aritra barua wrote:x^2+xy+y^2=(x+y+3)^3/27,find all(x,y)
It is obvious that $x+y$ is divisible by $3$ . Put $x+y$ = $3k$ .
So, the equation becomes,
$$(3k)^2 - x(3k-x) = (k+1)^3 $$
=> $$ x^2 - 3kx + 9k^2 = k^3 + 3k^2 + 3k +1 $$
=> $$ x^2 - 3kx - k^3 + 6k^2 - 3k -1 = 0 $$
As we have to find the integer roots of this equation, the discriminant must be a perfect square.
Discriminant = $$ 9k^2 + 4k^3 - 24k^2 + 12k +4 $$ = $$ (k-2)^2 (4k+1) $$
So, $4k+1$ must be a perfect square => $$k = n^2 + n$$
Substituting this, we get 2 pairs of $(x,y)=(n^3+3n^2-1,-n^3+3n+1)$ and $(-n^3 + 3n +1,n^3 + 3n^2 -1)$