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It's very known that 1729 is the smallest number that can be expressed as the sum of two cubes in two different ways, but is there a proof that not uses brute force?

If we let $n$ be the smallest number such that $n=a^3+b^3=c^3+d^3$ with $a\neq c$, $b \neq d$ and $a,b,c,d > 0$, then:

$n = (a+b)(a^2-ab+b^2)$

$n=(c+d)(c^2-cd+d^2)$

But I don't have any idea how to follow from here. If we take a look at 1729:

1729=13·19·7

It seems obvious that if a number is the product of only 2 primes then it can't be expressed as the sum of two cubes in two different ways, but I don't know how to prove either that the least number which such property must be product of 3 primes or more.

Does anyone has any idea of how to approach this?

A Lawliet
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    Don't check every number from 1 to 1729. Check a from 1 to 11, b from 1 to 11. Does any sum less than 1729 appear twice ? – Empy2 Jun 02 '18 at 12:16
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    This paper https://cs.uwaterloo.ca/journals/JIS/VOL6/Broughan/broughan25.pdf might help. – B. Goddard Jun 02 '18 at 13:34
  • You might want to research the Taxicab number. – farruhota Jun 02 '18 at 13:51
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    @Michael In fact check only $b$ from $1$ to $a$, and if $a$ is even, $b$ must be odd (or else $c,d$ must be even, contradicting minimality) – Hagen von Eitzen Jun 02 '18 at 14:19
  • It does indeed seem to be true that a taxicab number must have more than 2 distinct prime factors. But this is true only if the cube summands must be positive. $91=6^3-5^3=4^3+3^3$, $152=6^3-4^3=5^3+3^3$, $189=6^3-3^3=5^3+4^3$ have only 2 distinct prime factors each. – Rosie F Jun 03 '18 at 09:49
  • Where $n=a^3+b^3$, let $s=a+b$, $q=n/s=a^2-ab+b^2$ as you stated. Then if $3\mid s$ then $3\mid q$ so $9\mid n$; and if $3\nmid s$ then $q\bmod 6=1$ and $3\nmid n$. This rules out $3\mid n$ except if $9\mid n$. It suggests that two sorts of integers will yield results: $pqr$ where $p, q$ are primes of the form $6k+1$ and $3\nmid r$; and multiples of 9. – Rosie F Jun 03 '18 at 09:55
  • ... In the former case, $n$ must be within 2 of a multiple of 9. In any case, $n$ must be within 2 of a multiple of 7. – Rosie F Jun 03 '18 at 10:03

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