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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
2 years, 6 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.

- 2 years, 6 months ago

Thanks.

- 2 years, 6 months ago

To prove $$\left(\frac{-3}{p}\right)\equiv p\pmod{\! 3}$$, do this: $$\left(\frac{-3}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)=(-1)^{\frac{p-1}{2}}(-1)^{\frac{p-1}{2}}\left(\frac{p}{3}\right)=\left(\frac{p}{3}\right)\equiv p\pmod{\! 3}$$

- 2 years, 6 months ago

This is equivalent to whether -3 is a quadratic residue mod p.Using the quadratic reciprocity theorem, it follows easily.

- 2 years, 6 months ago