I am stuck at the following exercsise:
We know that $1729$ is the smallest number that can be expressed as the sum of two cubes in two different ways. Show that $1729$ is the only number $n$ with the following two properties:
$n$ can be expressed as the sum of two cubes in two distinct ways.
$n$ is of the form $ n = (6k+1)(12k+1)(18k+1)$ where all three factors are primes.
I do not know how I could start with this exercise, could you give me a hint?
Trivial part:
First verify that 1. and 2. hold for $n=1729=12^3+1^3=10^3+9^3=7\cdot 13\cdot 19$ (so $k=1$ in 2.).
The non-trivial part:
Assume $n$ has properties 1. and 2., say $$n=a^3+b^3=c^3+d^3=(6k+1)(12k+1)(18k+1)$$ with $\{a,b\}\ne\{c,d\}$ and $p_1:=6k+1,p_2:=12k+1,p_3:=18k+1$ prime. The positive divisors of $n$ are, in ascending order (because $p_3<3p_1$), $$ 1,\;p_1,\;p_2,\;p_3,\;p_1p_2,\;p_1p_3,\;p_2p_3,\;n.$$ Note that $p_1^3< \frac n2$. Also, $p_2>\sqrt[3]n$ and hence $p_1p_2>2\sqrt[3]n$. We have $$n=a^3+b^3=(a+b)(a^2-ab+b^2)$$ where $\frac n2\le\max\{a^3,b^3\}\le n$ and therefore $$\sqrt[3]{\frac n2}\le a+b\le 2\sqrt[3]n.$$ We conclude that $a+b\in\{p_2,p_3\}$. Similarly, $c+d\in\{p_2,p_3\}$.
If we know the value of $a+b$, then we also know $(a+b)^2-\frac n{a+b}=3ab$, hence find $\{a,b\}$ as the roots of $X^2-(a+b)X+ab$. We conclude that $a+b\ne c+d$. So wlog. $$a+b=p_2,\qquad c+d=p_3.$$
In the example of $n=1729$, we happen to have have $|c-d|=1$. So we might try to express $|c-d|$ in terms of the $p_i$ and see if that produces a severe restriction in the general case. A similar strategy with $|a-b|$ and based on the obaservation $\min\{a,b\}=1$ for $n=1729$ might also work but appears to be less simple.
Now we have $$ cd =\frac{(c+d)^2-(c^2-cd+d^2)}3= \frac{p_3^2-p_1p_2}3$$ and $$ (c-d)^2=(c^2-cd+d^2)-cd = p_1p_2-cd=\frac{4p_1p_2-p_3^2}3=-12k^2+12k+1.$$ This is negative for $k>1$ whereas we have a square on the left. We conclude that $k\le1$.