3

a, b, c, and d are all different positive integers. I've tried googling optimisation and constrained optimisation, but I've not found anything applicable as yet.

EDIT:

I tried using Lagrange mulipliers, but hit a wall:

$$f(a,b,c,d) = a+b+c+d$$

$$g(a,b,c,d) = a^3+b^3-c^3-d^3$$

Lagrange gives the system:

$$f_i = \lambda g_i, i = a,b,c,d$$

$$1 = \lambda(3i^2), i=a,b \\ 1 = \lambda(-3i^2), i=c,d$$

(Equations $1-4$)

$$a^3+b^3-c^3-d^3=0$$

(Eq $5$)

This would give $a=b$ and $c=d$, but this would break the constraint that they are all different, so don't know if Lagrange can help with this.

EDIT: (in progress)

$1729 = 1^3+12^3 = 9^3 + 10^3$ is the solution, as pointed out by Macavity, but I am seeking a proof of this.

George Tomlinson
  • 1,346
  • 8
  • 11

1 Answers1

4

If you have heard of a rather famous anecdote about the Ramanujan and the number $1729$, you would know that this is the least number expressible as the sum of two positive cubes in two different ways, i.e. $1729 = 1^3+12^3 = 9^3 + 10^3$. So the answer you seek is $1+12+9+10 = 32$.

Apart from enumerating and testing numbers, I see no fast way of finding this answer...

Macavity
  • 46,381
  • 1
    It's clear that this minimizes $a^3+b^3+c^3+d^3$. Is it clear that it minimizes $a+b+c+d$? My guess is that you'll not only need the $1729$ anecdote, but a bound on the next such solution. – awwalker Nov 05 '13 at 17:51
  • @AWalker Well the next one is $4104 = (2^3 + 16^3) = (9^3 + 15^3)$, but somehow this enumeration approach doesn't seem satisfying. From http://oeis.org/A001235 – Macavity Nov 05 '13 at 17:56
  • 1
    Other than enumeration, I am not sure if you will find a way. The general form you postulate doesn't work for $20683 = (10^3 + 27^3) = (19^3 + 24^3)$ or for $39312 = (2^3 + 34^3)=(15^3 + 33^3)$. I am sure if you test more solutions, you will find many more counter-examples. – Macavity Nov 05 '13 at 18:41
  • Thanks for the first counter-example (+1). The second isn't a counter-example, as the $4104$ solution fits my postulate and is a lower solution for $n=2$. – George Tomlinson Nov 05 '13 at 18:51
  • It turns out the question has been posed before in a different form, with the same answer: http://math.stackexchange.com/questions/487537/proof-that-1729-is-the-smallest-taxicab-number – George Tomlinson Nov 05 '13 at 19:15
  • @CalvinLin Yes, good catch, it has cropped up at least twice that I have seen, with different parts in each question. People getting more creative, and harder to spot these i suppose. – Macavity Nov 25 '20 at 11:56