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These are some examples.

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

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

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

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

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

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

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

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

How can i prove? Please help me, please... (ㅠㅠ)

José Carlos Santos
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Ryu Dae Sick
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1 Answers1

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Note that$\newcommand{\om}{\omega}$ $$a^2+ab+b^2=(a-\om b)(a-\om^2 b)$$ where $\om=\frac12(-1+i\sqrt3)$ Just as you can understand sums of two squares by using unique factorisation in the ring $\Bbb Z[i]$, you can prove your assertion by using unique factorisation in $\Bbb Z[\om]$.

Call the gcd of $a-b\om$ and $p$ in $\Bbb Z[\om],$ $x-y\om$ etc.

Angina Seng
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  • Thanks for youre answer. But I can't understand why your answer means that $\exists$ nonzero $a,b$ such that $a^2 + ab + b^2 \equiv 0 \pmod{p} \implies \exists x,y $ such that $x^2 + xy + y^2 = p$, exactly. I ask this question twice because $\exists x,y $ such that $x^2 + xy + y^2 = p$, NOT $p \equiv 1 \pmod{3} \implies \exists$ nonzero $a,b$ such that $a^2 + ab + b^2 \equiv 0 \pmod{p}$. Can you please help me? Please... – Ryu Dae Sick Jul 16 '17 at 07:45