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My question looks quite obvious, but I'm looking for a strict proof for this. (At least, I assume it's true what I claim.)

Why can't the sum of two cube roots of positive non-perfect cubes be an integer?

For example: $\sqrt[3]{100}+\sqrt[3]{4}$ isn't an integer. Well, I know this looks obvious, but I can't prove it...

For given numbers it will be easy to show, by finding under- and upper bounds for the roots (or say take a calculator and check it...).

Any work done so far:

Suppose $\sqrt[3]m+\sqrt[3]n=x$, where $x$ is integer. This can be rewritten as $m+n+3x\sqrt[3]{mn}=x^3$ (by rising everything to the power of $3$ and then substituting $\sqrt[3]m+\sqrt[3]n=x$ again) so $\sqrt[3]{mn}$ is rational, which implies $mn$ is a perfect cube (this is shown in a way similar to the well-known proof that $\sqrt2$ is irrational.).

Now I don't know how to continue. One way is setting $n=\frac{a^3}m$, which gives $m^2+a^3+3amx=mx^3$ but I'm not sure whether this is helpful.

Maybe the solution has to be found similar to the way one would do it with a calculator: finding some bounds and squeezing the sum of these roots between two well-chosen integers. But this is no more then a wild idea.

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  • $\begingroup$ These kinds of problems are incredibly difficult. Look at Fermat's Last Theorem. It took hundreds of years. $\endgroup$ – Fly by Night Jul 6 '13 at 20:47
  • $\begingroup$ Well $mn$ must be a perfect cube, but $400 = 2^4 \cdot 5^2$ is not, so that concludes the proof for your example, doesn't it? Or are you wondering about a general idea? $\endgroup$ – Patrick Da Silva Jul 6 '13 at 20:55
  • $\begingroup$ Indeed, I am asking for a general proof. $\endgroup$ – punctured dusk Jul 6 '13 at 20:57
  • $\begingroup$ There is a very general proof for this given here: ac.els-cdn.com/0022314X81900408/…. $\endgroup$ – Cocopuffs Jul 6 '13 at 20:59
  • $\begingroup$ Thanks, I'll check that one. In the meanwhile I've edited my question, the numbers have to be positive. $\endgroup$ – punctured dusk Jul 6 '13 at 21:04
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Suppose $a+b=c$, so that $a+b-c=0$, with $a^3, b^3, c$ all rational.

Then we have $-3abc=a^3+b^3-c^3$ by virtue of the identity $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$$ (take $c$ with a negative sign)

Hence $a+b$ and $ab$ are both rational, so $a$ and $b$ satisfy a quadratic equation with rational coefficients.

There are lots of ways of completing the proof from here.

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  • $\begingroup$ Guys, thanks for comments - both spot on, both reflected in edits. $\endgroup$ – Mark Bennet Jul 6 '13 at 21:20
  • $\begingroup$ I completed it like this: Let $a^2=sa-p$ where $s=a+b$ and $p=ab$, then $a^3=(s^2-p)a-sp$ so $a$ is rational unless both $a^3+sp$ and $s^2-p$ equal $0$ which implies $a=b=0$. (I've just noticed that you already assume $c$ not to be $0$ when concluding that $ab$ is rational, therefore I was a bit confused when trying to complete the proof.) But anyway thanks, also for leaving the rest of the proof for me, which made it an extra exercise. $\endgroup$ – punctured dusk Jul 6 '13 at 21:48

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