# 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 $p$.

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.

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 $$\left.p^2\ \middle|\ a^2+ab+b^2\right.\tag9$$

• Yes, this is much better. – Aqua May 5 at 8:59

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

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,

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

• Sorry, I could not understand from "and -3 is a quadratic residue..." – Swadhin Jul 14 '14 at 23:06
• Do you mean the notation ? – Rene Schipperus Jul 14 '14 at 23:08
• No sir, the whole of it from that part. – Swadhin Jul 14 '14 at 23:10
• Do you know the quadratic residue symbol $\left(\frac{q}{p}\right)$ ? – Rene Schipperus Jul 14 '14 at 23:12
• I think that I may not have seen it before. – Swadhin Jul 14 '14 at 23:13

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!)

• Thank you, but I cannot understand how to show that $a\equiv b\mod p$, sorry if I am missing something obvious. – Swadhin Jul 14 '14 at 23:09
• 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.) – Bruno Joyal Jul 14 '14 at 23:11
• I am sorry sir, but I cannot seem to use the identity as you suggested. Will help me see that? – Swadhin Jul 14 '14 at 23:15

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 $$3b^2=-3bpk-p^2k^2+pl\implies p|3b^2\implies p|b\implies p|a$$.