proof: p|b and p|a if $p|a^2+b^2+ab$ and p = 6k+5

Question

Let a,b be two natural numbers and p a prime number and

$p=6k+5 $ (k is an unfixed integer)

Prove that if $$ p|a^2+ab+b^2$$ Then $p|b$ and $p|a$

I found this problem in a discrete math book in the number theory section. I tried solving it by multiplying $a^2+ab+b^2$ by $a-b$ which results in $$p|a^3-b^3$$

But i don't know how to proceed.

Answer 1

You've already found that $p$ divides $a^3-b^3$, i.e. that $a^3\equiv b^3\pmod{p}$. In particular, if $p\mid a$ then clearly also $p\mid b$ and vice versa. So suppose $p\nmid a,b$. Given that $p\equiv5\pmod{6}$ the order of the group $(\Bbb{Z}/p\Bbb{Z})^{\times}$ is not divisible by $3$. This means it contains no elements of order $3$, and so the map $$(\Bbb{Z}/p\Bbb{Z})^{\times}\ \longrightarrow\ (\Bbb{Z}/p\Bbb{Z})^{\times}:\ n\ \longmapsto\ n^3,$$ is injective. This means that $a^3\equiv b^3\pmod{p}$ implies $a\equiv b\pmod{p}$, that is to say $p\mid(a-b)$. Then $p$ also divides $$(a^2+b^2+ab)-(a-b)^2=3ab,$$ and so $p$ divides either $a$ or $b$, or both, contradicting the assumption that $p\nmid a,b$.

Answer 2

The following is assuming OP's original p|a-b. Otherwise, disregard.

$p|a-b \implies p|(a-b)^{2} = a^{2}-2ab+b^{2}$.

Since $p|a^2+ab+b^2, p|(a^2+ab+b^2)-(a^{2}-2ab+b^{2})=3ab \implies p |ab$ as $p = 6k+5 \implies p \neq 3 \implies p \nmid 3$.

As $p$ is prime $p|ab \implies p|a$ or $p|b$. Using $p|a^2+ab+b^2$, you will find that WLOG if $p|a, p|a(a+b)=a^{2}+ab$ so that $p|b^{2} \implies p|b$. Hence $p|a,b$.

Answer 3

As $p|(a^2+ab+b^2)$

if $p|a\iff p|b$

For odd $p,$

$$p|(a^2+ab+b^2)\iff p|(2a+b)^2+3b^2$$

If $(p,ab)=1,$ $$\left(\dfrac{2a+b}b\right)^2\equiv-3\pmod p$$

Now use For what odd prime p is -3 a quadratic residues? Non-residue? OR

$-3$ is a quadratic residue iff $p \equiv 1 \pmod 3$ OR

Prove that $-3$ is a quadratic residue mod an odd prime $>3$ if and only if $p$ is of the form of $6n+1$