The question was asked in a Twitter interview.
For given integers $a$ and $b$, find all reals $x$, $y$ such that $1\leq x\leq a$ , $1\leq y\leq b$ and $(x^{1/3} + y^{1/3})^3$ is an integer.
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1 Answer
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By continuity, $(x^{1/3}+y^{1/3})^3$ assumes every real value between $(1^{1/3}+1^{1/3})^3=8$ and $(a^{1/3}+b^{1/3})^3$, including every integer in between. No other integers are achievable.
Once the integer is chosen, the set of $(x,y)$ values that yield that integer forms a curve between the $x=1$ and $y=1$ axes, that bulges toward the origin.
There isn't really a cute or nice answer to this question, except perhaps a picture.
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$\begingroup$ Thanks for the insightful answer. I never thought about it like this. $\endgroup$– xperienOct 8, 2015 at 4:52