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Proof

p is a prime number of the form 3k+2.Also p divides \(a^2+ab+b^2\).Prove that p divides a and b.

Note by Lawrence Bush
1 year, 10 months ago

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\(p\mid b\,\Rightarrow\, p\mid a\)

\(p\nmid b\,\) gives contradiction, as follows:

mod \(p\): \(\ a^2+ab+b^2\equiv 0\stackrel{\cdot 4}\iff (2a+b)^2\equiv -3b^2\iff \left(\frac{2a+b}{b}\right)^2\equiv -3\),

which is impossible. Using only quadratic reciprocity we can prove this interesting lemma: \(\left(\frac{-3}{p}\right)\equiv p\pmod{\! 3}\)

from which in this case by \(\left(\frac{-3}{p}\right)\equiv p\equiv 3k+2\equiv -1\pmod{\! 3}\) follows \(\left(\frac{-3}{p}\right)=-1\), so contradiction. Mathh Mathh · 1 year, 10 months ago

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@Mathh Mathh Thanks. Lawrence Bush · 1 year, 10 months ago

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@Lawrence Bush To prove \(\left(\frac{-3}{p}\right)\equiv p\pmod{\! 3}\), do this: \(\left(\frac{-3}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)=(-1)^{\frac{p-1}{2}}(-1)^{\frac{p-1}{2}}\left(\frac{p}{3}\right)=\left(\frac{p}{3}\right)\equiv p\pmod{\! 3}\) Mathh Mathh · 1 year, 10 months ago

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This is equivalent to whether -3 is a quadratic residue mod p.Using the quadratic reciprocity theorem, it follows easily. Bogdan Simeonov · 1 year, 10 months ago

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