A sketch: let the maximum exponent of p dividing the sequence be $k$ and let $p^k || ca^m - db^m$. Then show that if $p|a^i - b^i$, then $p^k|a^i -b^i$ else we can lift to get $n$ such that $p^{k+1} | ca^m - db^n$. The conclusion follows by choosing i to be the smallest such exponent aka the order of $\frac{b}{a} \pmod p$.

The key to the question is the power of $p$ that divides $a^i-b^i$ as above. Suppose otherwise, then we have $p^x||ca^k-db^k$,$p^y||ca^l-db^l$, where $x\neq y, x,y>0$, and $p^y$ is the largest possible power that divides any $ca^m-db^m$. We can make $l>k$ by adding multiple $\phi(p^y)$, $p^y$ will still divide $ca^l-db^l$, hence $p^y||ca^l-db^l$. Now let $ca^k=db^k+ep^x$, then $p\nmid e$. Using this in $ca^l-db^l$, we get $p^y||db^k(a^{l-k}-b^{l-k})+ea^{l-k}p^x$, hence $p^x||a^{l-k}-b^{l-k}$, as $db^k,ea^{l-k}$ are coprime with $p$. From LTE, we get $p^y||a^{(l-k)p^{y-x}}-b^{(l-k)p^{y-x}}$. Let $a^{(l-k)p^{y-x}}=b^{(l-k)p^{y-x}}+rp^y$, $p\nmid r$. Hence $a^{(l-k)p^{y-x}t}\equiv b^{(l-k)p^{y-x}t}+tb^{(l-k)p^{y-x}(t-1)}rp^y(mod p^{y+1})$. Let $ca^l=db^l+fp^y$, $p\nmid f$. Hence substituting $m=(l-k)p^{y-x}t+l$, we get $ca^m-db^m=db^l(a^{m-l}-b^{m-l})+fa^{m-l}p^y\equiv db^l\cdot tb^{(l-k)p^{y-x}(t-1)}rp^y+fa^{(l-k)p^{y-x}}p^y\equiv p^y(rdb^{l+(l-k)p^{y-x}(t-1)}t+fa^{(l-k)p^{y-x}})(mod p^{y+1})$. We let $t=(p-1)t'$ for positive integer $t'$, then $rdb^{l+(l-k)p^{y-x}(t-1)}$ and $fa^{(l-k)p^{y-x}}$ are fixed and non-zero $(modp)$, so let them be $\alpha,\beta$ respectively. Now take $t\equiv-\frac{\beta}{\alpha}(modp)$ and large enough, we get $p|rdb^{l+(l-k)p^{y-x}(t-1)}t+fa^{(l-k)p^{y-x}}$, so $p^{y+1}|ca^m-db^m$, contradiction.

Some comments: 0) This problem was inspired by a failed approach to ISL 2012 N6. 1) It's in fact not hard to show that $T$ only depends on $p,c,d$: we have $T = v_p(c^{p-1} - d^{p-1})$. 2) Brian Lawrence came up with a more interesting problem along these lines: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=57&t=598357

Redefine $M$ as $M+1$. Then let's see all of this modulo $p^M$. Let $g$ be a primitive root and $a/b=g^x,d/c=g^y$. Then $v_p(ca^n-db^n)$ is equal to the maximal $\ell$ such that $p^{\ell-1}(p-1) | xn-y$. Notice that $\ell \ge 1$ is achieved at least once and therefore $gcd(p-1,x)|y$. Notice that $M=\ell$ never satisfies said equation, and so $xn=y$ (mod $p^{M-1}$) is impossible (since if it were, we could achieve by CRT $xn=y$ (mod $p^{M-1}(p-1)$), therefore $v_p(x)>v_p(y)=:T$. Then if $p^{\ell} || ca^n-db^n$ and $\ell>0$ then $p^{\ell-1}(p-1) | xn-y$ is the maximal $\ell$ that satisfies this, then $p-1|xn-y$ and $p^T|xn-y$ and $p^{T+1}|xn$ but not $y$, therefore our answer is $T$.

We use $p$-adic numbers. We're interested in the numbers $v_p(ca^N-db^N)=v_p((a/b)^N-(d/c))$. If none of these numbers are positive, we are done. Else let $N_0$ be minimal such that $v_p((a/b)^{N_0}-(d/c))$,and let $d=\text{ord}_p(a/b)$. Then the $N$-th term is nonzero iff $N\equiv N_0\pmod{d}$, so if we set $q=(a/b)^d$, $r=(d/c)/(a/b)^{N_0}$, our goal is to show that if $p|q-1$, $p|r-1$, then either the numbers $v_p(q^n-r)$ ($n=0,1,\dots$) are unbounded, or they are all equal. Case 1. $v_p(q-1)>v_p(r-1)$. In this case, \[ v_p(q^n-r)=v_p((q^n-1)-(r-1))=v_p(r-1) \]for all $n\ge 1$, so the sequence $v_p(q^n-r)$ is constant. $\blacksquare$ Case 2. $v_p(q-1)\le v_p(r-1)$; we'll show that $v_p(q^n-r)$ is unbounded. To do this, let $|\cdot |$ be the $p$-adic absolute value, represent $q,r$ are $p$-adic integers, and recall the $p$-adic $\log$, defined by \[ \log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}\mp\dots \]for $|x|<1$ in $\mathbb{Q}_p$. Moreover, for odd $p$, it is clear that $|\log(1+x)|=|x|$, whence \[ \left|\frac{\log r}{\log q}\right|=\frac{|r-1|}{|q-1|}\le 1, \]which means $\nu:=\frac{\log r}{\log q}$ is not just in $\mathbb{Q}_p$, but also in $\mathbb{Z}_p$. Hence, there is a sequence of positive integers $n_1,n_2,\dots$ such that ($p$-adically) $\lim_{k\to\infty}n_k=\nu$, and thus by continuity, \[ \lim_{k\to\infty}|q^{n_k}-r|=\lim_{k\to\infty}|\exp(n_k\log q)-r|=|\exp(\nu \log q)-r|=0, \]i.e., $v_p(q^{n_k}-r)\to\infty$. $\blacksquare$

