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Let $p$ be a prime of the form $3k+2$ that divides $a^2+ab+b^2$ for some integers $a,b$. Prove that $a,b$ are both divisible by $p$.

My attempt:

$p\mid a^2+ab+b^2 \implies p\mid (a-b)(a^2+ab+b^2)\implies p\mid a^3-b^3$
So, we have, $a^{3k}\equiv b^{3k}\mod p$ and by Fermat's Theorem we have, $a^{3k+1}\equiv b^{3k+1}\mod p$ as $p$ is of the form $p=3k+2$.

I do not know what to do next. Please help. Thank you.

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Suppose that $p=3k+2$ is prime and $$ \left.p\ \middle|\ a^2+ab+b^2\right.\tag1 $$ then, because $a^3-b^2=(a-b)\left(a^2+ab+b^2\right)$, we have $$ \left.p\ \middle|\ a^3-b^3\right.\tag2 $$ Case $\boldsymbol{p\nmid a}$

Suppose that $p\nmid a$, then $(2)$ says $p\nmid b$. Furthermore, $$ \begin{align} a^3&\equiv b^3&\pmod{p}\tag3\\ a^{3k}&\equiv b^{3k}&\pmod{p}\tag4\\ a^{p-2}&\equiv b^{p-2}&\pmod{p}\tag5\\ a^{-1}&\equiv b^{-1}&\pmod{p}\tag6\\ a&\equiv b&\pmod{p}\tag7\\ \end{align} $$ Explanation
$(3)$: $\left.p\ \middle|\ a^3-b^3\right.$
$(4)$: modular arithmetic
$(5)$: $3k=p-2$
$(6)$: if $p\nmid x$, then $x^{p-2}\equiv x^{-1}\pmod{p}$
$(7)$: modular arithmetic

Then, because of $(1)$ and $(7)$, $$ \begin{align} 0 &\equiv a^2+ab+b^2&\pmod{p}\\ &\equiv 3a^2&\pmod{p}\tag8 \end{align} $$ which, because $p\nmid 3$, implies that $p\mid a$, which contradicts $p\nmid a$ and leaves us with

Case $\boldsymbol{p\mid a}$

If $p\mid a$, then $(2)$ says $p\mid b$ and we get $$ \left.p^2\ \middle|\ a^2+ab+b^2\right.\tag9 $$

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  • $\begingroup$ Yes, this is much better. $\endgroup$ – Aqua May 5 at 8:59
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Use quadratic reciprocity.

$$4(a^2+ab+b^2)=(2a+b)^2+3b^2$$

so if $$a^2+ab+b^2\equiv 0 \pmod p$$ then $$(2a+b)^2\equiv -3b^2 \pmod p$$ and $-3$ is a quadratic residue so

$$\left(\frac{-3}{p}\right)=1.$$ However by reciprocity,

$$\left(\frac{-3}{p}\right)=\left(\frac{p}{-3}\right)=\left(\frac{2}{3}\right)=-1$$

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  • $\begingroup$ Sorry, I could not understand from "and -3 is a quadratic residue..." $\endgroup$ – Swadhin Jul 14 '14 at 23:06
  • $\begingroup$ Do you mean the notation ? $\endgroup$ – Rene Schipperus Jul 14 '14 at 23:08
  • $\begingroup$ No sir, the whole of it from that part. $\endgroup$ – Swadhin Jul 14 '14 at 23:10
  • $\begingroup$ Do you know the quadratic residue symbol $\left(\frac{q}{p}\right)$ ? $\endgroup$ – Rene Schipperus Jul 14 '14 at 23:12
  • $\begingroup$ I think that I may not have seen it before. $\endgroup$ – Swadhin Jul 14 '14 at 23:13
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You have shown that $a^3 \equiv b^3 \pmod p$. Remark that $(3, p-1)=1$ because $p=3k+2$. Thus we can write $3m + (p-1)n = 1$ for some integers $m, n$. Use this to show that $a\equiv b \pmod p$, so that $a^2+ab+b^2 \equiv 3a^2 \equiv 3b^2 \equiv 0 \pmod p$, and conclude from there.

(P.S. I have to commend you for posting your work!)

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  • $\begingroup$ Thank you, but I cannot understand how to show that $a\equiv b\mod p$, sorry if I am missing something obvious. $\endgroup$ – Swadhin Jul 14 '14 at 23:09
  • $\begingroup$ Dear @Swadhin, you are welcome. Use the identity $3m + (p-1)n = 1$ to rewrite the equation $a^{3m} \equiv b^{3m} \pmod p$... (You might want to assume at the beginning that $(a,p)=(b,p)=1$, so that raising to a negative exponent makes sense - then, at the end, conclude by contradiction.) $\endgroup$ – Bruno Joyal Jul 14 '14 at 23:11
  • $\begingroup$ I am sorry sir, but I cannot seem to use the identity as you suggested. Will help me see that? $\endgroup$ – Swadhin Jul 14 '14 at 23:15
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By lil' Fermat, $a^{3k+1}\cong b^{3k+1}\bmod p\implies a\cdot b^{3k}\cong b\cdot b^{3k}\bmod p$. Thus $\begin{align} a\cong b\bmod p\implies a=b+pk \implies a^2+ab+b^2=(b+pk)^2+(b+pk)b+b^2=3b^2+3bpk+p^2k^2\end{align}$. But this expression is divisible by $p$. So $3b^2=-3bpk-p^2k^2+pl\implies p|3b^2\implies p|b\implies p|a$.

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