Can we find a polynomial satisfying certain conditions? Let $P(x,y)=ax^2+bxy+cy^2+ex+fy+g\in\mathbb Z[x,y]$ be a quadratic polynomial.
We say that $P'(x,y)=a'x^2+b'xy+c'y^2+e'x+f'y+g'\in\mathbb Z[x,y]$ is represented by $P$ if there exists an affine transformation $\tau(x,y)=(a_1x+a_2y+a_3,b_1x+b_2y+b_3)$ such that $P(\tau(x,y))=P'(x,y)$ ( $a_1,a_2,a_3,b_1,b_2,b_3\in\mathbb Z$).
We say that $P$ is equivalent to $P'$ if $P$ is represented by $P'$ and conversely.
The question is:
If $(a,b,c)=1$, can we find a $P'$ such that it is equivalent to $P$ and $(b',a'c')=1$?
My approach:
$P$ is equivalent to $P'$ iff $a_1b_2-a_2b_1=\pm1$.
$a'=aa_1^2+ba_1b_1+cb_1^2,$
$b'=2aa_1a_2+b(a_1b_2+a_2b_1)+2cb_1b_2,$
$c'=aa_2^2+ba_2b_2+cb_2^2.$
 A: I figure it out after a few days. However, my method is very complicated. I'll be appreciated if someone post simpler method.
Since $\Delta=b^2-4ac$ is an invariant of equivalent classes of equivalent quadratic polynomial, it suffice to show that $(\Delta,a'c')=1$.
By $(a,b,c)=1$ we have $(b,(a,c))=1$, which implies $(\Delta,(a,c))=1$. Now we consider the unimodular affine transformation $\tau(x,y)=(x+ny,y)$, where $n$ will be determined later. Thus $c'=an^2+bn+c$.
Since $(\Delta,(a,c))=1$, we can partition the prime factors of $\Delta$ into three types. Types $\mathrm{I}$: primes divide $a$; Types $\mathrm{II}$: primes divide $c$; Types $\mathrm{III}$: other primes. Let
$$n=tPR:=t\prod_{p\in type\ \mathrm{I}}p\prod_{r\in type\ \mathrm{III}}r,$$
where $t$ will be determined later. Thus $c'=tPR(aPRt+b)+c$. Now we have $(c',PR)=1$. It suffice to select proper $t$ such that $(c',Q)=(Q,t(aPRt+b))=1$, where $Q:=\prod_{q\in type\ \mathrm{II}}q$.
If $2|Q$, then $2|\Delta\Rightarrow2|b$. Since $2\nmid aPR$ we select $t$ such that $t\equiv1\ (\mathrm{mod}\ 2)$.
For each odd prime $q_i|Q$ we define $m_i$ by
$$m_i=\begin{cases} 1,&\text{if }b\not\equiv1\ (\mathrm{mod}\ q_i), \\ 2,&\text{if }b\equiv1\ (\mathrm{mod}\ q_i), \end{cases}$$
and solve the following system of $t$ by Chinese Remainder Theorem,
$$aPRt+b\equiv m_i\ (\mathrm{mod}\ q_i)\Leftrightarrow t\equiv (m_i-b)(aPR)^{-1}\ (\mathrm{mod}\ q_i).$$
Finally we get $n$ for which $(c',PQR)=1$ so $(c',\Delta)=1$. Thus $(b'^2,c')=1\Rightarrow(a',b',c')=1$. Notice that the above discussion don't change $a$, so we can apply same discussion for $a$ without change of $c$. Consequently, we get an equivalent quadratic polynomial such that $(\Delta,a'c')=1\Rightarrow(b',a'c')=1$.
