Numbers that can be expressed as the sum of two cubes in exactly two different ways It seems known that there are infinitely many numbers that can be expressed as a sum of two positive cubes in at least two different ways (per the answer to this post: Number Theory Taxicab Number).
We know that
$$1729 = 10^3+9^3 = 12^3 + 1^3,$$
and I am wondering if there are infinitely many numbers like this that can be expressed as the sum of two positive cubes in exactly two ways?
In fact, are there even any other such numbers?
EDIT:
As provided by MJD in the comments section, here are other examples:
$$4104 = 2^3+16^3 = 9^3+15^3,$$
$$13832 = 20^3+18^3=24^3+2^3,$$
$$20683 = 10^3 +27^3 = 19^3 +24^3.$$ 
 A: As for the question regarding whether or not there are infinitely many numbers that can be expressed as the sum of two cubes in two different ways or not, there's a very quick and simple way to prove this. Since you have 
$$ 1729 = 10^3 + 9^3 = 12^3 + 1^3 $$ 
Multiply both sides by $ n^3 $ where $ n $ is a positive integer to get 
$$ 1729 n^3 = (10n)^3 + (9n)^3 = (12n)^3 + n^3 $$ 
and there you have it, plug in any value for $ n $ and and you have your proof for the infinitude of such numbers. Obviously this doesn't include all the solutions but it proves that there certainly are an infinite number of them 
A: In the paper Characterizing the Sum of Two Cubes, Kevin Broughan gives the relevant theorem,
Theorem: Let $N$ be a positive integer. Then the equation $N = x^3 + y^3$ has a solution in positive integers $x,y$ if and only if the following conditions are satisfied:


*

*There exists a divisor $m|N$ with $N^{1/3}\leq m \leq (4N)^{1/3}.$ 

*And $\sqrt{m^2-4\frac{m^2-N/m}{3}}$ is an integer.
The sequence of integers $F(n)$,
$$\begin{aligned}
F(n) 
&= a^3+b^3 = (2 n + 6 n^2 + 6 n^3 + n^4)^3 + (n + 3 n^2 + 3 n^3 + 2 n^4)^3\\
&= c^3+d^3 = (1 + 4 n + 6 n^2 + 5 n^3 + 2 n^4)^3 + (-1 - 4 n - 6 n^2 - 2 n^3 + n^4)^3
\end{aligned}$$
for integer $n>3$ apparently is expressible as a sum of two positive integer cubes in exactly and only two ways. 
$$\begin{aligned}
F(4) &= 744^3+756^3 = 945^3+15^3\\
F(5) &= 1535^3+1705^3 = 2046^3+204^3\\
&\;\vdots\\
F(60) &=14277720^3+26578860^3 = 27021841^3+12506159^3
\end{aligned}$$
Using Broughan's theorem, I have tested $F(n)$ from $n=4-60$ and, per $n$, it has only two solutions $m$, implying in that range it is a sum of two cubes in only two ways.  Can somebody with a faster computer and better code test it for a higher range and see when (if ever) the proposed statement breaks down? Incidentally, we have the nice relations,
$$a+b = 3n(n+1)^3$$
$$c+d = 3n^3(n+1)$$
Note: $F(60)$ is already much beyond the range of taxicab $T_3$ which is the smallest number that is the sum of two positive integer cubes in three ways.
$$T_3 \approx 444.01^3 = 167^3+436^3 = 228^3+423^3 = 255^3+414^3$$
(Using the theorem, this yields 3 values for $m$.)
A: I found a number recently which can be expressed as a sum of two cubes in exactly two different ways.
  65673928=164³+394³=103³+401³

