I have been looking at binary quadratic forms for a question on MSE, number 3985889, about forms representing a prime (not dividing the discriminant) and primitively representing its cube. I calculated that the order of such a form must be one or two or four. The group under Gauss composition, mostly using Dirichlet's description.

When such a form has nice coefficients as $f=ax^2 + mc xy + ac y^2,$ triple $\langle a,mc,ac \rangle,$ some pleasant things happen. The duplicate form has coefficients $\langle a^2,mc,c \rangle$ and the fourth power is evidently the identity, as $c | mc.$ This is what Dickson calls "ambiguous" as a representative of the class. We define $ X =-ax^3 + 3acx y^2 + mc^2 y^3 \, , \; \; \; $ $ Y = mx^3 + 3ax^2 y -ac y^3 \, , \; \; \; $ after which $ F = a X^2 + mcXY + ac Y^2 $ is identically equal to $f^3$ as polynomials in $x,y.$

I have a single strange example so far, $\langle 14, 8, 29 \rangle.$ It is of order four, and with $ f = 14 x^2 + 8 xy+29y^2, $ $ u= 6x^3 + 60x^2y - 3xy^2 -42y^3, $ $ v = 8x^3- 18x^2y - 60xy^2 +y^3,$ $ h = 14 u^2 + 8 uv+29v^2 $ cause $h=f^3$ identically as polynomials. $u,v$ are coprime when $x,y$ are coprime, $y \neq 0 \pmod 2,\; \; \; y \neq x \pmod 3 ,\; \; \; x+y \neq 0 \pmod 5 ,\; \; \; y-3x \neq 0 \pmod {13}$

However, I was unable to put the form $\langle 14, 8, 29 \rangle$ into the desired shape $\langle a,mc,ac \rangle.$ For a favorable shape, the duplicate $\langle 10, 0, 39 \rangle$ would need to have a representative either $\langle 10, 20v, 39 + 10 v^2 \rangle$ or $\langle 39, 78v, 10 + 39 v^2 \rangle$ where the final coefficient is to be a square, in particular the square of something represented by $\langle 14, 8, 29 \rangle.$ But that does not happen; the proof involves a half dozen Pell type equations.

Question. Why does $\langle 14, 8, 29 \rangle$ have no equivalent expression as $\langle a,mc,ac \rangle,$ and where might we find other examples?

I have been looking at binary quadratic forms for a question on MSE, If a binary quadratic form primitively represents $n$ and $n^3$, must it be the identity form?, about forms representing a prime (not dividing the discriminant) and primitively representing its cube. I calculated that the order of such a form must be one or two or four. The group under Gauss composition, mostly using Dirichlet's description.

When such a form has nice coefficients as $f=ax^2 + mc xy + ac y^2$, triple $\langle a,mc,ac \rangle,$ some pleasant things happen. The duplicate form has coefficients $\langle a^2,mc,c \rangle$ and the fourth power is evidently the identity, as $c \mid mc$. This is what Dickson calls "ambiguous" as a representative of the class. We define \begin{align*} X ={} & -ax^3 + 3acx y^2 + mc^2 y^3 \\ Y ={} & mx^3 + 3ax^2 y -ac y^3 \end{align*} after which $ F = a X^2 + mcXY + ac Y^2 $ is identically equal to $f^3$ as polynomials in $x$, $y$.

I have a single strange example so far, $\langle 14, 8, 29 \rangle$. It is of order four, and with \begin{align*} f ={} & 14 x^2 + 8 xy+29y^2 , \\ u={} & 6x^3 + 60x^2y - 3xy^2 -42y^3, \\ v ={} & 8x^3- 18x^2y - 60xy^2 +y^3, \\ h ={} & 14 u^2 + 8 uv+29v^2 \end{align*} cause $h=f^3$ identically as polynomials. $u$, $v$ are coprime when $x$, $y$ are coprime and \begin{align*} y \neq{} & 0 \pmod 2, \\ y \neq{} & x \pmod 3 , \\ x+y \neq{} & 0 \pmod 5 , \\ y-3x \neq{} & 0 \pmod {13}. \end{align*}

However, I was unable to put the form $\langle 14, 8, 29 \rangle$ into the desired shape $\langle a,mc,ac \rangle$. For a favorable shape, the duplicate $\langle 10, 0, 39 \rangle$ would need to have a representative either $\langle 10, 20v, 39 + 10 v^2 \rangle$ or $\langle 39, 78v, 10 + 39 v^2 \rangle$ where the final coefficient is to be a square, in particular the square of something represented by $\langle 14, 8, 29 \rangle$. But that does not happen; the proof involves a half dozen Pell type equations.

