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Suppose $$f(x,y)=\sum_{i,j=0\\i+j\in\{0,2\}}^2a_{ij}x^{i}{y^j}$$ and $$g(x,y)=\sum_{i,j=0\\i+j\in\{0,2\}}^2b_{ij}x^{i}{y^j}$$ are two bivariate quadratics over $\mathbb Z[x,y]$.

What are the necessary and sufficient conditions for their resultants to be perfect squares over $\mathbb Z[x]$ and $\mathbb Z[y]$?

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  • $\begingroup$ @MaxAlekseyev Yes. $\endgroup$
    – Turbo
    Commented May 12 at 3:56
  • $\begingroup$ I've updated the question to say resultants rather than resultant. $\endgroup$
    – Max Alekseyev
    Commented May 12 at 16:30

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At very least it is necessary that each of the following three numbers is a square: $$a_{02}^2 b_{20}^2-a_{02} a_{11} b_{11} b_{20}-2 a_{02} a_{20} b_{02} b_{20}+a_{02} a_{20} b_{11}^2+a_{11}^2 b_{02} b_{20}-a_{11} a_{20} b_{02} b_{11}+a_{20}^2 b_{02}^2,$$ $$a_{00}^2 b_{20}^2-a_{00} a_{10} b_{10} b_{20}-2 a_{00} a_{20} b_{00} b_{20}+a_{00} a_{20} b_{10}^2+a_{10}^2 b_{00} b_{20}-a_{10} a_{20} b_{00} b_{10}+a_{20}^2 b_{00}^2,$$ $$a_{00}^2 b_{02}^2-a_{00} a_{01} b_{01} b_{02}-2 a_{00} a_{02} b_{00} b_{02}+a_{00} a_{02} b_{01}^2+a_{01}^2 b_{00} b_{02}-a_{01} a_{02} b_{00} b_{01}+a_{02}^2 b_{00}^2.$$ They represent the common leading coefficient and the trailing coefficients of the resultants.

ADDED. Under the constraints $a_{10}=a_{01}=b_{10}=b_{01}=0$, the resultants represent biquadratic polynomials and their trailing coefficients are squares. Hence, the necessary and sufficient condition is that the discriminants of the underlying quadratics of resultants are zero.

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  • $\begingroup$ Hi Max, maybe correct your solution so that the first and third lines are not identical... $\endgroup$
    – Derek
    Commented May 12 at 16:07
  • $\begingroup$ @Derek: Indeed, it was a copy-n-paste error, now corrected. Thanks for noticing! $\endgroup$
    – Max Alekseyev
    Commented May 12 at 16:26
  • $\begingroup$ @MaxAlekseyev I simplified the equations so that there are no linear terms ($a_{10}=a_{01}=b_{01}=b_{10}=0$). Will this yield any necessary and sufficient conditions? $\endgroup$
    – Turbo
    Commented May 12 at 20:27
  • $\begingroup$ @Turbo: In that case trailing coefficients are algebraic squares, and all we want are zero discriminants. See update. $\endgroup$
    – Max Alekseyev
    Commented May 16 at 17:00
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    $\begingroup$ @Turbo: For example, the resultant with respect to $y$ equals $u\cdot x^4 + v\cdot x^2 + w^2$, where $u,v,w$ are polynomials in $a$'s and $b$'s, and we want it to have zero discriminant, i.e. $v^2 - 4uw^2=0$. And similarly for the resultant w.r.t. $x$. $\endgroup$
    – Max Alekseyev
    Commented May 16 at 18:57

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