Primes which are the sum of two cubes

I was recently reminded of Fermats result on when a prime number can be expressed as the sum of two squares. This then got me thinking on when a prime can be wrote as the sum of two cubes. I start by considering
\[p=a^3+b^3, a,b\in\mathbb{N}\]
I then consider negative values of $b$ later.
Suppose that $p=a^3+b^3$ and let $a>b.$ Then note that
\[p=(a+b)(a^2-ab+b^2)\]
which factorises $p.$ Therefore $p=a+b$ and $a^2-ab+b^2=1.$ It is striaghtforward to see that $a+b=a^3+b^3$ only for $a=b=1$ so $p=2$ is the solution in this case.

As promised I now look at the case when $b$ is negative. This is the same as looking at the problem of when

\[p=a^3-|b|^3, a\in\mathbb{N},b\in\mathbb{Z}.\]

Which is equivalent to
\[p=a^3-b^3, a,b\in\mathbb{N}.\]

So we now solve when a prime number can be the difference of two cubes. We once again factorise to get
\[p=(a-b)(a^2+ab+b^2).\]
So $p=(a^2+ab+b^2)$ and $a-b=1.$ eliminating $b$ we obtain that $p=3a^2-3a+1.$ This is the form that Hexagonal numbers take and so an odd prime can be expressed as the sum of two cubes if and only if it is a hexagonal number.

True story.