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Is there an integer which is the sum of the cubes of two rational numbers, but not the sum of the cubes of two integers?

This is not possible if we consider cubes in place of squares (Davenport-Cassels theorem, see here, here), but what about my situation? Notice that $7$, see here, is not the sum of the cubes of two rational numbers, but every rational number is the sum of the cubes of three rational numbers.

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    $\begingroup$ cubes of two integers do the integers have to be positive? $\endgroup$ – TheSimpliFire Jul 17 '18 at 16:03
  • $\begingroup$ @TheSimpliFire : I think negative integers / rational are allowed $\endgroup$ – Alphonse Jul 17 '18 at 16:08
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If $n$ is any integer then $n^3 \equiv -1,0,1 \pmod 9$. Hence if $a,b$ are any integers:

$$a^3 + b^3 \equiv -2,-1,0,1,2 \pmod 9$$ Conversely if any integer $k \equiv 3,4,5,6 \pmod 9$, it cannot be a sum of two cubes of integers. Searching among small integers meeting this condition, I found:

$$\Big(\frac{2}{3}\Big)^3 + \Big(\frac{7}{3}\Big)^3 = 13$$

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$$(\frac{17}{21})^3+(\frac{37}{21})^3=6$$ which is neither the sum nor the difference ot two integer-cubes.

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  • $\begingroup$ This looks nice, but why does $6 = a^3+b^3 = a^3 - (-b)^3$ have no solution in the integers? $\endgroup$ – Alphonse Jul 17 '18 at 16:18
  • $\begingroup$ The smallest possible difference (if some cube with absolute value $8$ or more is included), is $7$ $\endgroup$ – Peter Jul 17 '18 at 16:24
  • $\begingroup$ Thanks. How did you find this answer? By using a computer,looking for rational cubes giving $6$ as a sum? (Just our of curiosity) $\endgroup$ – Alphonse Jul 17 '18 at 16:26
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    $\begingroup$ Of course with a computer : I searched for some solutions with PARI/GP and noticed that $6$ is a suitable integer. I used just brute force :) $\endgroup$ – Peter Jul 17 '18 at 16:27

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