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

Question

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.

Hint: what's the discriminant of $t^2 - t + 1$? Note that this is the dehomogenization of $x^2 - xy + y^2$. — Viktor Vaughn, Jun 24 '22 at 04:36
Does this answer your question? Primes congruent to 1 mod 6 - found using an Approach0 search. There are also other related ones, e.g., proof: p|b and p|a if $p|a^2+b^2+ab$ and p = 6k+5. — John Omielan, Jun 24 '22 at 04:48
Hint: $x^2-xy+y^2\equiv(x+y)^2\pmod3$; what are the quadratic residues mod $3$? — robjohn, Jun 24 '22 at 05:16
With RHS = 1 (mod 3) , the "Odd Prime Criteria" is unnecessary or a distraction. When P = 2(mod 3) , there is no Solution in Integers, whether Prime or not. Hint by @robjohn was very useful to me. — Prem, Jun 24 '22 at 05:40
Multiplying by $4$ gives $4x^2-4xy+4y^2 \equiv 0 \pmod{p}$. This can be rearranged as $4x^2-4xy+y^2 \equiv -3y^2 \pmod{p} ; \to ; (2x-y)^2 \equiv -3y^2 \pmod{p}$. Using that $p \not\mid y$ gives that $y$ has a multiplicative inverse $y^{-1}$, so multiplying both sides by $y^{-2}$ gives $(y^{-1}(2x-y))^2 \equiv -3 \pmod{p}$. Thus, $-3$ must be a quadratic residue modulo $p$, which means that $\left(\frac{-3}{p} \right) = 1$. — John Omielan, Jun 24 '22 at 08:09

score 0 · Answer 1 · answered Jun 24 '22 at 06:01

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.

Mod $6$ doesn’t seem needed. If $x\equiv 1$ or $2,$ and $y\equiv 0\pmod3,$ then $x^2-xy+y^2\equiv1\not\equiv2\pmod3$ — J. W. Tanner, Feb 18 '24 at 09:48

score 0 · Answer 2 · answered Jun 24 '22 at 09:06

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 }}

score 0 · Answer 3 · answered Jun 24 '22 at 10:00

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.

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

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