^2+x =0
=> 2y^2+(x^2-3x)y +(3x^2+x) =0
For y to be an integer, the discriminant must be a perfect square.
Here a=2, b=x^2-3x, c=3x^2+x
Then D = b^2-4ac
= (x^2-3x)^2 -(4)(2)(3x^2+x)
= x^4-6x^3+9x^2-24x^2-8x
= x(x^3-6x^2-15x-8)
Let f(x) = x^2-6x^2-15x-8
Now, f(-1) =0
So, (x+1) is a factor of f(x)
f(x) = (x+1)^2*(x-8),,,,,,,,,(*), so that:
D= x*(x+1)^2*(x-8)
If D =0, then from(*), we havex=-1,8
Now, D is a perfect square, so that:
x(x-8) = z^2(say)
=> (x-4)^2 - 16 = z^2 .......(**)
We know that, 4^2-16=0 and, 5^2-16=9 ( we could have solved (**) by regular procedure to arrive at the same result)
So, x-4=+/-4, +/-5
=> x= 0,8,-1,9
Since we are asked for nonzero solutions, we will skip the zero value of x.
Case 1: x= -1
Then, 2y^2+4y+2=0
=> 2(y+1)^2=0
=> y= -1
Case 2: x=8
2y^2+40y+200=0
=> 2(y+10)^2=0
=> y= -10
Case 3: x=9
2y^2+54y+252=0
=> y^2+27y+126=0
=> (y+21)((y+6)=0
=> y = -21, -6
Hence, (x,y) = (-1,-1), (8,-10), (9,-21), and (9,-6) constitutes all possible solutions to the given problem.
Edited on June 2, 2022, 9:06 am
Explanation to Puzzle Answer: Method 2