Question. Why does $\langle 14, 8, 29 \rangle$ have no equivalent expression as $\langle a,mc,ac \rangle$, and where might we find other examples?

I have been looking at binary quadratic forms for a question on MSE, number 3985889, about forms representing a prime (not dividing the discriminant) and primitively representing its cube. I calculated that the order of such a form must be one or two or four. The group under Gauss composition, mostly using Dirichlet's description.

When such a form has nice coefficients as $f=ax^2 + mc xy + ac y^2,$ triple $\langle a,mc,ac \rangle,$ some pleasant things happen. The duplicate form has coefficients $\langle a^2,mc,c \rangle$ and the fourth power is evidently the identity, as $c | mc.$ This is what Dickson calls "ambiguous" as a representative of the class. We define $ X =-ax^3 + 3acx y^2 + mc^2 y^3 \, , \; \; \; $ $ Y = mx^3 + 3ax^2 y -ac y^3 \, , \; \; \; $ after which $ F = a X^2 + mcXY + ac Y^2 $ is identically equal to $f^3$ as polynomials in $x,y.$

I have a single strange example so far, $\langle 14, 8, 29 \rangle.$ It is of order four, and with $ f = 14 x^2 + 8 xy+29y^2, $ $ u= 6x^3 + 60x^2y - 3xy^2 -42y^3, $ $ v = 8x^3- 18x^2y - 60xy^2 +y^3,$ $ h = 14 u^2 + 8 uv+29v^2 $ cause $h=f^3$ identically as polynomials. $u,v$ are coprime when $x,y$ are coprime, $y \neq 0 \pmod 2,\; \; \; y \neq x \pmod 3 $

However, I was unable to put the form $\langle 14, 8, 29 \rangle$ into the desired shape $\langle a,mc,ac \rangle.$ For a favorable shape, the duplicate $\langle 10, 0, 39 \rangle$ would need to have a representative either $\langle 10, 20v, 39 + 10 v^2 \rangle$ or $\langle 39, 78v, 10 + 39 v^2 \rangle$ where the final coefficient is to be a square, in particular the square of something represented by $\langle 14, 8, 29 \rangle.$ But that does not happen; the proof involves a half dozen Pell type equations.

Question. Why does $\langle 14, 8, 29 \rangle$ have no equivalent expression as $\langle a,mc,ac \rangle,$ and where might we find other examples?

I have been looking at binary quadratic forms for a question on MSE, number 3985889, about forms representing a prime (not dividing the discriminant) and primitively representing its cube. I calculated that the order of such a form must be one or two or four. The group under Gauss composition, mostly using Dirichlet's description.

When such a form has nice coefficients as $f=ax^2 + mc xy + ac y^2,$ triple $\langle a,mc,ac \rangle,$ some pleasant things happen. The duplicate form has coefficients $\langle a^2,mc,c \rangle$ and the fourth power is evidently the identity, as $c | mc.$ This is what Dickson calls "ambiguous" as a representative of the class. We define $ X =-ax^3 + 3acx y^2 + mc^2 y^3 \, , \; \; \; $ $ Y = mx^3 + 3ax^2 y -ac y^3 \, , \; \; \; $ after which $ F = a X^2 + mcXY + ac Y^2 $ is identically equal to $f^3$ as polynomials in $x,y.$

I have a single strange example so far, $\langle 14, 8, 29 \rangle.$ It is of order four, and with $ f = 14 x^2 + 8 xy+29y^2, $ $ u= 6x^3 + 60x^2y - 3xy^2 -42y^3, $ $ v = 8x^3- 18x^2y - 60xy^2 +y^3,$ $ h = 14 u^2 + 8 uv+29v^2 $ cause $h=f^3$ identically as polynomials. $u,v$ are coprime when $x,y$ are coprime, $y \neq 0 \pmod 2,\; \; \; y \neq x \pmod 3 ,\; \; \; x+y \neq 0 \pmod 5 ,\; \; \; y-3x \neq 0 \pmod {13}$

However, I was unable to put the form $\langle 14, 8, 29 \rangle$ into the desired shape $\langle a,mc,ac \rangle.$ For a favorable shape, the duplicate $\langle 10, 0, 39 \rangle$ would need to have a representative either $\langle 10, 20v, 39 + 10 v^2 \rangle$ or $\langle 39, 78v, 10 + 39 v^2 \rangle$ where the final coefficient is to be a square, in particular the square of something represented by $\langle 14, 8, 29 \rangle.$ But that does not happen; the proof involves a half dozen Pell type equations.

Question. Why does $\langle 14, 8, 29 \rangle$ have no equivalent expression as $\langle a,mc,ac \rangle,$ and where might we find other examples?