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.

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

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

\(p\nmid b\,\) 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.

Log in to reply

Thanks.

Log in to reply

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

Log in to reply

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

Log in to reply