# Cube roots don't sum up to integer

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.

• These kinds of problems are incredibly difficult. Look at Fermat's Last Theorem. It took hundreds of years. Jul 6, 2013 at 20:47
• 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? Jul 6, 2013 at 20:55
• Indeed, I am asking for a general proof. Jul 6, 2013 at 20:57
• Thanks, I'll check that one. In the meanwhile I've edited my question, the numbers have to be positive. Jul 6, 2013 at 21:04

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.
• 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. Jul 6, 2013 at 21:48