This problem is from Eisenstein primes, I already proved, that if $p$ is natural prime, then it can be decomposed into $a^2-ab+b^2$ or it is Eisenstein prime. So we have to prove, that if $p \, \text{ mod } \, 3 =1$ then $p$ is not Eisenstein prime. Also, I found out, that $p$ is Eisenstein prime if and only if the equation $x^2+x+1=0$ has no solutions $\text{ mod } \, p$. So, $x^2+x+1=0$ the same as $(2x+1)^2=-3$. Please, suggest any hints to solve this.
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do you knowwhat the Legendre symbol is? – Will Jagy May 29 '18 at 18:11
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only heard about it, i’ll try to use it, thanks – Ilya May 29 '18 at 18:14
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See this duplicate. – Dietrich Burde May 29 '18 at 21:16
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You can use the theory of quadratic forms. Since $p\equiv1\pmod3$ is equivalent (by quadratic reciprocity) to $-3$ being a square modulo $p$, then $p$ is represented by a positive definite integer quadratic form of discriminant $-3$. But every such form is equivalent to the form $x^2-xy+y^2$, by, for instance, the theory of reduced forms. So the prime $p\equiv1\pmod3$ entails $p$ being represented by $x^2-xy+y^2$.
Angina Seng
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