Proof

p is a prime number of the form 3k+2.Also p divides a2+ab+b2a^2+ab+b^2.Prove that p divides a and b.

Note by Lawrence Bush
6 years ago

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Comments

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

Bogdan Simeonov - 6 years ago

p∣b ⇒ p∣ap\mid b\,\Rightarrow\, p\mid a

p∤b p\nmid b\, gives contradiction, as follows:

mod pp: \ 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: (−3p)≡p(mod ⁣3)\left(\frac{-3}{p}\right)\equiv p\pmod{\! 3}

from which in this case by (−3p)≡p≡3k+2≡−1(mod ⁣3)\left(\frac{-3}{p}\right)\equiv p\equiv 3k+2\equiv -1\pmod{\! 3} follows (−3p)=−1\left(\frac{-3}{p}\right)=-1, so contradiction.

mathh mathh - 6 years ago

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Thanks.

lawrence Bush - 6 years ago

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To prove (−3p)≡p(mod ⁣3)\left(\frac{-3}{p}\right)\equiv p\pmod{\! 3}, do this: (−3p)=(−1p)(3p)=(−1)p−12(−1)p−12(p3)=(p3)≡p(mod ⁣3)\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 - 6 years ago
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