Resultant and Frobenius

Let odd and non-square $n(b)$ $=$ $\prod_{i=1}^{i=s} (k_ib+1)^{e_i}$ where $\gcd(k_1,...,k_s)$ $=$ $1$ be the product of $s$ distinct prime factors with their respective multiplicity of $e_i$. Let $f(b)$ $=$ $n(b)$ $+$ $1$. Let $g(b)$ $=$ $\prod_{i=1}^{i=s}(k_ib+2)^{e_i}$. Let $H$ $=$ $\gcd(n+1,$ polresultant$(f(x),g(x))$. Then $x^{n+1}$ $\equiv$ $1$ $\pmod{n,x^2-ax+1}$ implies $x^{H}$ $\equiv$ $1$ $\pmod{n,x^2-ax+1}$. (Unproven) This is quite remarkable because the resultant is independent of $b$.

By way of an example: Let $n(b)$ $=$ $(b+1)(2b+1)$. Then $H$ $=$ $\gcd(n+1,16)$. So the maximum for the gcd is $16$. But $x^{16}$ $\pmod{x^2-ax+1}$ has $a^{14}$ as the largest term for $B$ where the result is $Ax$ $+$ $B$. I.e. $B$ $=$ $-a^{14}$ $+$ $13a^{12}$ $-$ $66a^{10}$ $+$ $165a^8$ $-$ $210a^6$ $+$ $126a^4$ $-$ $28a^2$ $+$ $1$ and this has to be $1$ for pseudoprime. However the most one would have to search for in order to find a suitable $a$ such that $a^2-4$ has a Jacobi symbol of $-1$ over $n$ is no more than $\log(n)\log(\log(n))$. When $b$ gets large $B-1$ will be dominated by $|\log(n)\log(\log(n))|^{14}$ and will become less than $n$ so it can not be $0$ $\pmod{n}$. Thus seeking a suitable $a$ becomes unnecessary as part of a test.