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In the book of Ireland/Rosen, A Classical Introduction of Modern Number Theory, there is no proof of this Proposition 8.3.2 (Existence and Uniqueness), page 96:

If [$p$ is prime and] $p \equiv 1 \mod 3$, then there are integers $A$ and $B$ such that $4p = A^2 + 27B^2$. In this representation of $4p$, $A$ and $B$ are uniquely determined up to a sign.

Can anyone give a proof of this Proposition? Thanks for help :)

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    $p = 10$ doesn't work, Does $p$ must be prime ? – 2017-01-12
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    maybe he means p to be a prime. and 4*19 = 7^2 +27*1^2. – 2017-01-12
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    yes, p must be prime. – 2017-01-12
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    Take a look at https://en.wikipedia.org/wiki/Cubic_reciprocity – 2017-01-12

1 Answers 1

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Evidently Ireland and Rosen prove this beginning with Proposition 8.3.1, on page 95, then add some detail. See how, at the top of page 96, they say

Thus $3 | a-b,$ and we have

This does not need cubic reciprocity. The form $$ x^2 + xy + 7 y^2 $$ has discriminant $-27,$ also class number one. It integrally represents every prime $p \equiv 1 \pmod 3.$ Compared with $x^2 + xy + y^2,$ it only misses the single prime $3.$

$$ p = u^2 + uv + 7 v^2 $$ $$ 4p = 4 u^2 + 4uv + 28 v^2, $$ $$ 4p = (2u+v)^2 + 27 v^2, $$ or $$ 4p = x^2 + 27 y^2. $$ Notice also that $$ 4p = 4 u^2 + 2u(2v) + 7 (2v)^2, $$ or $$ 4p = 4 x^2 + 2xy + 7 y^2. $$

The stuff about cubic reciprocity tells us about $p$ itself rather than $4p:$ if $p \equiv 1 \pmod 3$ and $2$ is a cubic residue $\pmod p,$ then we get the stronger $p = x^2 + 27 y^2.$ If $p \equiv 1 \pmod 3$ but $2$ is not a cubic residue $\pmod p,$ then we get $p = 4 x^2 \pm 2xy + 7 y^2.$ This is Proposition 9.6.2 on page 119 in the second edition.

The first few primes $x^2 + 27 y^2,$ where my program includes $1,$ are

  1,     31,     43,    109,    127,    157,    223,    229,    277,    283,
307,    397,    433,    439,    457,    499,    601,    643,    691,    727,
733,    739,    811,    919,    997,

The first few primes $4 x^2 + 2xy + 7 y^2,$

  7,     13,     19,     37,     61,     67,     73,     79,     97,    103,
139,    151,    163,    181,    193,    199,    211,    241,    271,    313,
331,    337,    349,    367,    373,    379,    409,    421,    463,    487,

I don't see that I+R discuss the latter case. However, every prime that does not divide a discriminant $\Delta,$ and for which $(\Delta|p) = 1,$ is represented by some (primitive) form of that discriminant, this is from ordinary Gauss reduction. Here are the three classes of forms of discriminant $\Delta = -108,$ just one genus:

jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./classGroup
Absolute value of discriminant? 
108
Impr 3 0 9
Impr 2 2 14
Impr 6 6 6
Discr  -108 = 2^2 * 3^3  class  number  3

 all  
     108:  < 1, 0, 27>    Square       108:  < 1, 0, 27>
     108:  < 4, -2, 7>    Square       108:  < 4, 2, 7>
     108:  < 4, 2, 7>    Square       108:  < 4, -2, 7>

 squares  
     108:  < 1, 0, 27>
     108:  < 4, -2, 7>
     108:  < 4, 2, 7>
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    But there is no proof of the uniqueness of this proposition. This is what I'm interested about. – 2017-01-13