Prove that $m-n$ divides $p$ Let $a_1,a_2,a_3$ be a non constant arithmetic Progression of integers with common difference $p$ and $b_1,b_2,b_3$ be a geometric Progression with common ratio $r$. Consider $3$ polynomials $P_1(x)=x^2+a_1x+b_1, P_2(x)=x^2+a_2x+b_2,P_3(x)=x^2+a_3x+b_3$. Suppose there exist integers $m$ and $n$ such that $gcd(m,n)=1$ and $P_1(m)=P_2(n),P_2(m)=P_3(n)$ and $P_3(m)=P_1(n)$. Prove that $m-n$ divides $p$.
I made the $3$ equations but am unable to get the result from them. How should I manipulate them. Thanks.

(Too long for a comment)
The claim is not true. A counterexample is
$$m=5,\quad n=3,\quad p=3,\quad r=\frac 35,\quad a_1=-11,\quad b_1=\frac{75}{2}.$$
In the following, I'm going to write about how I figured out these values and about that the claim is true if $b_1$ is an integer.
We have
$$P_1(m)=P_2(n)\iff m^2+a_1m+b_1=n^2+a_2n+b_2$$
$$\iff m^2+a_1m+b_1=n^2+(a_1+p)n+b_1r\tag 1$$
$$P_2(m)=P_3(n)\iff m^2+a_2m+b_2=n^2+a_3n+b_3$$
$$\iff m^2+(a_1+p)m+b_1r=n^2+(a_1+2p)n+b_1r^2\tag 2$$
$$P_3(m)=P_1(n)\iff m^2+a_3m+b_3=n^2+a_1n+b_1$$
$$\iff m^2+(a_1+2p)m+b_1r^2=n^2+a_1n+b_1\tag 3$$
From $(2)-(1)$, we have
$$pm+b_1(r-1)=pn+b_1r(r-1)\tag4$$
From $(3)-(2)$, we have
$$pm+b_1r(r-1)=-2pn+b_1(1-r^2)\tag5$$
Now $(4)-(5)$ gives
$$b_1(r-1-r^2+r)=3pn+b_1(r^2-r-1+r^2)\iff pn=b_1(-r^2+r)\tag6$$
Also $-2\times (4)-(5)$ gives
$$pm=b_1(-r+1)\tag 7$$
Hence, from $(6)(7)$, we have
$$pmr=pn,$$
i.e.
$$r=\frac nm.$$
Now from $(7)$, we have
$$pm=b_1\left(-\frac nm+1\right)\Rightarrow pm^2=b_1(m-n)\tag8$$
Note here that $\gcd(m^2,m-n)=1$.
Hence, if $b_1$ is an integer, then we have that $m-n$ divides $p$.
By the way, from $(3)$, we can have $a_1=-m-n-p$.
