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One can start showing the following:

The integers $n$ which are of the form $x^2+xy+y^2$, for two relatively prime integers $x,y$ are precisely those positive integers occurring as divisors of $m^2+m+1$, for some integer $m$.

In other words, the polynomial $f(x)=x^2+x+1$ has the property that the positive divisors of the integers it represents are precisely those integers that can be properly represented by its homogenization $F(x,y)=x^2+xy+y^2$ ("properly" here refers to the condition $(x,y)=1$).

The proof uses the fact that the imaginary quadratic order $\mathbf{Z}[x]/f(x)$ has class number one, as anticipated by Elkies. It goes as follows:

Let $n$ be a positive divisor of $m^2+m+1$, for some integer $m$. Consider the quadratic form in $x,y$ given by:

$Q(x,y)=\frac{m^2+m+1}{n}x^2-(2m+1)xy+ny^2$;

it is with has integer coefficients, positive definite, and has discriminant equal to $-3$ (in particular it is primitive).

Since $h(-3)=1$ h(-3)=1$, there is only one reducedquadratic form, positive definite and with quadratic form of discriminant$-3$. Which This is$E(x,y)=x^2+xy+y^2$. Therefore$Q$and$E$are properly equivalent (that is there is a determinant one determinant-one change of variable variables taking one into the other), and since$Q$certainly properly represents$n$, so does$E$. The converse is similar and uses the fact that if a quadratic form$Q$properly represents an integer$n$, then$Q$is properly equivalent to a form of the type$nx^2+bxy+cy^2$(this is lemma 2.3 of Cox's wonderful book "Primes of the form x^2+ny^2"). Of course once Once you understand the positive integers$n$that are properly represented by$E$, then you can get them all, after scaling by squares. The original problem is then reduced to understanding those integers$n$for which$x^2+x+1$has a zero in$\mathbf{Z}/(n)$. Using the Chinese Remainder Theorem, this can be reduced to the case where$n$is a prime power$p^s$. Then for$p\neq 3$Hensel's Lemma tells you that your equation has a solution mod$p^s$if and only if it has a solution mod$p$. With quadratic reciprocity you can conclude that any prime divisor of$n$has to be congruent to$1$mod$3$. I am at the moment missing how you can solve the equation$x^2+x+1=0$in$\mathbf{Z}/3^s$, but I think that a version of Hensel's Lemma applies. I'll think about it. [EDIT: The equation$x^2+x+1=0$has a solution in$\mathbf{Z}/(3^s)$, with$s\geq 1$, if and only if$s=1$. (Therefore$x^2+xy+y^2$represents properly only$3$and$1$as powers of$3$.) One can see this by checking that there is no solution for$s=2$, and therefore for$s>2$. More generally, if$p>2$then$x^{p-1}+x^{p-2}+\ldots+x+1=0$has no solutions in$\mathbf{Z}/(p^s)$with$s>1$. The$p$--adic valuation of an integer of the form$x^{p-1}+x^{p-2}+\ldots+x+1=(x^p-1)/(x-1)$is either zero or one.] (Suggested reading. Cox's book . It is a wonderful one. quoted above and Serre's paper:$\Delta=b^2-4ac$) 3 added 511 characters in body One can start showing the following: The integers$n$which are of the form$x^2+xy+y^2$, for two relatively prime integers$x,y$are precisely those positive integers occurring as divisors of$m^2+m+1$, for some integer$m$. In other words, the polynomial$f(x)=x^2+x+1$has the property that the positive divisors of the integers it represents are precisely those integers that can be properly represented by its homogenization$F(x,y)=x^2+xy+y^2$("properly" here refers to the condition$(x,y)=1$). The proof uses the fact that the imaginary quadratic order$\mathbf{Z}[x]/f(x)$has class number one, as anticipated by Elkies. It goes as follows: Let$n$be a positive divisor of$m^2+m+1$, for some integer$m$. Consider the quadratic form in$x,y$given by:$Q(x,y)=\frac{m^2+m+1}{n}x^2-(2m+1)xy+ny^2$; it is with integer coefficients, positive definite, and has discriminant equal to$-3$(in particular it is primitive). Since$h(-3)=1$there is only one reduced quadratic form, positive definite and with discriminant$-3$. Which is$E(x,y)=x^2+xy+y^2$. Therefore$Q$and$E$are properly equivalent (that is there is