I cannot assure everything is right; however... Let $a_{n}=qn+A$ and $b_{n}=rn+A$ with $n\geq 0$ . Then taking $n=m(AB)^{2}$ we have that $a_{m(AB)^{2}}=A(mAB^{2}q+1)$ has the same set of exponents as $b_{m(AB)^{2}}=B(mA^{2}Br+1)$ for all $m\geq 0$. Since $A=a_{0}$ and $B=b_{0}$ have the same set of exponents and no prime dividing $A$ ($B$) divides $mAB^{2}q+1$ ($mA^{2}Br+1$) this means that also $mAB^{2}q+1$ and $mA^{2}Br+1$ have the same set of exponents for all $m\geq 0$. Let $AB^{2}q=Q$ and $A^{2}Br=R$. Suppose $Q$ and $R$ are different and without loss of generality, suppose $Q<R$. $Q=1$ leads to $A=B=q=1$, which is easy to verify that doesn't work with $Q<R$ (take $m=R+2$ and $m=R-2$ and see that the condition implies the existance of two positive squares at distance 4 from each other), so suppose $Q>1$. Take a $x>1$ such that $hcf(x,\phi(R))=1$ which means that modulo $Q$ and modulo $R$ the only solution to $a^{x}\equiv1$ is $a\equiv1$. When $m=\frac{(Q+1)^{x}-1}{Q}$, $a_{m}=(Q+1)^{x}$, so in order to have the same exponent set, $b_{m}$ should be an $x$th power too. However: $b_{m}=\frac{(Q+1)^{x}-1}{Q}R+1<\frac{(R+1)^{x}-1}{R}R+1=(R+1)^{x}$ and $(R+1)^{x}$ is the least $x$th power $>1$ and $\equiv1$ modulo R, so we must admit $Q=R$, which means $AB^{2}q=A^{2}Br \rightarrow \; \frac{A}{B}=\frac{q}{r}=\frac{a_{n}}{b_{n}}$, for all $n$.

Mimoide wrote: When $m=\frac{(Q+1)^{x}-1}{Q}$, $a_{m}=(Q+1)^{x}$, I don't think this is right!

barasawala wrote: Mimoide wrote: When $m=\frac{(Q+1)^{x}-1}{Q}$, $a_{m}=(Q+1)^{x}$, I don't think this is right! So how will you fix it?

Perhaps this is simpler: Let $a_{n}=qn+A$, $b_{n}=rn+B$, and take $n=m(AB)^{2}$ for all $m$. Then we get the sequences $c_{m}=qmA^{2}B^{2}+A=A(qAB^{2}m+1), d_{m}=B(rA^{2}Bm+1)$. Now let $p$ be a prime which is coprime to $q, r, A, B$. Then of course we can find arbitrarily large $k$s such that $p^{k}\equiv 1 \pmod{qAB^{2}}$, and for each such $k$ there exists $m$ such that $c_{m}=Ap^{k}$. Now, since $A$ and $B$ have the same set of exponents and they are coprime to $qAB^{2}m+1$ and $rA^{2}Bm+1$, this means that $p^{k}$ and $rA^{2}Bm+1$ must have the same set of exponents too. However, the set of exponents of $p^{k}$ is just $\{k \}$, so $rA^{2}Bm+1=s^{k}$ for some prime $s$. However, if we take a large enough $k$, then for any $s$ not equal to $p$ the ratio between $Ap^{k}$ and $Bs^{k}$ will be too large to be asymptotically equal to the ratio between $q$ and $r$, since it will get bigger the bigger $k$ is. Then for infinitely many $m$ we have $c_{m}=Ap^{k}$ and $d_{m}=Bp^{k}$. This implies that $q/r=A/B$, and we're done.

Severius wrote: then for any $s$ not equal to $p$ the ratio between $Ap^{k}$ and $Bs^{k}$ will be too large to be asymptotically equal to the ratio between $q$ and $r$, What does this mean?

