<s>[INST] Integers representable as binary quadratic forms 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 $m=3x^2+2xy+4y^2$?
 [/INST] you don't seem to be mentioning Gauss composition. You have a genus of forms,  equivalent to $\langle 1,8,27 \rangle,$ then $\langle 3,8,29 \rangle,$ 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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