Mathematics Asked on November 23, 2020

Find all pairs of integers $(x, y)$ such that $$x^3+y^3=(x+y)^2.$$

Since $x^3+y^3 = (x+y)(x^2-xy+y^2)$ we get that $$x^2-xy+y^2=x+y$$

this can be expressed as $$x^2-(y-1)x+y^2-y=0.$$

Since we want integers we should probably look at when the discriminant is positive?

$$Delta = (y-1)^2-4(y^2-y)=-3y^2+6y+1$$

so for $Delta geqslant 0$

$$-frac{2sqrt3}{3}+1 leqslant y leqslant frac{2sqrt3}{3}+1$$

only possible solutions are $y=0,1,2.$ However I don’t see how this is helpful at all here. What should I do?

You are almost there. Substitute $y = 0, 1, 2$ and solve for $x$ in each case.

When $y=0$, the equation is $x^3 = x^2$. The two solutions for $x$ are $0, 1$.

When $y = 1$, the equation is $x^3+1 = (x+1)^2$. Expanding and rearranging gets $x^3-x^2-2x=0$, and the solutions are $x = -1, 0, 2$.

When $y = 2$, the equation is $x^3+8 = (x+2)^2$. Expanding and rearranging gets $x^2-x^2-4x+4 = 0$, and the solutions are $-2, 1, 2$. (You could use RRT to get the solutions.)

So far, we have eight pairs, namely $$(0, 0), (1, 0), (-1, 1), (0, 1), (2, 1), (-2, 2), (1, 2), (2, 2).$$

However, also note that when $x = -y$, the equation is satisfied, since $$(-y)^3+y^3 = ((-y)+y)^2 rightarrow 0 = 0$$

Therefore, all possible solutions are $$(0, 1), (1, 0), (1, 2), (2, 1), text{ and } (x, -x).$$

Correct answer by FruDe on November 23, 2020

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