Okay... suppose $c_{m}=Ap^{k}$ and $d_{m}=Bs^{k}$ where $s$ is a prime not equal to $p$. Then $\frac{c_{m}}{d_{m}}=\frac{A}{B}( \frac{p}{s})^{k}$ (*). Let $p_{1}< p_{2}< \ldots$ be the sequence of all primes and $p=p_{l}$; then if $s<p$, we have $s \leq p_{l-1}$, while if $s>p$, we have $s \geq p_{l+1}$. Now, look at $\frac{c_{m}}{d_{m}}$ again. Evidently, that ratio should tend to $\frac{q}{r}$ as $m$ tends to infinity. However, for those $m$ such that $c_{m}=Ap^{k}$ and $d_{m}=Bs^{k}$, if $p \neq s$ then $( \frac{p}{s})^{k}$ will either be $\geq ( \frac{p}{p_{l-1}})^{k}$ or $\leq ( \frac{p}{p_{l+1}})^{k}$. If we take $k$ large enough, these numbers will be really big and really small respectively, so that they can't tend to a nonzero constant as $m$ tends to infinity.

A bit overpowered but the problem is anyway not an easy one. Very nice! Let us write $$a_n=g_1(c_1n+d_1)$$and $$b_n=g_2(c_2n+d_2)$$where $g_1,g_2$ are non zero integers and $c_1,d_1$ and $c_2,d_2$ are relatively prime integers, with $c_1$ and $c_2$ non zero. Let $M>0$ be any natural number and $q$ be any sufficiently large prime number. Define the prime number $p$ (using the notion of inverse congruence) as follows: $$p \equiv \, d_1 \, (\bmod \, c_1)$$and $$p \equiv \, d_1-\frac{c_1}{c_2}\cdot d_2 \, (\bmod \,q^M)$$and observe that such a prime number $p$ exists by the Dirichlet's Theorem and the Chinese Remainder Theorem. Let $p=c_1n+d_1$ and note that $a_n=g_1p$ and $$b_n=\frac{g_2}{c_1}\cdot \left(c_2(p-d_1)+c_1d_2\right)$$and we get that $q^M \mid b_n$. In particular, a number $v_q(b_n) \ge M$ belongs in the set of exponents of $b_n$. However, the largest element in the set of exponents of $a_n$ in this case, is bounded by the largest exponent of a prime dividing $g_1$ and we get a contradiction as $M$ sufficiently large violates our hypothesis. Thus, either $d_1=0$ or $\frac{c_1}{c_2}=\frac{d_1}{d_2}=l$; in the former case, set $n=p$ satisfying $$p \equiv \, -\frac{c_1}{c_2}\cdot d_2 \, (\bmod \, q^M)$$and repeat the same argument. In the latter case, our conclusion follows by setting $k=\frac{g_2}{g_1l}$, $b_n=ka_n$ for all $n$. Other trivial possibilities, where either of the sequences is constant are ignored and can be resolved in a fairly easy way following the same argument.

