1729, and related questions I just read this paragraph: (written by G. H. Hardy, on Ramanujan)

I remember once going to see him when he was lying ill at Putney. I
  had ridden in taxi cab number 1729 and remarked that the number seemed
  to me rather a dull one, and that I hoped it was not an unfavorable
  omen. ‘No,’ he replied, ‘it is a very interesting number; it is the
  smallest number expressible as the sum of two cubes in two different
  ways.’

Was Ramanujan right?
What are other numbers having such property (expressible as the sum of two cubes in two different ways)?
Are there infinite number of them?
And, on the other hand:
What if the word "cubes" is replaced by "5-degree power"? Would such numbers exist? If yes, what would be the smallest?

Another SO question related to 1729: Proof that 1729 is the smallest taxicab number
 A: If negative numbers are allowed, then $91=3^3+4^3=6^3+(-5)^3$.
A: Theorem 412 of Hardy and Wright, An Introduction to the Theory of Numbers, 6th edition, page 442, says, 
"Whatever $r$, there are numbers that are representable as sums of two positive cubes in at least $r$ different ways."
The proof is quite elementary, but involves a bit more typing than I am keen to do. The notes say the proof was found by Fermat, but there was one place in the argument where he just assumed something that actually needed proof; Mordell was the first to write a complete proof, but didn't publish it. So I suppose the hardy & Wright book was the first place a full proof was published. 
For the question about 5th powers, see my comment on the original question. 
A: Very late for this party, but yes, there is an infinite number of taxicab numbers. The complete solution in positive integers to,
$$x_1^3+x_2^3 = x_3^3+x_4^3$$
was given by Choudhry's On Equal Sums of Cubes (1998). For positive integers $a,b,c$,
$$\begin{aligned}
d\,x_1 &= (a^2 + a b + b^2)^2 + (2a + b)c^3\\
d\,x_2 &= (-a^3 + b^3 + c^3)c\\
d\,x_3 &= (a^2 + a b + b^2)^2 - (a - b)c^3\\
d\,x_4 &= (a^3 + (a + b)^3 + c^3)c\end{aligned}$$
where,
$$(a^3-b^3)^{1/3}<\,c\,<\frac{(a^3-b^3)^{2/3}}{a-b}$$
and $d=1$, or chosen such that $\text{GCD}(a,b,c)=1$.
P.S. For Choudhry's complete solution in positive integers to $x_1^3+x_2^3+ x_3^3=x_4^3$, see this post.
