Hardy-Ramanujan Number/Examples/1729

Theorem

The $2$nd Hardy-Ramanujan number $\map {\operatorname {Ta}} 2$ is $1729$:

\(\ds 1729\)

\(=\)

\(\ds 12^3 + 1^3\)

\(\ds \)

\(=\)

\(\ds 10^3 + 9^3\)

Proof

We wish to demonstrate that $1729$ is the $2$nd Hardy-Ramanujan number making it the smallest positive integer which can be expressed as the sum of $2$ cubes in $2$ different ways.

To accomplish this, we will need to inspect $66$ sums of $a^3 + b^3$ starting with $1^3 + 1^3$ and ending with $11^3 + 11^3$

Recall from Integer Addition is Commutative, $a^3 + b^3 = b^3 + a^3$, therefore we only need to inspect one of these pairs.

Without loss of generality, we will choose to inspect $a \ge b$ in the table below.

The reason that we don't need to inspect $a \ge 12$ is that $12^3 + 1^3 = 1729$ and adding anything larger than $1$ to $12^3$ will be larger than $1729$ and the cube of any number larger than $12$ will be larger than $1729$

The reason there are $66$ sums to inspect is that the $1$st row will have $1$ sum, the $2$nd row will have $2$ sums, all the way up to the $11$th row having $11$ sums.

From Closed Form for Triangular Numbers, this sum is known to be:

$\ds T_{11} = \sum_{i \mathop = 1}^{11} i = \frac {11 \paren {11 + 1} } 2 = 66$

Finally, we note that the only other strictly positive integer(s) that "almost" took this honor from $1729$ are $854$ and $855$ which are only apart by $1$.

$\blacksquare$