Let $x_i = ca^i - db^i$, and let $\nu_p (x) = v(x)$ for convenience. Let $N = \max_i v(x_i)$; i.e. $p^N$ is the largest power of $p$ that divides some $x_i$. (We know that $N$ exists due to the problem statement.) Let $\mathbb{Z}_k$ denote $\mathbb{Z}/k\mathbb{Z}$. For all $k \ge 1$, we will define $q_k, r_k \in \mathbb{Z}_{p^k}$ such that \[q_k \equiv \frac{a}{b} \pmod{p^k}\]and \[r_k \equiv \frac{d}{c} \pmod{p^k}.\] Now let $m,n$ be positive integers, and consider the statement $S$ that $p^n \mid x_m$. Note that $S$ is equivalent to \[ca^m \equiv db^m \pmod{p^n}.\]This can be rearranged as $\frac{a^m}{b^m} \equiv \frac{d}{c} \pmod{p^n}$, or equivalently \[ q_n^m \equiv r_n \pmod{p^n}. \]For fixed $n$, let $S_n$ be the set of $m$ that make $S$ true. Note that $S_n$ can be empty. If $S_n$ is not empty, it consists of all positive integers congruent to $j_n$ modulo $d_n$, where $d_n$ is the order of $q_n$ modulo $p^n$. Here, $j_n$ is some integer dependent on $n$. Let us characterize $d_n$. Note that $d_n$ is the smallest positive integer $k$ such that $a^k \equiv b^k \pmod{p^n}$, or equivalently \[v(a^k - b^k) \ge n.\]Let $s$ be the smallest positive integer such that $a^s \equiv b^s \pmod{p}$. It's obvious that $p$ divides $a^k - b^k$ if and only if $s$ divides $k$, so let us assume that $s$ divides $k$. Now \[v(a^k - b^k) = v((a^s)^{k/s} - (b^s)^{k/s}) = v(a^s - b^s) + v(\frac{k}{s})\]by LTE. But $s \mid p-1$, so $v(s) = 0$; thus \[v(\frac{k}{s}) = v(k).\]Let $h = v(a^s - b^s)$; now we need the smallest $k$ divisible by $s$ such that \[h + v(k) \ge n.\]We finally have our characterization for $d_n$: If $n \le h$, we have $d_n = s$. If $n \ge h$, we have $d_n = s \cdot p^{n-h}$. Consider $S_N$. We know that \[q_N^m \equiv r_N \pmod{p^N}\]for all $m \equiv j_N \pmod{d_N}$. Furthermore, $S_{N+1}$ is empty by assumption, so $q_{N+1}^m \equiv r_{N+1} \pmod{p^{N+1}}$ has no solutions in $m$. We know that $q_N \equiv q_{N+1} \pmod{p^N}$ and that $r_N \equiv r_{N+1} \pmod{p^N}$, so we know that \[ q_{N+1}^m \equiv r_{N+1} \pmod{p^N}\]holds for all $m \equiv j_N \pmod{d_N}$. Assume for the sake of contradiction that $d_{N+1} = pd_N$. Then we can write \[q_{N+1}^{j_N + kd_N} \equiv r_{N+1} + z_k \cdot p^N \pmod{p^{N+1}}\]for all $0 \le k \le p-1$, and $z_k \in \mathbb{Z}_p$. Since the order of $q_{N+1}$ modulo $p^{N+1}$ is $d_{N+1} = pd_N$, we know that as $k$ varies, $q_{N+1}^{j_N + kd_N}$ takes on $p$ distinct values modulo $p^{N+1}$. This means that $z_k$ takes on $p$ distinct values modulo $p$, so there exists $k_0$ such that $z_{k_0} \equiv 0 \pmod{p}$. Then \[q_{N+1}^{j_N + k_0d_N} \equiv r_{N+1} \pmod{p^{N+1}}, \]contradicting the fact that $S_{N+1}$ is empty. Thus we conclude that $d_{N+1} = d_N$. This means that \[d_1 = d_2 = \cdots = d_N.\]We will note that $S_1 \supseteq S_2 \supseteq \cdots \supseteq S_N$. But this implies \[ j_1 \equiv j_2 \equiv \cdots \equiv j_N \pmod{d_1} \]and so \[S_1 = S_2 = \cdots = S_N.\]We see immediately that if $i \not\in S_1$, then $v(x_i) = 0$, and if $i \in S_1$, then $v(x_i) = N$. This means that $N$ is our desired $T$ and the proof is complete. $\blacksquare$

numbertheorist17 wrote: Suppose we have distinct positive integers $a, b, c, d$, and an odd prime $p$ not dividing any of them, and an integer $M$ such that if one considers the infinite sequence \begin{align*} ca &- db \\ ca^2 &- db^2 \\ ca^3 &- db^3 \\ ca^4 &- db^4 \\ &\vdots \end{align*}and looks at the highest power of $p$ that divides each of them, these powers are not all zero, and are all at most $M$. Prove that there exists some $T$ (which may depend on $a,b,c,d,p,M$) such that whenever $p$ divides an element of this sequence, the maximum power of $p$ that divides that element is exactly $p^T$. Assume to the contrary. Pick a huge $N$ and solve $a \equiv bx \pmod{p^N}$ and $d \equiv cy \pmod{p^N}$. Now we conclude $\left(v_p(ca^n-db^n)\right)_{n \ge 1}=\left(v_p(x^n-y)\right)_{n \ge 1}$. Remark that $x,y$ are coprime to $p$. Let $d$ be the order of $x\pmod{p}$ and define $$m \overset{\text{def}}{:=} \max \left \{ v_p(x^n-y) \, n \ge 1 \right \}$$then we prove the following claim. Lemma. $v_p(x^d-1) \le m$ (Proof) Suppose $v_p(x^d-1)>m$. Pick $u$ and $v$ such that $v_p(x^u-y)=a$ and $v_p(x^v-y)=b$ for $0<a<b$. Now $v_p(x^u-x^v)=a$ hence $p \mid x^{|u-v|}-1$. Thus, $v_p(x^{|u-v|}-1) \ge v_p(x^d-1)$ as $d \mid |u-v|$. Consequently, $m \ge b>a \ge v_p(x^d-1)$ proving the claim. $\blacksquare$ Pick $n$ such that $v_p(x^n-y)=m$. By the lemma and LTE, there exists $k \equiv 0 \pmod{d}$ with $v_p(x^k-1)=m$. Observe that $$\frac{x^{jk}-1}{x^k-1}=(x^k)^{j-1}+\dots+(x^k)+1 \equiv j \pmod{p^m}$$so $x^{jk}-1=(x^k-1) \cdot (i\cdot p^m+j)$. Let $x^k-1=p^m \cdot \ell$ and $x^n-y=p^m \cdot s$. Then $x^{jk}-1=p^m\ell \cdot (ip^m+j) \equiv j\ell p^m \pmod{p^{m+1}}$. Now we observe that $$x^{n+jk}-y=(x^n-y)+x^n(x^{jk}-1) \equiv p^m (s+j(\ell x^n)) \pmod{p^{m+1}}$$thus we may pick $j$ suitably to obtain $v_p(x^{n+jk}-y) \ge m+1$; contradicting the maximality of $m$. $\blacksquare$

Sorry this looks way too simple, can someone please point out where's my mistake ?

transformation, we can WLOG assume that $c = b = 1$ and $0 < \nu_p(a^n - d) < M+1$ for all $n$ and a constant $M$. Let $g$ be a primitive root modulo $p^{M+1}$, and write $a \equiv g^r \mod p^{M+1}$, $d \equiv g^s \mod p^{M+1}$. Now note that for any $m \leq M+1$, we have $p^m | a^n - d \Leftrightarrow g^{rn} \equiv g^s \mod p^m \Leftrightarrow nr \equiv s \mod \phi(p^m) \Leftrightarrow p^{m-1}(p-1) | nr - s$ (call this $\smiley$) Note that since $p$ divides $a^n - d$ for all $n$, by $\smiley$, we have $p-1 | nr - s$ for all $n$ (call this $\spadesuit$). Also note that the congruence $nr \equiv s \mod p^m$ don't have a solution for fixed $r,s$ with $n$ varying iff $\nu_p(r) > \nu_p(s)$ AND $\nu_p(s) < m$. Since $p^{M+1}$ never divides $a^n - d$ for any $n$, using $\smiley$ we have $p^M(p-1)$ doesn't divide $nr-s$ for any $n$. Using $\spadesuit$, we see for all $n$, $nr \not \equiv s \mod p^M$. By the previous para, this implies $\nu_p(r) > \nu_p(s)$ and $\nu_p(s) < M$. Let $t = \nu_p(s)$. Now I claim that $\nu_p(a^n - d) = t+1$ for all $n$. Indeed, by $\smiley$ and using $\nu_p(a \pm b) = \min(\nu_p(a), \nu_p(b))$ when $\nu_p(a) \neq \nu_p(b)$, we have $\nu_p(nr-b) = t$ and the conclusion follows by $\smiley$ and $\spadesuit$.

Kayak wrote: Note that since $p-1$ divides $a^n - d$ for all $n$, by $\smiley$. Sorry if I am missing something But How does this follow from $\smiley$ ?

Sorry that was a typo. I mean Quote: Note that since $p$ divides $a^n - d$ for all $n$, by $\smiley$, we have $p-1 | nr - s$ for all $n$ (call this $\spadesuit$). Not Quote: Note that since $p-1$ divides $a^n - d$ for all $n$, by $\smiley$, we have $p-1 | nr - s$ for all $n$ (call this $\spadesuit$).

