$p\equiv 2\pmod 3$ is an odd prime. Prove that there are no integers $x$, $y$ satisfying $p=x^2-xy+y^2$.

Question

$p\equiv 2\pmod 3$ is an odd prime. Prove that there are no integers $x$, $y$ satisfying $p=x^2-xy+y^2$.

The textbook says because when $p\equiv 2\pmod 3$ is an odd prime, $\left( \frac{-3}{p} \right) =-1$. But I don't know how to use $\left( \frac{-3}{p} \right) =-1$ to prove the result.

Answer 1

Not certain why $(-3/p) = -1$ matters?

0 and 1 are the QRs modulo $3$. Assume $x$ and $y$ exist such that $x^2 -xy +y^2 = p \equiv 0$ is prime.

If $x^2 \equiv y^2 \equiv 0 \pmod 3$, then $3 \mid p$ and it's not prime.

If $x^2 \equiv y^2 \equiv 1$, then $xy \equiv 1$ or $2$, and $p \equiv 0$ or $1$, which is not $2$.

Hence if they exist, $x \equiv 1$ or $2, y \equiv 0$ WLOG. But for a prime, $p \equiv 2 \pmod 3 \iff p \equiv 5 \pmod 6$. Then in modulo $6$, $x^2 \equiv 1$ or $4, y^2 \equiv 0$ or $3$, and $xy \equiv 0$ or $3$.

Adding $0$s or $3$s to $1$ or $4$ will never get us to $5 \pmod 6$. Therefore we have contradiction, and the $x$ and $y$ we sought do not exist.

Answer 2

Here $RHS = x^2-xy+y^2 \equiv x^2-xy+y^2+3xy \equiv x^2+2xy+y^2 \equiv (x+y)^2 \equiv 1 \pmod3$

Given :
$LHS = P \equiv 2 \pmod3$

Hence, there can be no Solution in Integers [[ I have not used the "Odd Prime Criteria", hence it is true even in case $ P = 2,5,8,11,14,17,20,23,26 ....$ ]]

{{ This is the expansion of my earlier comment based on the Hint by user @robjohn }}

Answer 3

I'll give you a solution which uses the theory of quadratic forms on the field of $p$-adic numbers $\mathbb Q_p$. If we are able to prove that there are no solutions of the equation $p=x^2-xy+y^2$ in $\mathbb Q_p$ then, clearly, there are no integers $x,y$ satisfying the equation, since $\mathbb Z\subset \mathbb Q_p$. The first thing we have to do is to find an equivalent diagonal expression of the quadratic form $f(x,y)=x^2-xy+y^2$, in order to easily compute the local invariants of $f$. With respect to the basis $\mathcal B=\{e_1=(1,0),e_2=(1,2)\}$ we can see that the expression of $f$ is diagonal. More precisely $f$ is equivalent to $g(x,y)=x^2+3y^2$. The following theorem holds:

Theorem: Given $a\in\frac{\mathbb Q_p^\times}{\mathbb Q_p^{\times,2}}$ and $f$ a non-degenerate quadratic form of rank $n$, discriminant $d$ and Hasse-invariant $\epsilon$, the equation $f(x_1,\ldots,x_n)=a$ has a solution over $\mathbb Q_p$ if and only if:

• $n=1$ and $a=d$
• $n=2$ and $(a,-d)=\epsilon$
• $n=3$ and $a\neq -d$ or $a=-d$ and $(-1,-d)=\epsilon$
• $n\geq 4$ always

The discriminant and the Hasse-invariant are the same for equivalent forms, so we have that: $d_f=3$ and $\epsilon_f=(1,3)$. Since $p\in\frac{\mathbb Q_p^\times}{\mathbb Q_p^{\times,2}}$ and $n=2$, if we prove that $(p,-d_f)\neq \epsilon_f$ we have done. It's not too much difficult to prove that $(p,-3)=\left(\frac{-3}{p}\right)$, and using the hint of your book: $(p,-3)=-1$ for an odd prime $p$ such that $p\equiv 2$ mod $3$. While we have that $\epsilon_f=(1,3)=1$, since $1$ is a square.

So there are no solutions $x,y\in\mathbb Q_p$ to the equation $f(x,y)=p$, and therefore there are no solutions over integers.