If $A\equiv 1\pmod{3}$, then $4p=A^2+27B^2$ uniquely determines $A$. If $p\equiv 1\pmod{3}$, it's well know that $p$ can be expressed as
$$
p=\frac{1}{4}(A^2+27B^2).
$$
In this letter by Von Neumann, he mentions that Kummer determined that $A$ is in fact uniquely determined by the additional condition $A\equiv 1\pmod{3}$.
The reference that is listed is Kummer's De residuis cubicis disquisitiones nonnullae analyticae, but I'm having a hard time getting my hands on the original source. Is there an elelementary proof of why $A$ is uniquely determined under these two conditions? Thanks.
 A: It looks like you're doing calculation in the number field $\mathbb{Q}(\sqrt{-3})$. In particular, the typical algebraic integer is of the form
$$ u + v \frac{1 + \sqrt{-3}}{2} $$
or put differently, as
$$ \frac{A + B \sqrt{-3}}{2} $$
where $A$ and $B$ have the same parity. The norm of this element is
$$ N = \frac{A + B \sqrt{-3}}{2} \frac{A - B \sqrt{-3}}{2} = \frac{1}{4}(A^2 + 3 B^2)$$
The unit group of (the ring of integers of) this number field is simply the set of sixth roots of unity.
Assuming $N$ is prime, there are only 12 elements whose norm is $N$: the six multiples of this element by the sixth roots of unity, and their complex conjugates:


*

*$\frac{1}{2} (A + B \sqrt{-3}) $

*$\frac{1}{2} (\frac{A-3B}{2} + \frac{A+B}{2} \sqrt{-3}) $

*$\frac{1}{2} (\frac{-A-3B}{2} + \frac{A-B}{2} \sqrt{-3}) $

*$\frac{1}{2} (-A - B \sqrt{-3}) $

*$\frac{1}{2} (\frac{-A+3B}{2} + \frac{-A-B}{2} \sqrt{-3}) $

*$\frac{1}{2} (\frac{A+3B}{2} + \frac{-A+B}{2} \sqrt{-3}) $

*$\frac{1}{2} (A - B \sqrt{-3}) $

*$\frac{1}{2} (\frac{A-3B}{2} + \frac{-A-B}{2} \sqrt{-3}) $

*$\frac{1}{2} (\frac{-A-3B}{2} + \frac{-A+B}{2} \sqrt{-3}) $

*$\frac{1}{2} (-A + B \sqrt{-3}) $

*$\frac{1}{2} (\frac{-A+3B}{2} + \frac{A+B}{2} \sqrt{-3}) $

*$\frac{1}{2} (\frac{A+3B}{2} + \frac{A-B}{2} \sqrt{-3}) $


If $N$ is a prime integer, then $(A + B \sqrt{-3})/2$ is actually a prime element of $\mathbb{Z}[(1 + \sqrt{-3})/2]$. If we also require $N \neq 3$, then $A,B$ are both odd and $A$ is not divisible by $3$.
It's easy to check that exactly four of these elements have a coefficient on $\sqrt{-3}$ equivalent to $0$ modulo $3$. WLOG, we may assume it is the four $(\pm A \pm B \sqrt{-3})/2$. Among these, exactly two have the first term equivalent to $1$ modulo $3$. WLOG assume it is the two $(A \pm B \sqrt{-3})/2$. Finally, WLOG we may assume $B$ is positive, and we're down to a single possibility.
A: If $n=a^2+kb^2=c^2+kd^2$ with $b\ne\pm d$, then a little algebra shows that $$n={(ad+bc)(ad-bc)\over(d+b)(d-b)}$$ From there, it's not hard to show that if $4n=a^2+27b^2=c^2+27d^2$ with $a\ne\pm c$ then $n$ is not prime. The contrapositive says if $p$ is prime then $4p$ has at most one representation, up to the signs of $a$ and $b$, as $a^2+27b^2$. Now, $a$ can't be a multiple of $3$, so exactly one of the two numbers $\pm a$ must be $1\pmod3$. 