Lemma: Suppose $y$ and $\varepsilon$ are rational numbers that are well defined mod $p$ that satisfy $1\le\nu_p(y-1)\le\nu_p(\varepsilon-1)$. Then, $\nu_p(y^m-\varepsilon)$ can be made arbitrarily large while varying the positive integer $m$. Proof of Lemma: Let $y=1+kp^u$ and $\varepsilon=1+\ell p^v$ where $u\le v$ and $p\nmid k,\ell$. We see then that \[y-\varepsilon = kp^u-\ell p^v,\]so $\nu_p(y-\varepsilon)\ge u=\nu_p(y-1)$. We claim that if $\nu_p(y^n-\varepsilon)\ge\nu_p(y-1)$, then there exists some other positive integer $m$ such that \[\nu_p(y^m-\varepsilon)>\nu_p(y^n-\varepsilon).\]The lemma will then follow by induction. Suppose \[y^n-\varepsilon = tp^e\]for $\gcd(t,p)=1$. Recall that $y=1+kp^u$, and pick $\tau\in\mathbb{N}$ such that $\tau\equiv -t/k\pmod{p}$. Furthermore, let \[m=n+\tau p^{e-u}.\]We see that \begin{align*} y^m-\varepsilon &= (y^m-y^n)+(y^n-\varepsilon) \\ &= y^n((1+kp^u)^{\tau p^{e-u}}-1)+tp^e \\ &= y^n(k\tau p^e+(\mathrm{integer})\cdot p^{e+u})+tp^e. \end{align*}Since $u\ge 1$, we have \[y^m-\varepsilon p^e(k\tau y^n+t)+(\mathrm{integer})p^f\]for $f>e$. But $k\tau y^n+t\equiv 0\pmod{p}$ so $\nu_p(y^m-\varepsilon)\ge e+1$, so we're done. $\blacksquare$ We'll now solve the problem. Let \[f(n)=\nu_p(ca^n-db^n)=\nu_p((a/b)^n-d/c).\]Since $f(n)\ge 1$ for some $n$, pick the smallest $r\in\mathbb{N}$ such that \[(a/b)^r\equiv(d/c)\pmod{p}.\]Let $\ell$ be the order of $a/b$ mod $p$, and let $y=(a/b)^\ell$. We see that \[f(n)=\nu_p((a/b)^r y^{\frac{n-r}{\ell}}-d/c)\]if $f(n)\ge 1$. Let $m=\frac{n-r}{\ell}$, so the nonzero elements are \[g(m)=\nu_p(y^m-(d/c)(b/a)^r).\]We WTS that if $g(m)$ is bounded then it is constant. Let $\varepsilon=(d/c)(a/b)^r\equiv 1\pmod{p}$. We see that \[g(m)=\nu_p(y^m-\varepsilon).\]By the lemma, we have $u=\nu_p(y-1)>\nu_p(\varepsilon-1)=v$. Let $y=1+kp^u$ and $\varepsilon=1+\ell p^v$. We see that \[g(m)=\nu_p(1+mkp^u+(\mathrm{integer})\cdot p^{2u}-1-\ell p^v)=\nu_p(-\ell p^v+(\mathrm{integer})\cdot p^u)=v,\]so $g(m)$ constant as desired. $\blacksquare$

numbertheorist17 wrote: Suppose we have distinct positive integers $a, b, c, d$, and an odd prime $p$ not dividing any of them, and an integer $M$ such that if one considers the infinite sequence \begin{align*} ca &- db \\ ca^2 &- db^2 \\ ca^3 &- db^3 \\ ca^4 &- db^4 \\ &\vdots \end{align*}and looks at the highest power of $p$ that divides each of them, these powers are not all zero, and are all at most $M$. Prove that there exists some $T$ (which may depend on $a,b,c,d,p,M$) such that whenever $p$ divides an element of this sequence, the maximum power of $p$ that divides that element is exactly $p^T$. Sorry but I don't really understand this question. Are we asked to prove that if $p$ divides an element of this sequence, then $\nu_p$ of that element is $T$, in other words all of the elements in the sequence which is divisible by $p$ has $\nu_p$ of $T$?

@above Your interpretation is correct. Not sure if you caught it, but it is crucial that all the $\nu_p$ values are bounded above by $M$, otherwise the problem is false. Stated again, the problem tells us that the sequence of $\nu_p$ values of the numbers is bounded above, and asks us to show that in fact all nonzero terms in the sequence are the same.

Here's a solution by Aryan-23 and me: numbertheorist17 wrote: Suppose we have distinct positive integers $a, b, c, d$, and an odd prime $p$ not dividing any of them, and an integer $M$ such that if one considers the infinite sequence \begin{align*} ca &- db \\ ca^2 &- db^2 \\ ca^3 &- db^3 \\ ca^4 &- db^4 \\ &\vdots \end{align*}and looks at the highest power of $p$ that divides each of them, these powers are not all zero, and are all at most $M$. Prove that there exists some $T$ (which may depend on $a,b,c,d,p,M$) such that whenever $p$ divides an element of this sequence, the maximum power of $p$ that divides that element is exactly $p^T$. Write $A=\tfrac{a}{b}$ and $B=\tfrac{d}{c},$ so that we have $0 \le \nu_p(A^k-B) < M$ for all $k.$ Define $\nu(x):=\nu_p(A^x-B).$ Assume on the contrary $x,y$ such that $0<\nu(x)<\nu(y) \leqslant M.$ Let $\ell$ be the order of $A$ modulo $p.$ Clearly, $\ell \mid y-x.$ Now, we present a lemma: Lemma: Consider $\ell^\ast$ to be the order of $A$ modulo $p^{\nu(y)}.$ Then we can find a $k$ such that $\ell \mid k\ell^\ast,$ and for any $N,$ we have \begin{align*} \nu(y+k{\ell^\ast}) &\ne \nu(x) \\ \nu_p\left(y-x+k\ell^\ast \right) &\geqslant N. \end{align*}Proof: Since $\ell \mid p-1,$ hence $(\ell,p)=1.$ Hence the two equations \begin{align*} k &\equiv 0 \pmod{\ell} \\ y-x &\equiv -k\ell^\ast \pmod{p^N}. \end{align*}have a solution by CRT. Next, we want $\nu(y+k{\ell^\ast}) \ne \nu(x).$ Now $$A^{y+k\ell^\ast} \equiv A^y \equiv B \pmod{p^{\nu(y)}}.$$Hence, $\nu(y+k{\ell^\ast}) \geq \nu(y)>\nu(x),$ so done. $\square$ For any $k,$ since $\nu(y+k{\ell^\ast}) \ne \nu(x),$ hence (note that $p \nmid A)$ $$\nu_p(A^{y-x+k\ell^\ast}-1)=\nu_p((A^{y+k\ell^\ast}-B)-(A^x-B))=\min\{\nu(y+k{\ell^\ast}),\nu(x)\}<M.$$However, then using LTE, since $\ell \mid y-x+k\ell^\ast,$ \begin{align*} M&>\nu_p(A^{y-x+k\ell^\ast}-1) \\ &=\nu_p(A^\ell-1)+\nu_p\left( \frac{y-x+k\ell^\ast}{\ell}\right) \\ &=\left(\nu_p(A^\ell-1)-\nu_p(\ell)\right)+\nu_p\left(y-x+k\ell^\ast \right) \\ &=c+N, \end{align*}where $c=\nu_p(A^\ell-1)-\nu_p(\ell)$ is a constant. For large enough $N,$ this gives a contradiction. $\blacksquare$

