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If $p\equiv 1\pmod{3}$, it's well know that $p$ can be expressed as $$ p=\frac{1}{4}(L^2+27M^2). $$ A handful of places, like wikipedia, say that in this representation $L$ and $M$ are unique up to sign, but none actually has a proof.

Is there a elementary proof that $L$ and $M$ are unique up to sign? Thanks.

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  • $\begingroup$ where in wikipedia? To be specific, it really is well known that such $p = u^2 + 3 v^2$ only one way up to sign. Your version seems to me to require two proofs, (A) when 2 is a cube $\pmod p,$ and (B) when 2 is not a cube $\pmod p.$ $\endgroup$
    – Will Jagy
    Feb 5, 2013 at 22:22
  • $\begingroup$ @WillJagy Here in the article on cubic reciprocity in the subsection of $p\equiv 1\pmod{3}$. I think I might change the question as I've been digging around since I posted it. $\endgroup$ Feb 5, 2013 at 22:35

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After your comment, the correct thing to say is that there is essentially one representation $$ p = x^2 + 3 y^2, $$ where in this case the word essentially can be taken to mean up to $\pm$ sign.

Meanwhile, if you multiply the discriminant by 9, there is essentially one way to write $$ p = u^2 + 27 v^2, $$ $$ p = 4 u^2 + 2 u v + 7 v^2 $$ $$ p = 4 u^2 - 2 u v + 7 v^2. $$ It is the first of these if 2 is a cube $\pmod p,$ otherwise it is the second and third. We distinguish the $\pm$ cases in the second and third forms so as to get the correct group under Gauss composition. As always, "essentially" means "up to integral automorph." Note that $$ 4 (u^2 + 27 v^2) = (2u)^2 + 27 (2v)^2, $$ $$ 4( 4 u^2 + 2 u v + 7 v^2) = (4u+v)^2 + 27 v^2, $$ which leads to pretty much what you want.

This is treated in Ireland and Rosen, for one. An encyclopaedic treatment in David A. Cox, Primes of the form $x^2 + n y^2,$ in any way to describe the problem you might like. Still, I think your best bet is Hudson and Williams 1991 at MMMEEEEEEE

For, well, culture, see my question What numbers are integrally represented by $4 x^2 + 2 x y + 7 y^2 - z^3$

This all can be written in the language of imaginary quadratic fields. I've never cared enough to work that out.

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  • $\begingroup$ sorry, I changed the question. I'll roll it back to prevent confusion, and post the new question separately. $\endgroup$ Feb 5, 2013 at 22:53

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