Solution. I began by considering \[ \left (\frac {x+y}{3} + 1\right )^3 = A \] for some integer \(A\). My thought was that it’s harder for the right-hand side to be an integer given the division so that makes for a natural starting place.
Rearrangement gives us \[ x+y = 3\left (A^{1/3} - 1\right ) \] Since we want integer solutions \((x,y)\), then we would need that right-hand side to remain an integer. That means we can choose some integer \(k\) such that \(k = A^{1/3} - 1\). So let’s suggest that \(x+y = 3k\) for some integer \(k\). Notice the left-hand side \(x^2 + xy + y^2\) is almost \((x+y)^2 = 9k^2\). Since \(x+y=3k\), we may take \(y\) as dependent: \(y = 3k - x\).
Now we have \[ x^2 + xy + y^2 = 9k^2 - xy = (k+1)^3 \] This gives us \[ xy = 3kx - x^2 = 9k^2 - (k+1)^3 = -k^3 + 6k^2 - 3k - 1 \] Observe that this is a quadratic polynomial in \(x\). Form the discriminant of this polynomial: \[ \Delta = 9k^2 + 4(k^3 -6k^2 + 3k + 1) = 4k^3 -15k^2+12k+4 \] By inspection, we can surmise that if \(\Delta \) has a rational root \(\frac {p}{q}\), then both \(p\) and \(q\) divide \(4\) so \(p,q = 1, 2,\,\text {or}\, 4\). We’re interested in integer solutions, so we’ll presume that \(p/q = 1, 2,\,\text {or}\, 4\). After trying the possibilities, the only root that gives a cleanly factorizes the discriminant is \(k=2\) with multiplicity \(2\).
Working out the polynomial long division of \((k-2)^2\) into \(4k^3 -15k^2+12k+4\) gives the remaining factor: \[ 4k^3 -15k^2+12k+4 = (k-2)^2(4k+1). \]
Recalling the quadratic formula, the numerator works out to be \[ 3k \pm \sqrt {4k^3 -15k^2+12k+4} = 3k \pm (k-2)\sqrt {4k+1} \] We now require, if we want integer solutions, that \(\sqrt {4k+1} = aw+b\) for some integers \(a,b\) and \(w\). Squaring gives us \[ 4k+1 = a^2w^2 + 2abw + b^2 \] Let’s choose \(b=1\) so that after a little algebra we get \[ k = \frac {a^2w^2}{4} + \frac {aw}{2} \] As \(k\) is an integer, we must have \(2|aw\) and \(4 | a^2w^2\). Since \(w\) is presumed fixed, the clear choice is picking \(a=2\). Now we have \(4k+1 = (2w+1)^2\) for some integer \(w\).
Expanding \(4k+1 = (2w+1)^2\), we can now write \[ k = w^2 + w. \]
Let’s return to the numerator of the quadratic formula and plug in our new expression for \(k\): \[ 3k \pm \sqrt {\Delta } = 3(w^2 + w) \pm (2w+1)(w^2+ w - 2) \]
Consider the parity of the parts of this expression. If \(w\) is even, then \(w^2 + w\) is even and \(w^2 + w - 2\) is also even. In that case, \(3k \pm \sqrt {\Delta }\) is even. Now if \(w\) is odd, then \(w^2 + w\) is even and \(w^2 + w - 2\) is again even. In this case too we have that \(3k \pm \sqrt {\Delta }\) is even. It follows then that the roots \[ x_{\pm } = \frac {3k\pm \sqrt {\Delta }}{2} \] are integers.
From \(y = 3k - x\), we obtain the corresponding integer \(y\) values.