Nice! Here's my solution: For brevity, let $x \equiv ab^{-1} \pmod{p^r}$ and $y \equiv cd^{-1} \pmod{p^r}$ for some $r>M^{100000}$. Let $M=\max\{\nu_p(x^k-y)\}$ as $k$ varies on the set of natural numbers, and denote $e=\text{ord}_p(x)$. FTSOC assume there exist $m,n$ with $\nu_p(x^n-y)=N$ and $\nu_p(x^m-y)=M$ such that $0<N<M$, and so that $N$ is the minimal positive power of $p$ dividing any term of the given sequence. Note that if $p \mid x^u-y$, then we have $u \equiv n \pmod{e}$. Keeping this in mind, the crucial Lemma is as follows- LEMMA If such $N,M$ exist, then $\nu_p(x^e-1)=N$. (Proof) Since $x^{n \pm e} \equiv x^n \pmod{p}$, so $p$ must also divide $x^{n \pm e}-y$. By the minimality of $N$, we have $p^N \mid x^{n \pm e}-y$, which gives $$x^{n \pm e} \equiv y \equiv x^n \pmod{p^N} \Rightarrow p^N \mid x^e-1$$Now, suppose $\nu_p(x^e-1)>N$. Then $$x^{n \pm e} \equiv x^n \not \equiv y \pmod{p^{N+1}} \Rightarrow \nu_p(x^{n \pm e}-y)=N$$But then repeating this argument, we will get that $\nu_p(x^s-y)=N$ for all $s \equiv n \pmod{e}$, which is a contradiction to the existence of $m$. Return to the problem. Let $z=p^{M-N}e$, so that our Lemma and LTE implies $\nu_p(x^z-1)=M$. Consider an integer $0<k<p$ satisfying $$k \equiv -(x^{-1})^m \left(\frac{x^m-y}{p^M} \right) \left(\frac{x^z-1}{p^M} \right)^{-1} \pmod{p} \Leftrightarrow x^m(1+k(x^z-1)) \equiv y \pmod{p^{M+1}}$$Due to our assumption of $M$ being maximal, the above choice of $k$ ensures that $$x^{gz} \not \equiv 1+k(x^z-1) \pmod{p^{m+1}} \quad \forall g \in \mathbb{N}$$This gives $$(x^z-1)(1+x^z+x^{2z}+\dots +x^{(g-1)z})=x^{gz}-1 \not \equiv k(x^z-1) \pmod{p^{M+1}}$$Since $p^M \mid x^z-1$, so the above statement is equivalent to $$k \not \equiv 1+x^z+x^{2z}+ \dots +x^{(g-1)z} \pmod{p} \Leftrightarrow k \not \equiv g \pmod{p}$$where the last equality follows from the fact that $p \mid x^z-1$. However, this is a clear contradiction, since $g$ can take any positive integral value, so we can always choose $g \equiv k \pmod{p}$. Thus, such $M,N$ cannot exist, proving the existence of a single $T$, as desired. $\blacksquare$ EDIT: $700^{\text{th}}$ post on HSO

Observe that \[v_{p}(ca^{r} - bd^{r}) = v_{p}\Big(\frac{a^{r}c - b^{r}d}{b^{r}c}\Big) = v_{p}\Big(\frac{a^{r}}{b^{r}} - \frac{d}{c}\Big) = v_{p}\left(\Big(\frac{a}{b}\Big)^{r} - \frac{d}{c}\right)\]Let $x = \frac{a}{b}, y = \frac{d}{c}$. We now have the following problem: given that $v_{p}(x^{r} - y)$ is bounded, show that if $v_{p}(x^{r} - y) \neq 0$, then $v_{p}(x^{r} - y)$ is a constant. Redefine $M = v_{p}(x^{p-1} - 1)$. First, I claim that there does not exist $t$ such that $v_{p}(x^{t} - y) > M$. This is because, assume some $t$ exists such that $v_{p}(x^{t} - y) = K > M$. Then, we have \[d = ord_{p^{M}}(x) | ord_{p^{K}}(x) = d\cdot q\]\[\Rightarrow K\leq v_{p}(x^{d\cdot q} - 1) = v_{p}(x^{d} - 1) + v_{p}(q) = M + v_{p}(q) \Rightarrow p^{K-M} | q\]Then, since the order is minimal, we have $p^{M - k} = q$. Observe that the order of $x\mod p^{k+1}$ is $d\cdot p^{K-M + 1}$. Now, consider \[x^{d\cdot p^{K-M}}, x^{2d\cdot p^{K-M}}, \ldots x^{(p-1)d\cdot p^{K-M}} \equiv 1\mod p^{K}\]The above statement is true since $x^{d\cdot p^{K-M}}\equiv 1\mod p^{K}$. Furthermore, I claim the above is a permutation of \[1, p^{k} + 1, 2\cdot p^{k} + 1, \ldots (p-1)\cdot p^{k} + 1\mod p^{k+1}\]This is because they all must be $1\mod p^{k}$, so the above are the only possibilities. In addition, if $r_{1} < r_{2}$ exist that are equivalent $\mod p^{K-M+1}$, then we get: \[x^{r_{1}d\cdot p^{K-M}} \equiv x^{r_{2}d\cdot p^{K-M}}\mod p^{K-M+1} \Rightarrow x^{d(r_{2} - r_{1})\cdot p^{K-M}}\equiv 1\mod p^{K-M+1}\]but this is a contradiction by the minimality of order. Thus, since there are $p$ possibilities, $p$ different expressions, and no two expressions can be the same, this means those expressions must be a permutation. Now, let $c$ be a number such that $x^{t}\equiv y + c\cdot p^{k}\mod p^{K + 1}$, where $p\not| c$. This exists since $v_{p}(x^{t} - y) = k$. Consider some $n$ such that $p | yn + c$, and let $l$ be a number such that \[x^{l\cdot d \cdot p^{K-M}} \equiv n\cdot p^{k} + 1\]Now, we have \[x^{t + l\cdot d\cdot p^{K-M}} \equiv x^{t}\cdot (n\cdot p^{k} + 1) \equiv (y + c\cdot p^{k})(n\cdot p^{k} + 1) \equiv (yn + c)p^{k} + y\equiv y\mod p^{k+1}\]This means there exists some $s = t + l\cdot d\cdot p^{K-M}$ such that $v_{p}(x^{s} - y) \geq k+1$. Using the same procedure, we can find a new $s$ such that $v_{p}(x^{s} - y) \geq k+2$, and so on, which means the sequence is unbounded, a contradiction. This means that for all $t$, $v_{p}(x^{t} - y)\leq M$. Now, consider the smallest $r$ such that $p | x^{r} - y$. By order, the only $s$ where $p | x^{s} - y$ is when $ord_{p}(x) | r-s$. However, since $ord_{p}(x) = ord_{p^{M}}(x)$, this also means for all $s$ such that $p | x^{s} - y$, we must have $p^{M} | x^{s} - y$. Furthermore, we've already proven that $p^{M}$ is the maximal power of $p$ that can divide it, so $v_{p}(x^{s} - y) = M$, which proves the problem.

We will consider every thing $\pmod {p^{M+1}}$.Let $r$ be a primitive root $\pmod{p^{M+1}}$.So $\frac{a}{b}=r^{m}$ and $\frac{d}{c}=r^n$ for some $m,n\le \phi(p^{M+1})=p^M(p-1)$. Claim: $gcd(p^M(p-1),m)\nmid n$. $\emph{proof.}$ FTSOC,assume $gcd(p^M(p-1),m)\mid n$.By the theory of linear congruence,there exists $x$ such that:\[\phi(p^{M+1})=p^M(p-1)\nmid mx-n\]So we have: \[r^{mx-n}\equiv 1\pmod{p^{M+1}}\implies r^{mx}\equiv r^n\pmod{p^{M+1}}\]\[\implies (\frac{a}{b})^x\equiv \frac{d}{c} \pmod{p^{M+1}}\].Contradiction.$\qquad\square$ Claim: $gcd(p-1,m)\mid n$ $\emph{Proof.}$ We know that there exists $x\in\mathbb N$ such that $$(\frac{a}{b})^x\equiv \frac{d}{c}\pmod{p}\implies r^{mx-n}\equiv 1\pmod{p}$$. Also we have,$r^{(p-1)}\equiv 1\pmod{p}$.Using GCD tricks,we have $r^{gcd(p-1,mx-n)}\equiv 1\pmod{p}$ FTSOC assume $d=gcd(p-1,mx-n)<p-1$.Then by LTE, $r^{d\times p^M}\equiv 1\pmod{p^{M+1}}$.But $d\times p^M<\phi(p^{M+1})$.Which contradicts the property of primitive root $r$. Then if $g=gcd(p-1,m)$ then $g|p-1|mx-n$ and $g|m$ So $g|n$ as well$\qquad \square$ From the 2 claim it is easy to see that $gcd(p^M,m)\nmid n$. Then we have $v_p(m)>v_p(n)$ and define $v_p(n)=T-1$.So $T-1<M$. But then \begin{align*} (\frac{a}{b})^x\equiv \frac{d}{c}\pmod{p}&\implies\\ r^{mx-n}\equiv 1\pmod{p}&\implies\\ p-1|mx-n \end{align*} Then as $p^{T-1}||mx-n$ hence again by LTE,$$p^T||r^{mx-n}-1\implies p^T||(\frac{a}{b})^x-\frac{d}{c}$$$\blacksquare$

