Condition for a prime to divide $a^2+ab+b^2$.

Question

Any hints for proving that if $p$ is a prime of the form $3k+2$ and $p \mid a^2+ab+b^2$ then $p \mid a$ and $p \mid b$?

Answer 1

Here's a way to think about it.

We define the multiplicative order of a prime $p$ to a base $a$ and $b$, $ord_p(a,b)$, as the first positive integer $k$ such that $a^k-b^k\equiv 0\mod p$ (that is the first $k$ such that $p|a^k-b^k$). It is a well known result (Fermat's little theorem) that $p|a^{p-1}-b^{p-1}$ if $gcd(a,p)=gcd(b,p)=1$. Further it can be easily checked that if $gcd(a,p)=gcd(b,p)=1$, then $p|a^{m}-b^{m}$ if and only if $ord_p(a,b)|m$ (otherwise the minimality of order would be contradicted*). Importantly we have that $ord_p(a,b)|p-1$ when $gcd(a,p)=gcd(b,p)=1$.

Now here's the suggestive part: $a^3-b^3=(a-b)(a^2+ab+b^2)$. More specifically, from our discussion above, if $p|a^2+ab+b^2$ and $gcd(a,p)=gcd(b,p)=1$, then $ord_p(a,b)=3$ (it divides the 'primitive' part of $a^3-b^3$, not the $a-b$ part which divides, well, $a^1-b^1$). But we also showed that in this case $3=ord_p(a,b)|p-1$. This means $p-1\equiv 0\mod 3\Rightarrow p\equiv 1\mod 3\Rightarrow p=3n+1$ for some $n$. In other words we cant write p=3k+2.

So if we assume that $p|a^2+ab+b^2$ and $p=3k+2$, then we have a contradiction and thus our assumption that $gcd(a,p)=gcd(b,p)=1$ cannot be true. But then we can take wlog that $gcd(a,p)\neq 1$. Since $p$ is prime, $p|a$. But if $p|a$ and $p|a^2+ab+b^2$, then $p|b$ (otherwise $a^2+ab+b^2\equiv 0+0+b^2\not\equiv 0\mod p$). (One can alternatively intuit that if $p|a$ and $p\nmid b$ then there is no $ord_p(a,b)$, since $a^m-b^m\equiv -b^m\not\equiv 0\mod p$, and thus $p\nmid a^2+ab+b^2|a^3-b^3$)

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*Show that $g^k=e$ if and only if the order of $g$ is finite and divides $k$. this shows a brief proof of this statement for groups in general, and very similar rationale applies here

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Edit: Let $k$ be an integer. This outlook comes with the additional benefit that it is generalizable; for example

if $p\neq 5k+1$ and $p|a^4 + a^3 b + a^2 b^2 + a b^3 + b^4$ then $p|a$ and $p|b$

by the same rational. More generally, writing $\Phi_n(a,b)$ as the $n$-th cyclotomic polynomial,

if $p\neq nk+1$ and $p|\frac{\Phi_n(a,b)}{\gcd(n,\Phi_n(a,b))}$ then $p|a$ and $p|b$.