For a odd prime $p, \exists a,b$ such that $a^2 + ab + b^2 \equiv 0 \pmod{p} \iff \exists x,y $ such taht $x^2 + xy + y^2 = p$?

Question

We know that $p=x^2+xy+y^2$ if and only if $p \equiv 1 \pmod {3}$.

But I need $a^2+ab+b^2 \equiv 0 \pmod{p} $ if and only if $p \equiv 1 \pmod {3}$, more generalized theorem.

I think that i should proof below :

$\exists a,b$ such that $a^2 + ab + b^2 \equiv 0 \pmod{p} \iff \exists x,y $ such taht $x^2 + xy + y^2 = p$

Easy to prove ($\Longleftarrow$), but not ($\implies$).

These are some examples.

$2^2 + 2 \cdot 4 + 4^2 = 4( 1^2 + 1 \cdot 2 + 2^2) = 4 \cdot 7$

$3^2 + 3 \cdot 9 + 9^2 = 9( 1^2 + 1 \cdot 3 + 3^2) = 9 \cdot 13$

$7^2 + 7 \cdot 11 + 11^2 = 13( 2^2 + 2 \cdot 3 + 3^2) = 13 \cdot 19$

$5^2 + 5 \cdot 25 + 25^2 = 25( 1^2 + 1 \cdot 5 + 5^2) = 25 \cdot 31$

$10^2 + 10 \cdot 26 + 26^2 = 28( 3^2 + 3 \cdot 4 + 4^2) = 28 \cdot 37$

$6^2 + 6 \cdot 36 + 36^2 = 36( 1^2 + 1 \cdot 6 + 6^2) = 36 \cdot 43$

$13^2 + 13 \cdot 47 + 47^2 = 49( 4^2 + 4 \cdot 5 + 5^2) = 49 \cdot 61$

$29^2 + 29 \cdot 37 + 37^2 = 49( 2^2 + 2 \cdot 7 + 7^2) = 49 \cdot 67$

Answer 1

Given any sum with a particular prime factor $p$ we can isolate the factor $p$ via Eisenstein integer multiplication.

Take your example with $10, 26$ having the sum $28×37$. How to isolate the factor $37$?

First get rid of the common factor $2^2$ which is totally useless:

$5^2+(5×13)+13^2=7×37$

Next we associate thus quadratic form with the Eisenstein integer $5-13\omega$ whose squared norm matches the value of the quadratic form. Generally: $|a-b\omega|^2=a^2+ab+b^2$.

Now $7$ has several sums that similarly match up with Eisenstein integers having that squared norm:

$7=1^2+(1×2)+2^2 \rightarrow 1-2\omega$ $7=2^2+(2×1)+1^2 \rightarrow 2-\omega$ $7=1^2-(1×3)+3^2 \rightarrow 1+3\omega$ $7=3^2-(3×1)+1^2 \rightarrow 3+\omega$ $7=2^2-(2×3)+3^2 \rightarrow 2+3\omega$ $7=3^2-(3×2)+2^2 \rightarrow 3+2\omega$

All we need do is multiply our Eisenstein integer for $7×37$ above, $5-13\omega$, by the above multipliers, and look for a product that has a factor of $7$ which we drop out. We will then have a combination equalling $37$ with integer inputs.

$(5-13\omega)×(1-2\omega)=(5×1-13×2)-(5×2+13×1+13×2)\omega=-21-49\omega=-7×(3+7\omega)$

So $|3+7\omega|^2=37$ meaning $3^2-(3×7)+7^2=37$.

To get a solution with positive integers we can multiply $3+7\omega$ by $\omega$ to get a product with mixed signs:

$(3+7\omega)×\omega=-7-4\omega$

$(-7-4\omega)×\omega=4-3\omega$

Then $37=4^2+(4×3)+3^2$.

All these manipulations turn on the Euclidean domain property of the Eisenstein integers so we are sure they work.