Suppose that $a_n=a_0+nr,b_n=b_0+ns$ for all positive integer $n$ Consider $a_m,b_m$ where $m=la_0^2b_0^2$ we get $a_m=a_0(1+lb_0^2a_0r),b_m=b_0(1+la_0^2b_0s)$ Let $f(n)$ denote "set of exponent" of $n$ for all $n\in \mathbb{Z}^+$ Since $gcd(a_0,1+lb_0^2a_0r)=1$ we get that $|f(a_0)\cap f(1+lb_0^2a_0r)|=0$ WLOG that $a_0s\geq b_0r$, if $a_0s=b_0r$ we are done, so suppose that $a_0s>b_0r$ Since there exist infinitely many prime $p$ that $b_0^2a_0r\mid p-1$ Take $l=\frac{p-1}{b_0^2a_0r}$ we get that $f(a_m)=f(a_0)\cup \{ 1\} =f(b_0)\cup f(1+la_0^2b_0s) \Rightarrow 1+|f(a_0)|\geq |f(b_0)|+|f(1+la_0^2b_0s)|\geq |f(b_0)|+1$ So $|f(a_0)|\geq |f(b_0)|$, similarly $|f(b_0)|\geq |f(a_0)|$, so $|f(a_0)|=|f(b_0)|$ Take $l=\frac{p^z-1}{b_0^2a_0r},z>\max \{ t\mid t\in f(a_0)\cup f(b_0)\}$ we get that $f(a_m)=f(a_0)\cup \{ z\} =f(b_0)\cup f(1+la_0^2b_0s)$ So $f(a_0)=f(b_0)$, so $\{ z\} =f(1+la_0^2b_0s)$ This give us for each $z\in \mathbb{Z}^+$ there exist prime number $q$ such that $q^z=1+\frac{p^z-1}{b_0^2a_0r}\times a_0^2b_0s$ Consider $z>\log_{\frac{p+1}{p}}\left( \frac{a_0s}{b_0r}\right)$ we get $\frac{a_.s}{b_0r}<\left( \frac{p+1}{p}\right)^z\rightarrow \sqrt[z]{\frac{a_0s}{b_0r}} <\frac{p+1}{p}$ Since $q^z=1+\frac{p^z-1}{b_0^2a_0r}\times a_0^2b_0s$ we get $1< \frac{a_0s}{b_0r}=\frac{q^z-1}{p^z-1}$, so $q>p$ Since $q>p$ we easily get that $\frac{q^z-1}{p^z-1} \geq \frac{q^z}{p^z}$, so $\frac{a_0s}{b_0r} \geq \frac{q^z}{p^z}\rightarrow \sqrt[z]{\frac{a_0s}{b_0r}} \geq \frac{q}{p}$ but we have $ \sqrt[z]{\frac{a_0s}{b_0r}} <\frac{p+1}{p}$, so $p+1>q$ impossible Finally, we get $a_0s=b_0r$ and so the sequence $(a_n),(b_n)$ satisfy $\frac{a_n}{b_n}=\frac{a_0}{b_0}=\frac{r}{s}$ for all $n\in \mathbb{Z}^+$ and we are done.

Let $a_n=d(a_0+xn),\ b_n=e(b_0+yn)$ with $(a_0,x)=(b_0,y)=1$. By list I mean set of exponents. Note that the arithmetic progressions $\alpha_n=d(a_0+xa_0^2b_0^2den)$ and $\beta_n=e(b_0+ya_0^2b_0^2den)$ also have the same list for any $n$ as they are just corresponding subsequences of $(a_n)_{n\ge 0}$ and $(b_n)_{n\ge 0}$. Write $\alpha_n=da_0(1+kn),\ \beta_n=eb_0(1+\ell n)$ with $da_0|k$ and $eb_0|\ell$. We thus have that $(da_0,1+kn)=1$ for any $n$, so the list of of $\alpha_n$ is the reunion of the list of $da_0$ and the list of $1+kn$ (and the similar for $\beta_n$). Picking $n_0=\dfrac{(k+1)^{\varphi (k) t}-1}{k}$ with $t$ greater than anything in the list of $eb_0$, we have that the list of $\alpha_{n_0}$ is the reunion of the list of $da_0$ with $\varphi(k)t$. The list of $\beta_{n_0}$ is the reunion of the list of $eb_0$ and the list of $1+\ell n$. As the two lists have to be equal, we infer that the list of $da_0$ is greater or equal in length than the list of $eb_0$. Similarly we can get the reversed inequality, so the two lists actually have the same length. Nonetheless, $\varphi(k)t$ is greater than anything in the list of $eb_0$, so actually the lists of $eb_0$ and $da_0$ are the same.

that $\ell k=a^2, 2l=2ab, 1=b^2$, so $\ell=k$, or equivalently $xb_0=ya_0$. We are now done: $\dfrac{a_n}{b_n}=\dfrac{ d \left ( a_0+ \frac{ya_0}{b_0}n \right ) }{e(b_0+yn)}=\dfrac{da_0}{eb_0}$ for any $n$.