a determinant one change of variable taking one into the other), and since$Q$certainly properly represents$n$, so does$E$. The converse is similar and uses the fact that if a quadratic form$Q$properly represents an integer$n$, then$Q$is properly equivalent to a form of the type$nx^2+bxy+cy^2$(this is lemma 2.3 of Cox's wonderful book "Primes of the form x^2+ny^2"). Of course once you understand the positive integers$n$that are properly represented by$E$, then you can get them all, after scaling. The original problem is then reduced to understanding those integers$n$for which$x^2+x+1$has a zero in$\mathbf{Z}/(n)$. Using the Chinese Remainder Theorem, this can be reduced to the case where$n$is a prime power$p^s$. Then for$p\neq 3$Hensel's Lemma tells you that your equation has a solution mod$p^s$if and only if it has a solution mod$p$. With quadratic reciprocity you can conclude that any prime divisor of$n$has to be congruent to$1$mod$3$. I am at the moment missing how you can solve the equation$x^2+x+1=0$in$\mathbf{Z}/3^s$, but I think that a version of Hensel's Lemma applies. I'll think about it. [EDIT: The equation$x^2+x+1=0$has a solution in$\mathbf{Z}/(3^s)$, with$s\geq 1$, if and only if$s=1$. (Therefore$x^2+xy+y^2$represents properly only$3$and$1$as powers of$3$.) One can see this by checking that there is no solution for$s=2$, and therefore for$s>2$. More generally, if$p>2$then$x^{p-1}+x^{p-2}+\ldots+x+1=0$has no solutions in$\mathbf{Z}/(p^s)$with$s>1$. The$p$--adic valuation of an integer of the form$x^{p-1}+x^{p-2}+\ldots+x+1=(x^p-1)/(x-1)$is either zero or one.] (Suggested reading. Cox's book. It is a wonderful one. Serre's paper:$\Delta=b^2-4ac$) 2 deleted 8 characters in body One can start showing the following: The nonzero integers$n$which are of the form$x^2+xy+y^2$, for two relatively prime integers$x,y$are precisely those positive integers occurring as divisors of$m^2+m+1$, for some integer$m$. In other words, the polynomial$f(x)=x^2+x+1$has the property that the positive divisors of the integers it represents are precisely those integers that can be properly represented by its homogenization$F(x,y)=x^2+xy+y^2$("properly" here refers to the condition$(x,y)=1$). The proof uses the fact that the imaginary quadratic order$\mathbf{Z}[x]/f(x)$has class number one, as anticipated by Elkies. It goes as follows: Let$n$be a positive divisor of$m^2+m+1$, for some integer$m$. Consider the quadratic form in$x,y$given by:$Q(x,y)=\frac{m^2+m+1}{n}x^2-(2m+1)xy+ny^2$; it is with integer coefficients, positive definite, and has discriminant equal to$-3$(in particular it is primitive). Since$h(-3)=1$there is only one reduced quadratic form, positive definite and with discriminant$-3$. Which is$E(x,y)=x^2+xy+y^2$. Therefore$Q$and$E$are properly equivalent (that is there is a determinant one change of variable taking one into the other), and since$Q$certainly properly represents$n$, so does$E$. The converse is similar and uses the fact that if a quadratic form$Q$properly represents an integer$n$, then$Q$is properly equivalent to a form of the type$nx^2+bxy+cy^2$(this is lemma 2.3 of Cox's wonderful book "Primes of the form x^2+ny^2"). Of course once you understand the positive integers$n$that are properly represented by$E$, then you can get them all, after scaling. The original problem is then reduced to understanding those integers$n$for which$x^2+x+1$has a zero in$\mathbf{Z}/(n)$. Using the Chinese Remainder Theorem, this can be reduced to the case where$n$is a prime power$p^s$. Then for$p\neq 3$Hensel's Lemma tells you that your equation has a solution mod$p^s$if and only if it has a solution mod$p$. With quadratic reciprocity you can conclude that any prime divisor of$n$has to be congruent to$1$mod$3$. I am at the moment missing how you can solve the equation$x^2+x+1=0$in$\mathbf{Z}/3^s$, but I think that a version of Hensel's Lemma applies. I'll think about it. (Suggested reading. Cox's book. It is a wonderful one. Serre's paper:$\delta=b^2-4ac$)\Delta=b^2-4ac$)

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