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Prove/Disprove that there are infinitely many natural numbers satisfying the given property:

n $\in$ N can be expressed as the sum of cubes of 2 natural numbers in two different ways,

$$\mathsf {OR}$$

$$\mathsf {x_1^3 + x_2^3 = x_3^3 + x_4^3 = n, where \ x,n\in N}$$

Ayushmaan
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  • https://math.stackexchange.com/questions/469151/find-all-integer-solutions-to-diophantine-equation-x3y3z3-w3/776918#776918 – individ Sep 21 '17 at 10:17
  • I just wanted to know if the first number could be got only with hit and trial approach or was there a more methodological one. – Ayushmaan Sep 21 '17 at 10:24

1 Answers1

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If you found one example, it is very straightforward as

$$x_1 ^3 + x_2 ^3= x_3 ^3 +x_4 ^3 \Longrightarrow (kx_1)^3+(kx_2)^3= (kx_3)^3+(kx_4)^3$$

However, I haven't been able to find such an example.

Edit: found it!

$$1729=10^3+9^3=12^3+1^3$$

Sofía Marlasca Aparicio
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    If someone has watched The man who knew infinity he/she will immediately identify this example. – Vidyanshu Mishra Sep 21 '17 at 10:07
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    Here are more such numbers https://math.stackexchange.com/questions/1274816/numbers-that-can-be-expressed-as-the-sum-of-two-cubes-in-exactly-two-different-w and here https://en.wikipedia.org/wiki/Taxicab_number#Known_taxicab_numbers – Widawensen Sep 21 '17 at 12:11
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    This is the most extensive list https://gist.github.com/mjdominus/cc67be601f9e178b3ee7 – Widawensen Sep 21 '17 at 12:18