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It is known that odd prime $p$ can be represented as $p=x^2+y^2$ if and only if $p \equiv 1$ mod $4$, represented as $p=x^2+2y^2$ if and only if $p \equiv 1$ or $3$ mod $8$, represented as $p=x^2+3y^2$ if and only if $p \equiv 0$ or $1$ mod $3$, etc. The situation with representation as $p=x^2+ny^2$ becomes more complicated at $n=11$: if $p=x^2+11y^2$ then $p=0,1,3,4,5$ or $9$ mod $11$, (that is, $p=11$ or is the qudratic residue mod $11$) but the converse statement is not true. In fact, primes in these residues classes are represented as either $p=x^2+11y^2$ or $p=3x^2+2xy+4y^2$, and these sets of primes are disjoint.

Example $2^2 + 11 \cdot 1^2 = 15 = 3 \cdot 5$ shows that all prime factors of an integer representable as $x^2+11y^2$ can be of the form $3x^2+2xy+4y^2$. My question is whether the opposite can be true: do there exists any integer representable $3x^2+2xy+4y^2$ that has all its prime factors in the form $x^2+11y^2$?

Update inspired by Will Jagy's answer. The answer mentions polynomial $f(z)=z^3+z^2-z+1$. Ok, if $p$ is in the form $x^2+11y^2$ then $f(z)=0$ is solvable (in fact has $3$ solutions if $p\neq 11$) modulo $p$. If $p_1$ and $p_2$ are two distinct primes of this form, it follows that $f(z)=0$ is solvable modulo $m=p_1p_2$, and so on. But why $z^3+z^2-z+1=0$ cannot be solvable modulo a (not necessarily prime) integer $m$ representable as $m=3x^2+2xy+4y^2$?

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    $\begingroup$ No. All prime of the form $x^2 + 11y^2$ are in the principal class, and any product of such primes are also in the principal class. $\endgroup$
    – Stanley Yao Xiao
    Commented Mar 24, 2022 at 19:16
  • $\begingroup$ Thank you for the answer, but I thought that the principle class is the class of forms equivalent to the principal form. In this example, this is class of forms equivalent to $x^2+11y^2$, such as, for example, form $(x+y)^2+11y^2$, etc. When you write "primes are in the principle class", you probably mean some class of integers (not of forms), and this class is closed under the product operation. More details or reference would help. $\endgroup$
    – Bogdan Grechuk
    Commented Mar 24, 2022 at 19:47
  • $\begingroup$ $(a+b\sqrt{-11})(c+d\sqrt{-11})=(ac-11bd)+(ad+bc)\sqrt{-11}$, so $(a^2+11b^2)(c^2+11d^2)=(ac-11bd)^2+11(ad+bc)^2$. $\endgroup$
    – Gerry Myerson
    Commented Mar 24, 2022 at 22:17
  • $\begingroup$ A prime number $n$ is represented by a binary form with discriminant $d$ iff $4d$ is a quadratic residue modulo $4n$ ($\gcd(n,d)=1$). $\endgroup$
    – markvs
    Commented Mar 24, 2022 at 22:34
  • $\begingroup$ @Gerry Myerson - thanks, I know that if two integers are representable in the form $x^2+11y^2$ then so is their product. But how this implies that the product cannot be also representable as $3x^2+2xy+4y^2$? After all, $15$ is representable as $15=2^2+11 (1)^2$ and also as $15=3(-1)^2+2(-1)(2)+4 (2)^2$. $\endgroup$
    – Bogdan Grechuk
    Commented Mar 24, 2022 at 22:34

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you don't seem to be mentioning Gauss composition. You have a genus of forms, equivalent to $\langle 1,8,27 \rangle,$ then $\langle 3,8,9 \rangle,$ then $\langle 9,8,3 \rangle.$ These are convenient for Dirichlet's description of composition.

There is a cancellation for principal forms: if $x^2 + 8xy + 27 y^2$ represents both $p$ and $np,$ it also represents $n.$

You mention 15, $$ \left( 3x^2 + 8xy + 9 y^2 \right) \left( 9z^2 + 8zw + 3 w^2 \right) = \color{magenta}{ u^2 + 8 uv + 27 v^2,} $$ where $u = 3xw + 9 yz +8yw$ and $v=xz-yw.$

Lots more...A prime $p \neq 2,11$ with Legendre symbol $(-44|p) = 1$ is represented by $x^2 + 8xy + 27 y^2$ if and only if the polynomial $z^3 + z^2 - z + 1$ factors into three distinct linear factors $\pmod p.$ Cubic because class number $h(-44) = 3$

Oh, Dirichlet composition is available everywhere, I copied from D. A. Cox, Primes of the Form $x^2 + n y^2$

there is always more

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  • $\begingroup$ Yes, there is a big theory here, but I did not mentioned it, because I do not see how exactly it answers the question. Can product of primes in the form $x^2+11y^2$ be represented as $3x^2+2xy+4y^2$? How Gauss composition and/or cubic polynomial characterization for primes helps here? $\endgroup$
    – Bogdan Grechuk
    Commented Mar 25, 2022 at 7:54
  • $\begingroup$ @BogdanGrechuk it is all in zakuski.utsa.edu/~jagy/jagy_division.pdf $\endgroup$
    – Will Jagy
    Commented Mar 25, 2022 at 17:57
  • $\begingroup$ Thank you! The book D. A. Buell. Binary Quadratic Forms: Classical Theory and Modern Computations. indeed contains theorem (Theorem 4.26) that answers this question in general form. $\endgroup$
    – Bogdan Grechuk
    Commented Mar 25, 2022 at 19:56
  • $\begingroup$ @BogdanGrechuk the place this came up was in showing some ternary forms regular. One or two versions are in the 1928 dissertation of Burton Wadsworth Jones. He found the regular diagonal ternaries $a x^2 + b y^2 + c z^2.$ A decade later, he and Gordon Pall found many more (not diagonal), also the first examples of what is now called spinor regularity. Related to your questions at zakuski.utsa.edu/~jagy/inhom.cgi and ternaries at zakuski.math.utsa.edu/~kap $\endgroup$
    – Will Jagy
    Commented Mar 25, 2022 at 21:59

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