Show that if a prime $p ≠ 3$ is such that $p≡1$ (mod 3) then p can be written as $a^2-ab+b^2$ where a and b are integers.
I have no idea how to approach this question, so any help much appreciated! This is in the context of algebraic number theory, so I'm not sure if it's helpful to consider rings of integers or anything like that.
Given that $p = a^2 - ab + b^2$, we can factor over $\mathbb{Z}[\zeta_3]$ (ring of Eisenstein integers) to obtain $p = \left(a + \zeta_3b\right)\left(a + \zeta_3^2b\right)$.
Therefore the ideal $(p)$ is not inert over $\mathbb{Z}[\zeta_3]$.
Then by the Dedekind factorization theorem we must have that $X^2 + X + 1$ splits into linear factors in $\mathbb{F}_p[X]$, i.e. there are roots in $\mathbb{F}_p$.
But the roots are formally $\frac{-1 \pm \sqrt{-3}}{2}$, so $-3$ must be a perfect square mod $p$. By quadratic reciprocity, this means that $p \equiv 1 \pmod 3$ as desired.
The converse is also true: given that $p \equiv 1 \pmod 3$ we use quadratic reciprocity to show that $-3$ is a perfect square mod $p$. Then the roots $\frac{-1 \pm \sqrt{-3}}{2}$ of the minimal polynomial $X^2 + X + 1$ are in $\mathbb{F}_p$, so $X^2 + X + 1$ splits in $\mathbb{F}_p[X]$.
By the Dedekind factorization theorem, the ideal $(p)$ factors into a product of two prime ideals in $\mathbb{Z}[\zeta_3]$. Since $\mathbb{Z}[\zeta_3]$ is a PID, we can write $(p) = (\alpha)(\beta) = (\alpha\beta)$, then it follows that $p = u\alpha\beta$ for some $\alpha, \beta \in \mathbb{Z}[\zeta_3]$ nonunits and $u$ some unit.
Finally taking norms we must have $N_{\mathbb{Q}(\zeta_3)/\mathbb{Q}}(\alpha) = p$, so $p$ is the norm of $\alpha = a + b\zeta_3$, which is $a^2 - ab + b^2$.
This is just a matter of quadratic reciprocity law.
Let $\omega$ be a primitive cubic root of unity and $K=\mathbf Q(\omega)= \mathbf Q(\sqrt {-3})$ . It is known that the ring of integers of $K$ is $A=\mathbf Z[\omega]=\mathbf Z[\sqrt {-3}]$, which is a PID. Knowing that $N(a+b\omega)=a^2 - ab +b^2$ (where $N$ is the norm map defined by $N(z)=z.\bar z$), your question amounts to the characterisation of a rational prime $p$ which is a norm from $A$, i.e. $p$ splits completely in $A$, or equivalently, $p$ is odd and $-3$ is a quadratic residue mod $p$ (note that the case $p=2$ is excluded because $-3\neq 1$ mod $8$), i.e. $(\frac {-3}{p})=1$ . But $(\frac {-3}{p})=(\frac {3}{p})(\frac {-1}{p})=(\frac {3}{p})$ because $p$ is odd, and quadratic reciprocity asserts here that $(\frac {3}{p})=(\frac {p}{3})$ . Finally, the wanted $p$ is characterized by $(\frac {p}{3})=1=(\frac {-3}{p})$. The first equality implies that $p\equiv 1$ mod $3$, and the second that $p$ splits, as we have seen above.
COMMENT.- I fear it is a problem not too elementary. It can be solved using the theory of representation of integers by quadratic forms. In short, consider the discriminant of the form $x^2-xy+y^2$ which is equal to $\Delta=-3$.
One can use the following result: The integer $n$ is represented by a quadratic form of discriminant $\Delta$ if and only if $\Delta$ is a square modulo $4n$.
Here four examples.
►For $p=7$ one has $x^2=-3\pmod{28}$ since $-3=25$, obvio.
►For $p=13$ one has $-3$ is a square modulo $52$ because $-3=49$ obvio.
►For $p=19$ one has $-3=x^2\pmod{76}$ has, for example, the solution $x=15$(there are other solutions).
►For $p=31$ the equation $-3=x^2\pmod{124}$ has, for example,the solution $x=11$.
