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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 $\Delta \geqslant 0$

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

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

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    $\begingroup$ Don't forget the case $x + y = 0$! $\endgroup$ – Izaak van Dongen Aug 27 at 16:17
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    $\begingroup$ also, draw a careful picture of the ellipse on graph paper. Once you deal with $x+y = 0,$ the remaining points make an ellipse that is not large. Get some graph paper and draw a picture. Or print out printablepaper.net/category/graph People here often ask about help visualizing, the way to get better is to draw things yourself $\endgroup$ – Will Jagy Aug 27 at 16:26
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    $\begingroup$ All that remains (in addition to the case $x+y=0$) is to substitute $y=0,1,2$, and solve for $x$. $\endgroup$ – Théophile Aug 27 at 16:28
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    $\begingroup$ $x^3+y^3=(x+y)^2\implies (2 x - y - 1)^2 + 3 (y - 1)^2 = 4$. $\endgroup$ – Dmitry Ezhov Aug 27 at 16:29
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    $\begingroup$ @A-levelStudent It imples that $|y-1|\leq 1$ and $|2x-y-1|\leq 2$. There are not many cases to check (note that $x$ and $y$ are integers). $\endgroup$ – richrow Aug 27 at 18:31
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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 $$(0, 1), (1, 0), (1, 2), (2, 1), \text{ and } (x, -x).$$

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  • $\begingroup$ Thanks, didn't catch that @mjw. Edited :) $\endgroup$ – FruDe Aug 28 at 11:53
  • $\begingroup$ You are welcome. $\endgroup$ – mjw Aug 28 at 12:29

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