Problem:
Solve in integers the equation
x2+xy+y2=(3x+y​+1)3
Solution:
Let x+y=3k, with k∈Z. Then x2+x(3k−x)+(3k−x)2=(k+1)3, which reduces to
x2−(3k)x−(k3−6k2+3k+1)=0
Its discriminant Δ is
9k2+4(k3−6k2+3k+1)=4k3−15k2+12k+4
We notice the (double) root k=2, so Δ=(4k+1)(k−2)2. It follows that 4k+1=(2t+1)2 for some nonnegative integer t, hence k=t2+t and
x=21​(3(t2+t)±(2t+1)(t2+t−2))
We obtain (x,y)=(t3+3t2−1,−t3+3t+1) and (x,y)=(−t3+3t+1,t3+3t2−1), t∈{0,1,2,…}.
OR
One can also try to simplify the original equation as much as possible. First with k= 3x+y​+1 we get
x2−3xk+3x=k3−9k2+18k−9
But then we recognize terms from the expansion of (k−3)3 so we use s=k−3 and obtain
x2−3xs−6x=s3−9s−9
So again it becomes natural to use x−3=u. The equation becomes
u2−3su−s3=0
We view this as a quadratic in u, whose discriminant is s2(9+4s), and so 9+4s must be a perfect square, and because it is odd, it must be of the form (2t+1)2. It follows that s=t2+t−2, and so k=t2+t+1. We obtain the same family of solutions.
The problems on this page are the property of the MAA's American Mathematics Competitions