I really hope this is true$:$ Let us begin by choosing the largest positive integer $\ell,$ for which $p^{\ell}\mid ca^t-db^t$ for some positive integer $t.$ Fix the unique integers $u, v$ in $\{1, \dots, p^{l+1}-1\},$ for which $a\equiv bu\pmod {p^{\ell +1}}$ and $d\equiv cv\pmod {p^{\ell +1}}.$ ($u, v$ are well defined and not divisible by $p,$ since $p\nmid a, b, c, d$) Claim: $v_p(u^t-v)\leq \ell$ for all $t\in\mathbb{N}$ Proof: Suppose that $p^{\ell+1}\mid u^t-v$ for some $t.$ Then multiplying the corresponding congruence by $cb^t,$ we obtain $$ca^t\equiv db^t\pmod {p^{\ell+1}},$$a clear contradiction$.$ Let $t_0$ be the minimal positive integer$,$ for which $p^{\ell}\mid u^{t_0}-v$ and write $u^{t_0}-v=p^{\ell}q,$ where $q$ is a positive integer with $\gcd(q, p)=1.$ Denote $r=\text{ord}_p(u).$ We proceed by assuming the problem statement is false and we will try to reach a contradiction by constructing a $t,$ for which $u^t-v\equiv 0\pmod {p^{\ell+1}}.$ Let $t=t_0+d$ for some positive integer $d$ to be chosen later$.$ Now $$u^t-v=u^du^{t_0}-v=v(u^d-1)+u^dqp^{\ell}\hspace{1cm} (*)$$ We want to make $v_p(u^d-1)=\ell.$ Let's first examine what happens if $v_p(u^r-1)\geq \ell+1.$ It is usually the case that $$r=\text{ord}_p(u)\mid\text{ord}_{p^{\ell+1}}(u)$$But in our case we also have that $p^{\ell+1}\mid u^r-1,$ so it must be true that $\text{ord}_{p^{\ell+1}}(u)\mid r,$ so we conclude $r=\text{ord}_{p^{\ell+1}}(u).$ Let $s$ be such that $p\mid u^s-v.$ Now $p\mid u^{t_0}-v$ as well$,$ so $$p\mid u^{t_0}(u^{s-t_0}-1)$$and therefore $r\mid s-t_0,$ but now $$u^s-v=u^{t_0}u^{s-t_0}-v\equiv u^{t_0}-v\pmod {p^{\ell+1}}$$and so $v_p(u^s-v)=\ell$ for all such $s,$ but this is a contradiction$.$ Now suppose that $f:=v_p(u^r-1)\leq \ell.$ We will chose $d=re\hspace{0.2cm}(\square),$ where $e$ is some positive integer$.$ Note that by $\text{LTE}$ $$v_p(u^d-1)=v_p((u^r)^e-1)=f+v_p(e)$$Thus we can make $v_p(e)=\ell-f\geq 0$ by simply setting $e=p^{\ell-f}k$ for some positive integer $k\not\vdots p$ to be determined later$.$ Let's see how that helps us$.$ We have (due to $(*)$)$$u^{t}-v=v(u^d-1)+u^dqp^{\ell}\equiv v(u^d-1)+qp^{\ell}\pmod {p^{\ell+1}}$$The last expression being congruent to $0 \pmod {p^{\ell+1}}$ is equivalent to $$v\cdot\frac{u^d-1}{p^{\ell}}+q\equiv 0\pmod p$$Since $p\nmid v,$ if we can find $p-1$ numbers $d_1, d_2, \dots, d_{p-1}$ which are of the type $(\square)$ and for which $\frac{u^{d_1}-1}{p^{\ell}}, \dots, \frac{u^{d_{p-1}}-1}{p^{\ell}}$ are pairwise not congruent $\pmod p$ (and non-zero $\pmod p$)$,$ we win$,$ because these numbers $\frac{u^{d_i}-1}{p^{\ell}}$ span all non-zero residues modulo $p,$ in particular $-\frac{q}{v}$ $.$ Let us see when exactly it is true that $$\frac{u^{d_1}-1}{p^{\ell}}\equiv \frac{u^{d_2}-1}{p^{\ell}}\pmod p\hspace{1cm} (\triangle)$$for some $d_1=re_1=rp^{\ell-f}k_1$ and $d_2=re_2=rp^{\ell-f}k_2.$ The congruence $(\triangle)$ is equivalent to $$u^{d_1}\equiv u^{d_2}\pmod {p^{\ell+1}}\iff u^{d_2}(u^{d_1-d_2}-1)\equiv 0\pmod {p^{\ell+1}}$$$$\iff\text{ord}_{p^{\ell+1}}(u)\mid d_1-d_2\hspace{1cm} (\bigcirc)$$ We will first prove the following crucial Lemma: $$\text{ord}_{p^{\ell+1}}(u)=p^s\text{ord}_p(u)=p^sr \hspace{0.2cm}\text{ for some non-negative integer $s$}$$ Proof: By Euler we know that $\text{ord}_{p^{\ell+1}}(u)\mid p^{\ell}(p-1),$ so let $\text{ord}_{p^{\ell+1}}(u)=p^sr_1$ for some $r_1\mid p-1.$ Next observe that $$1\equiv u^{p^sr_1}\equiv u^{r_1}\pmod p,$$so $r\mid r_1.$ Recall that $f=v_p(u^r-1)\leq \ell.$ Now by $\text{LTE},$ $$v_p\bigg(u^{rp^{\ell+1-f}}-1\bigg)=v_p(u^r-1)+\ell+1-f=\ell+1,$$so $$p^{\ell+1}\mid u^{rp^{\ell+1-f}}-1$$and therefore $$r_1\mid p^sr_1=\text{ord}_{p^{\ell+1}}(u)\mid rp^{\ell+1-f}$$and since $\gcd(r_1, p)=1,$ we conclude $r_1\mid r,$ which is enough to conclude that $r_1=r$ and we are done$.$ $\square$ Now return to $\bigcirc.$ By the lemma$,$ we have that $\triangle$ is satisfied precisely when $$p^sr=\text{ord}_{p^{\ell+1}}(u)\mid d_1-d_2=rp^{\ell-f}(k_1-k_2)$$If we can show that $$\ell-f\leq s-1,$$it would follow that $\triangle$ is false whenever $k_1\not\equiv k_2\pmod p.$ Therefore we can just choose $$d_1=re_1=rp^{\ell-f}, d_2=re_2=2rp^{\ell-f}, \cdots, d_{p-1}=re_{p-1}=(p-1)rp^{\ell-f}.$$ But note that $$p^{\ell+1}\mid u^{\text{ord}_{p^{\ell+1}}(u)}-1=u^{p^sr}-1$$and so $$\ell+1\leq v_p(u^{p^sr}-1)=v_p(u^r-1)+v_p(p^s)=f+s,$$concluding the proof$.$ $\blacksquare$

Let $B$ and $D$ be integers such that $bB\equiv 1 \equiv dD ( mod p^{M+1})$. We have that $$v_p(ca^n-db^n)=v_p(cD(aB)^n-dD(bB)^n)=v_p(cD(aB)^n - 1)$$since we have that $v_p(ca^n-db^n)\leq M$. Now take $cD=k$ and $aB=x$. We would want to prove that $v_p(kx^n-1)$ is either $0$ or a fixed positive integer $T$. Claim If $g$ is a primitive root $mod $ $p^{m+1}$ then $g$ is also a primitive root $mod$ $p^m$. Proof: Assume that $ord_{p^{m}}(g)=t<p^{m-1}(p-1)$. Then by LTE $v_p(g^{pt}-1)=v_p(g^t-1)+1\geq m+1$, thus $ord_{p^{m+1}}(g)\leq pt < p^m(p+1)$, a contradiction. Now let $k\equiv g^a(mod p^{M+1})$ and $x\equiv g^b(mod p^{M+1})$ ,where $g$ is a quadratic residue $mod p^{M+1}$ ( Sry for repeating letters). From the claim if $g$ is a quadratic residue $mod$ $p^{M+1}$ then $g$ is a quadratic residue $mod $ $p^s$ for $s\leq M$. Now $kx^n=g^a(g^b)^n=g^{a+bn}$. We can't have $g^{a+bn}\equiv 1(mod p^{s})$ for $s\geq M+1$, so it is impossible to have $p^s(p-1)|a+bn$ for some $s>M$. Note that we also have $p|kx^n-1$ for some $n$ which implies that $(p-1)|a+bn$ has a solution$\Rightarrow $ $p^s(p-1)|a+bn$ for some $s>M$ is impossible only when $v_p(a)<v_p(b)$. Now note that we always have that if $(p-1)$ doesn't divide $a+nb$ then $v_p(kx^n-1)=0$, and if it does , then $$v_p(kx^n-1)=v_p(g^{a+bn}-1)=v_p(a+bn)+1=v_p(a)+1$$and thus it is fixed, and we're finally done!

