Primes congruent to 1 mod 6

I came across a claim that I found interesting, but can't seem to prove for some reason. I have the feeling it should be easy

a prime $p$ can be written in the form $p = a^2 -ab +b^2$ for some $a,b\in\mathbb{Z}$ if and only if $p\equiv 1\bmod{6}$

Solution 1:

$a^2 - ab + b^2 = (-a)^2 + (-a)b + b^2$. So it suffices to deal with $a^2 + ab + b^2$. Now, take a prime $p \equiv 1 \pmod{6}$. It is elementary to show their exists an integer $d$ such that $d^2 \equiv -3 \pmod{p}$, now take $z \equiv \frac{-1 + d}{2} \pmod{p}$ (so its a third root of unity modulo $p$). Now define $\mathcal L = \{(a,b) \in \mathbb{Z}^2 | a \equiv zb \pmod{p}\}$. It is straightfoward to check $\mathcal L$ is a lattice whose fundamental parallelogram has area $p$. Now by Minkowski's theorem one has $\mathcal L$ contains a nontrivial lattice point inside the ellipse $a^2 + ab + b^2 < 2p$. Call this point $(a,b)$. But then $a^2 + ab + b^2 \equiv 0 \pmod{p}$ based on the definition of the lattice, thus it must be $a^2 + ab + b^2 = p$. The if part follows.

For the "only if" part, just check modulo $3$ and note that $a^2 + ab + b^2 \equiv 0,1 \pmod{3}$. Note that the problem statement fails for $p=3$ due to that.

Solution 2:

Here is another solution for the "if" part, using algebraic number theory. Let $p$ be a prime satisfying $p \equiv 1 \pmod 3$ and consider the number field $K = \mathbb Q(\omega)$ where $\omega = (1\pm \sqrt{-3})/2$ is a primitive third root of unity. By quadratic reciprocity, $$\left( \frac{-3}{p} \right) = \left(\frac{-1}{p} \right) \left( \frac{3}{p} \right) = (-1)^\frac{p-1}{2} (-1)^\frac{p-1}{2} \left(\frac{p}{3}\right) = \left(\frac{1}{3}\right) = 1,$$ so $\mathbb Z/p\mathbb Z$ contains a square root of $-3$. Since $-3$ is the discriminant of $X^2+X+1$ (the minimal polynomial of $\omega$), the polynomial splits in $\mathbb F_p[X]$, therefore $p$ splits in $K$: $$(p) = \mathfrak p \overline{\mathfrak p}$$ for a prime $\mathfrak p$ of $K$. Since $K$ has class number one (the Minkowski bound is $<2$), $\mathfrak p$ is principal, say $\mathfrak p = (a+b\omega)$. So we have $$(p) = \mathfrak p \overline{\mathfrak p} = (a+b\omega)(a+b\omega^2) = (a^2+ab+b^2).$$ Since $K$ is imaginary quadratic, the only units in $\mathcal O_K = \mathbb Z[\omega]$ are $\pm 1$, so $$p = \pm (a^2+ab+b^2).$$ Since $a^2+ab+b^2$ is positive, we must in fact have "+": $$p = a^2+ab+b^2 = (-a)^2 - (-a)b + b^2.$$

Solution 3:

If $(a,b)=d,d^2\mid (a^2-ab+b^2)$

If $d>1,a^2-ab+b^2$ can not be prime $\implies d=1$

Now, if prime $p=a^2-ab+b^2\implies p\mid (a+b)(a^2-ab+b^2)\implies p\mid (a^3+b^3)$

So, $$a^3\equiv(-b)^3\pmod p\implies \left(-\frac ab\right)^3\equiv 1\pmod p\implies ord_p \left(-\frac ab\right)\mid 3$$

If $ord_p \left(-\frac ab\right)=1,p(=a^2-ab+b^2)\mid (a+b)$

If $(a^2-ab+b^2)\mid(a+b),(a^2-ab+b^2)\mid(a+b)^2$ $\implies (a^2-ab+b^2)\mid 3ab$ $\implies (a^2-ab+b^2)\mid 3$ as $(a^2-ab+b^2,a)=(b^2,a)=1$ as $(a,b)=1$

But, $a^2-ab+b^2>3,$ for $a,b>2$

$$\implies ord_p \left(-\frac ab\right)= 3\implies 3\mid \phi(p)\implies p\equiv1\pmod 3\equiv1,4\pmod 6$$

Hence, $p\equiv1\pmod 6$ as $p\equiv4\pmod 6$ is even and $p>2$

Solution 4:

Here is another, albeit non-elementary, solution for the "if" part due to Ireland and Rosen. If $p\equiv 1\pmod{3}$, there exists a multiplicative character $\chi$ of order $3$. Then, $\chi\in\{1,\omega,\omega^2\}$, where $\omega=e^{2\pi i/3}=\frac{1}{2}(-1+\sqrt{-3})$. Now, consider the Jacobi sum of two such characters: $$J(\chi, \chi)=\sum_{a+b=1}\chi(a)\chi(b)\in\mathbf{Z}[\omega].$$ Then, we can write $J(\chi, \chi)=a+b\omega$, for some $a,b\in\mathbf{Z}$. We have $$p=N(J(\chi, \chi))^2=N(a+b\omega)=a^2-ab+b^2,$$ as desired.