First 25 mohs problem hopefully this isn't a fakesolve... Let $k\ge 1$ be the maximum $v_p$ of the terms of the sequence. Choose integers $r\equiv \frac{a}{b}\pmod{p^{k+1}}$ and $s\equiv \frac{d}{c}\pmod{p^{k+1}}$, we have $ca^n - db^n\equiv r^n - s\pmod{p^{k+1}}$. Now let $n$ be the least positive integer such that $r^n\equiv s\pmod{p^k}$ and let $t = \text{ord}_{p^k}(r)$. Claim: If $r^m\equiv s\pmod{p^k}$ then $m = n + dt$ for some nonnegative integer $d$. Proof: first $n\le t$, since otherwise $r^{n-t}\equiv s\pmod{p^k}$ would contradict the minimality of $n$. Now suppose some $m=n' + d't$ with $1\le n'\le t$. Then $r^{|n-n'|}\equiv 1\pmod{p}$ gives $n=n'$ by the minimality of $t$, as desired. Now we know there exists no solution to the congruence $r^{n + dt}\equiv s\pmod{p^{k+1}}$. Let $r^n\equiv x\cdot p^k + s\pmod{p^{k+1}}$ and $r^t\equiv y\cdot p^k + 1\pmod{p^{k+1}}$. We have $r^{n + dt}\equiv r^n\cdot (r^t)^d\equiv (x\cdot p^k + s)(y\cdot p^k + 1)^d\equiv p^k(x + sdy) + s\pmod{p^{k+1}}$ so if $y\not\equiv 0\pmod{p}$ we can choose $d\equiv -\frac{x}{sy}\pmod{p}$ and arrive at a contradiction. Hence $y\equiv 0\pmod{p}$, so $r^t\equiv 1\pmod{p^{k+1}}$. Now let $t'$ be the order of $r$ mod $p$. Claim: $r^{t'}\equiv 1\pmod{p^{k+1}}$. Proof: This clearly holds if $k=1$ since then $t=t'$. Then assume the the contrary and assume $k\ge 2$. Let $e = v_p(r^{t'} - 1)$. It is well known that if $p | r^m - 1$ then $t' | m$. Now by LTE we have $v_p(r^{\ell t'} - 1) = e + v_p(\ell)$. In particular, it is easy to check that $\text{ord}_{p^k}(r) = p^{k-e}t' < p^{k-e+1}t' = \text{ord}_{p^{k+1}}(r)$ which contradicts $r^t\equiv 1\pmod{p^{k+1}}$ and the claim is proved. Finally, let $n'$ be the least positive integer such that $r^{n'}\equiv s\pmod{p}$ and let $r^{n'}\equiv z\pmod{p^{k+1}}$. By the same argument used for the first claim, if $r^m\equiv s\pmod{p}$ then $m = n' + dt'$ for some nonnegative integer $d$. Then for any $d$ $r^{n' + dt'}\equiv z(r^{t'})^d\equiv z\pmod{p^{k+1}}$ and since $n$ exists such that $r^n\equiv s\pmod{p^k}$, we must have $z\equiv s\pmod{p^k}$ and we are done.

Let $x = \frac{a}{b}, y = \frac{c}{d}$. We want to show if the sequence $\{v_p(x^k - y)\}_{k \ge 1}$ is bounded, then has only one positive value. Let $t = \text{ord}_p(x)$ and $e = v_p(x^t - 1)$. Claim: All $v_p(x^k - y)$ must be $<e$ (otherwise sequence would be unbounded). Proof: Pick a $k$ such that $v_p(x^k - y)$ is maximized. Says its $e_0$. FTSOC $e_0 \ge e$. Let $f = t \cdot p^{e_0 - e}$. Consider the numbers $x^k, x^{k+f},x^{k + 2f}, \ldots,x^{k + f(p-1)}$. By LTE they are all distinct modulo $p^{e_0+1}$ while all are same modulo $p^{e_0}$ (i.e. all are $y$ mod $p^{e_0}$). This means for some $k_0$ we have $x^{k_0} \equiv y \pmod{3^{e_0} + 1}$, contradiction. $\square$ Now pick $k_1,k_2$ such that $v_p(x^{k_1} - y) = e_1 \ge 1$ and $v_p(x^{k_2} - y) = e_2 \ge 1$. We have shown $e_1,e_2 < e$, and we want to show $e_1 = e_2$. FTSOC $e_1 < e_2$. Then $v_p(x^{k_1} - x_{k_2}) = e_1$. Then $v_p(x^{k_1 - k_2} - 1) = e_1$. As $e_1 \ge 1$ so this forces $t \mid k_1 - k_2$, implying $v_p(x^{k_1 - k_2} - 1) \ge e$. So $e_1 \ge e$, contradiction. $\blacksquare$

Redefine $M$ to be the largest positive integer such that $p^M$ divides some element of the sequence. Let $i$ be the least positive integer such that $p^M \mid a^i-b^i$, and let $n$ be the least $n$ with that we have $p^M \mid ca^n-db^n$. Then any $m$ with $p^M \mid ca^m-db^m$ is congruent to $n \pmod{i}$. For some $m=n+ki$, we have $$ca^m-db^m \equiv 0 \pmod{p^{M+1}} \iff (a/b)^m \equiv d/b \pmod{p^{M+1}} \iff (a/b)^{ki} \equiv sp^M+1 \pmod{p^{M+1}}\qquad (\heartsuit)$$for some $s$, with the last step from $(a/b)^n \equiv c/d \pmod{p^M}$. Since $(a/b)^i \equiv 1 \pmod{p^M}$, we can write $(a/b)^i \equiv rp^M+1 \pmod{p^{M+1}}$, so $$(a/b)^{ki} \equiv (rp^M+1)^k \equiv krp^M+1 \pmod{p^{M+1}},$$and thus $(\heartsuit)$ is true if and only if $$krp^M+1 \equiv sp^M+1 \pmod{p^{M+1}} \iff kr \equiv s \pmod{p},$$which there exists a solution $k=s/r$ for unless $p \mid r$, in which case we have $p^{M+1} \mid a^i-b^i$; that is, we do not have any $i$ with $\nu_p(a^i-b^i)=M$. Now let $j$ be the least positive integer with $p \mid a^j-b^j$, and let $N=\nu_p(a^j-b^j)$. By Exponent Lifting, if $M\geq N$, we have $$\nu_p\left(a^{jp^{M-N}}-b^{jp^{M-N}}\right)=\nu_p(a^j-b^j)+M-N=M,$$which contradicts our previous contradiction, hence $N>M$. Then for any $m$ (different from the previous definition) such that $p \mid ca^m-db^m$, we must actually have $m=n+kj$, in which case $$ca^m \equiv db^m \pmod{p^M} \iff (a/b)^{n+kj} \equiv d/c \pmod{p^M} \iff (a/b)^{kj} \equiv 1 \pmod{p^M},$$which is true as $p^M \mid p^N\mid a^{kj}-b^{kj}$. Combined with the fact that we cannot have $p^{M+1} \mid ca^m-db^m$, it follows that $\nu_p(ca^m-db^m)=M$ for all $m$ with $p \mid ca^m-db^m$, so we can take $T=M$ to finish. $\blacksquare$

Wow. It's very nice example of $\nu_{p}$ Since $\frac{a}{b} \not\equiv 0 (p)$, so the indices of the terms divisible by $p$ is an arithmetic sequence. Let $s, s + T, s + 2T, \dots$ be a indices of the terms divisible by $p$. Then $\nu_{p}(ca^{s + Tn} - db^{s + Tn}) = \nu_{p}((\frac{a^T}{b^T})^n - \frac{b^s d}{a^s c})$. Let $x = \frac{a^T}{b^T}$ and $y = \frac{b^s d}{a^s c}$. Then the problem showing that if $\nu_{p}(x^n - y) > 0$, then $\nu_{p}(x^n - y) = \alpha$ for some $\alpha > 0$. Let $\alpha$ be maximum value of the sequence $(\nu_{p}(x^n - y))_{n \ge 1}$. Assume the contrary, there exists some $0 < \beta < \alpha$ and $m > 0$ such that $\nu_{p}(x^m - y) = \beta$. Let $n$ is positive integer such that $\nu_{p}(x^n - y)$. Then $\nu_{p}(x^n - x^m) = \nu_{p}(x^n - y - (x^m - y)) = \beta$, so $\nu_{p}(x^n - x^m) = \beta$. Thus LTE yields $\nu_{p}(x - 1) + \nu_{p}(n - m) = \beta$, i.e $\nu_{p}(x - 1) \le \beta < \alpha$. Thus by LTE, we have $\nu_{p}(x^{p^{\alpha - \nu_{p}(x - 1)}} - 1) = \alpha$, therefore $\gcd(p, \frac{x^{p^{\alpha - \nu_{p}(x - 1)}} - 1}{p^{\alpha}}) = 1$. Let $k = p^{\alpha - \nu_{p}(x - 1)}$. Then $\frac{x^{kn} - 1}{p^{\alpha}} = \frac{x^k - 1}{p^{\alpha}} \cdot ((x^k)^{n - 1} + (x^k)^{n - 2} + \dots + 1) \equiv \frac{x^k - 1}{p^{\alpha}}n (p)$, so $\frac{x^{kn} - 1}{p^{\alpha}}$ can take arbitrary value modulo $p$. So we can take $t$ such that $x^k - x^n \equiv y - x^n (p^{\alpha + 1})$. This contradicts the sequence $(\nu_{p}(x^n - y))_{n \ge 1}$ is bounded. $\blacksquare$

I have no idea WHY I chose the symbols I did for solving this problem, but it was on the paper that I used to solve it, so... enjoy :3 We define $\nu_p$ for rationals easily. Observe that \[ \nu_p(ca^n - db^n) = \nu_p \left( \left( \frac{a}{b} \right)^n - \frac{d}{c} \right). \]Call $ \frac ab = Q$ and $\frac dc = P$. Also remark that the LTE lemma also holds for rational numbers $Q$, so long as $\nu_p(Q) = 0$. Now we start cooking. We will show the contrapositive. Suppose there exist two positive integers $\sigma$ and $\delta$ such that \[ \nu_p (Q^\sigma - P) = \alpha \qquad \nu_p(Q^\delta - P) = \beta, \qquad \beta > \alpha. \]Observe then that \[ \nu_p(Q^\delta - Q^\sigma) = \alpha, \]or more particularly there exists some minimal value $c$, with $c \ge 1$ such that \[ \nu_p(Q^c - 1) = \partial \ge 1. \]Note that $\partial \le \alpha$ as well. This will be important. By mod reasons if we ever have $\nu_p(Q^a - 1) \ge 1$, then $c \mid a$. We will let $\varepsilon= Q^c$. Thus by LTE, \[ \nu_p(\varepsilon^k - 1) = \nu_p(\varepsilon- 1) + \nu_p(k) = \partial + \nu_p(k). \] $\textbf{Claim.}$For every integer $l \ge \partial$ and for each $\psi$ from $1$ to $p-1$, we have that there exists an integer $\omega$ such that \[ Q^\omega - 1 \equiv \psi p^l \pmod{p^{l+1}}. \] $\textit{Proof.}$ Consider the expression \[ \nu_p(\varepsilon^{sp^{l - \partial}} - 1) = \partial + l - \partial = l. \]Here $s$ is an integer from $1$ to $p-1$. Now if \[ \varepsilon^{sp^{l - \partial}} - 1 \equiv \varepsilon^{s'p^{l-\partial}} - 1\pmod{p^{l+1}} \]for some $s' > s$ from $0$ to $1$, we simply remark that $\nu_p(\varepsilon) = 0$, move to one side, and find that \[ l+1 = \nu_p(\varepsilon^{(s'-s)p^{l - \partial}}) = \partial + l - \partial = l, \]contradiction. Thus we attain all such possible things. \endproof $\textbf{Claim.}$ Suppose that there is an integer $k$ such that $\nu_p(Q^k - P) = T \ge \partial$. Then there exists an integer $k'$ such that $\nu_p(Q^{k'} - P) \ge T+1$. $\textit{Proof.}$ Let $Q^k - P \equiv \varphi p^T \pmod{p^{T+1}}$. From the previous lemma we choose some integer $m$ such that \[ Q^m - Q^k \equiv -\varphi p^T \pmod{p^{T+1}}. \]Therefore, \[ Q^m - P \equiv 0 \pmod{p^{T+1}}. \square \]Therefore since $\alpha \ge \partial$, $\nu_p(Q^n - P)$ is unbounded.

Uhh I hope this is right I guess??? Since the $\nu_p$ are always at most $M$, then we can set $a$ to be $a/b$ mod $p^{M+1434}$ and set $d$ to be $d/c$ mod $p^{M+1434}$, and set $b=c=1$. Now rename $d$ to $r$. Now we have the sequence $x_k=a^k-r$ for positive integers $k$ such that $\nu_p(x_k)$ is bounded(since it couldn't pass $M$, it was not good mod $p^{M+1}$, so taking mod $p^{M+1434}$ won't change it being not good mod $p^{M+1}$). Since $p$ is supposed to divide some of these, it must divide one of the first $p-1$ things. Therefore, there is some power of $a$, say $a^t$ that will be able to get $r/a^t$ to be $1$ mod $p$. From here, we have a lemma: \lemma{1} Suppose that $x\equiv 1\pmod{p^k}$ but $x\not\equiv 1\pmod{p^{k+1}}$. Then $x,x^2,x^3,\dots,x^p$ cycles over all residues mod $p^{k+1}$ that are $1$ mod $p^k$. \proof{} Let $x=sp^k+1$. Then the binomial theorem immediately finishes. Therefore, if the first power of $a$ that is $1$ mod $p$ is not $1$ mod $p^2$, we can carry $r/a^t$ to be $1$ mod $p^2$ for some $t$. Then if the first power of $a$ that is $1$ mod $p^2$ is not $1$ mod $p^3$, we can carry it to be $1$ mod $p^3$ for some $t$. This goes on forever, so the only way to stop it is if the first power of $a$ that is $1$ mod $p^k$ is also $1$ mod $p^{k+1}$, and also at that point $r/a^t$ was $1$ mod $p^k$ but not $1$ mod $p^{k+1}$. This immediately stops the progress. \lemma{2} If $x\equiv 1\pmod{p^k}$ and $x\not\equiv 1\pmod{p^{k+1}}$, then $x^p\not\equiv 1\pmod{p^{k+2}}$. \proof{} Let $x=zp^{k+2}+sp^{k+1}+tp^k+1$. Then raising this to the $p$ power would cause everything involving $z$ to cancel. Also, we can never actually use $s$ since we would need $1$ of it and $p-1$ of $1$, but then it gets multiplied by $p$ in the multinomial theorem. Even with $tp^k$, you can only get it once at $tp^{k+1}$, so \[x^p\equiv tp^{k+1}+1\not\equiv 1\pmod{p^{k+2}},\]as desired. \textbf{Corollary 1.} Suppose that $x\equiv 1\pmod{p^t}$ and $x\not\equiv 1\pmod{p^{t+1}}$. Then the powers of $x$ reach every residue mod $p^{k+1}$ that is $1$ mod $p^k$, where $k$ is any positive integer greater than or equal to $t$. Therefore, basically, if powers of $a$ can ``get started" into any $1$ mod $p$ territory without ``leaving $d$ behind" in $\nu_p$ when subtracted by $1$, then the powers of $a$ will never ``speed ahead", causing the $\nu_p$ to be unbounded. Therefore, the only way that the $\nu_p$ is bounded is if the powers of $a$ cannot get started without leaving $d$ behind, but then, we can't get started anyways so $d$ is stuck with the power of $a$ you get before $a$ gets started. $\blacksquare$

Recall $v_p(\frac{x}{y})=v_p(x)-v_p(y)$ Now let $x=\frac{d}{c}$, $y=\frac{a}{b}$, $m_i=y^i-x$ Redefine $M$ to be $max(v_p(m_i))$ Let $a>b$ be natural numbers such that $v_p(m_a)=v_p(m_b)=M$, $d=a-b$. Then $m \leq v_p(m_a-m_b)=v_p(y^d-1)$. But then $v_p(m_{a+d}-m_a)=v_p(y^d-1) \geq m$, so $m_{a+d}$ is also divisible by $p^M$. Similarly, by induction, we get that $v_p(m_{a+kd})=M$ for every non-negative integer k. But then $v_p(m_{a+kd}-m_a)=v_p(y^{kd}-1)=v_p(y^d-1)+v_p(k)$. If $v_p(y^d-1)=M$, then for every $k=1,\ldots, p-1$ we have $v_p(m_{a+kd})=M$ and $v_p(m_{a+kd}-m_a)=M$, which is obviously impossible modulo $p^{M+1}$. So $v_p(y^d-1) \geq M+1$ Now let $n=ord_p(y)$ and $r=v_p(y^n-1)$. Suppose $r \leq M$. Then $v_p(y^{np^{M-r}}-1)=M$. Now let $f=a+np^{M-r}$. Obviously, $v_p(m_f)=M$, but $v_p(y^{f-a}-1)=M$, which is a contradiction with statements we proved above. So $r \geq M+1$. Now, to finish the solution, consider any natural number $V$. We need to prove, that $v_p(m_V)$ is either $0$, or $M$. If it is $0$, we are done. Now suppose it is positive. Then $1 \leq v_p(m_V-m_a)=v_p(y^{|V-a|}-1)$, so $V-a \vdots n$, from which $v_p(m_V-m_a) \geq r \geq M+1$, so $v_p(m_V)=v_p(m_a)=M$.

Let $g(n) = ca^n - db^n$ over positive integers $n$. We may assume that $p^M$ is the largest power of $p$ dividing a value of $g$ (where $M > 0$). It suffices to prove that $\nu_p(g(n)) \in \{0, M\}$ for all positive integers $n$. Notice that $g(n)$ is never $0\pmod{p^{M+1}}$ and at some point $0\pmod{p^M}$. Let $r$ be the remainder when $\frac ab$ is divided by $p^{M+1}$ and $s$ be the remainder when $\frac cd$ is divided by $p^{M+1}$. We see that $g(n) \equiv db^n(s r^n - 1)\pmod{p^{M+1}}$. Now let $f(n) = s r^n - 1$. We see that $\nu_p(db^n) = 0$ and $p^{M+1}$ doesn't divide $g(n)$, so $\nu_p(f(n)) = \nu_p(g(n))$ for all positive integers $n$. Let $D$ be the order of $r$ modulo $p$. We have that $p\mid f(n) \iff r^n \equiv \frac 1s \pmod p$. Let $t$ be some positive integer satisfying $\nu_p(f(t)) = M$. We see that $p\mid f(t)$, so $p\mid f(n) \iff n \equiv t\pmod D$. Claim: For any $m$ with $\nu_p(m-1) = M$, $m^d$ can take any residue modulo $p^{M+1}$ that is $1\pmod{p^M}$ as $d$ varies over positive integers. Proof: Let $m = k \cdot p^M + 1$. The only terms in the binomial expansion of $m^d$ that aren't divisible by $p^{M + 1}$ are $d\cdot k \cdot p^M $ and $1$. Hence $m^d \equiv d\cdot k \cdot p^M + 1 \pmod{p^{M + 1}}$. Since $k$ is relatively prime to $p$, $dk$ can take any residue modulo $p$, so $dk p^M$ can take any residue modulo $p^{M+1}$ that is divisible by $p^M$. $\square$ Claim: $r^D \equiv 1\pmod{p^{M+1}}$.

Now choose $k$ such that $(r^x)^k \equiv \frac{1}{s r^t} \pmod{p^{M+1}}$ (this is possible as $\frac{1}{s r^t} \equiv \frac{1}{1} = 1 \pmod{p^M}$. We have $f(t + xk) = s r^{t + xk} - 1 \equiv s r^t \cdot (r^x)^k - 1 \equiv 0\pmod{p^{M+1}}$, which is a contradiction to the fact that $\nu_p(f(n)) \le M$. $\square$ For any $n$ with $p\mid f(n)$, there exists an integer $c$ with $t + cD = n$, so $f(n) = s r^{t + cD } \equiv s r^t \pmod{p^{M+1}}$, and thus $\nu_p(f(n)) = M$. Hence $\nu_p(f(n)) \in \{0,M\}$ for all positive integers $n$, so $\nu_p(g(n)) \in \{0,M\}$ also.

Of course, since none of them are divisible by $p$, divide by $cdb^n$ to change the expression to $$(\frac{a}{b})^n-\frac{d}{c}$$which has the same $v_p$. We will show that, if $$v_p((\frac{a}{b})^n-\frac{d}{c})=k_1, v_p((\frac{a}{b})^m-\frac{d}{c})=k_2$$for $k_1>k_2>0$, then the $v_p$ of the sequence is unbounded, which would solve the problem. If we subtract these, we get that $$v_p((\frac{a}{b})^n-(\frac{a}{b})^m)=k_2,$$which of course reduces to $$v_p((\frac{a}{b})^{n-m}-1)=k_2.$$ We will use this fact to generate unbounded $v_p$. Let $T_h=(n-m)\cdot p^h$. By lifting the exponent, $$v_p((a/b)^{T_h}-1)=k_2+h.$$ We show that, for $k\geq k_2$, if there exists $n$ such that $$v_p((a/b)^n-d/c)=k,$$then there exists $N$ such that $$v_p((a/b)^N-d/c)\geq k+1,$$which of course suffices as this shows unbounded $v_p$ exist. Suppose that $$(\frac ab)^n\equiv \frac dc +rp^k\pmod{p^{k+1}}.$$Then, from the LTE conclusion for $h=k-k_2$, we have $$(a/b)^{T_{k-k_2}}\equiv 1\pmod{p^{k}}.$$Now, suppose $$(a/b)^{T_{k-k_2}}\equiv 1+sp^{k}\pmod{p^{k+1}}.$$Note that $r$ and $s$ are both nonzero. Then, $$(a/b)^{n+uT_{k-k_2}}=(\frac dc +rp^k)(1+usp^k)\equiv \frac dc + p^k(r+\frac dc us)\pmod{p^{k+1}}.$$Since we can pick $u$ such that $r+\frac dc us\equiv 0\pmod{p}$, since all variables are nonzero mod $p$, we can make the RHS equivalent to $\frac dc$, and thus we are done by picking $N=n+uT_{k-k_2}$. remark the "main idea" here is that, really one of the most directly ways to interpret a condition of the form "$v_p$'s differ" is by adding or subtracting here. Here, this is done to eliminate $d/c$. Then, the remaining part of the problem is really quite straightforward, using the new condition combined with LTE to increase the $v_p$ by 1.

Notice that all terms of this sequence which are divisible by $p$, if any, are of the form $x+y\ell$, where $\ell$ is $\operatorname{ord}_p(ab^{-1})$. We can then rewrite our terms as \[\left(\left(\frac ab\right)^{\ell}\right)^y - \frac{db^x}{ca^x} = \alpha^y - \beta,\] where $\alpha, \beta \equiv 1 \pmod p$. To finish, it suffices to show that if $u$, $v$ exist such that \[e_1 = v_p(\alpha^u-\beta) < v_p(\alpha^v-\beta) = e_2,\] then we can find a term $\alpha^{w(u-v)}-\beta$ divisible by $p^{e_2+1}$. We build $w$ into two steps, so suppose $w = w_1w_2$. Notice LTE allows us to choose $w_1 = p^{e_2-e_1}$ to give \begin{align*} e_1 + v_p(w) &= v_p((\alpha^u-\beta) - (\alpha^v-\beta)) + v_p(w) \\ &= v_p(\alpha^{u-v}-1) + v_p(w) \\ &= v_p(\alpha-1) + v_p(u-v) + v_p(w) \\ &= v_p(\alpha^{w_1(u-v)}-1) \\ &= e_2. \end{align*} We now construct $w_2$. Notice that we want \[\alpha^{w_1w_2(u-v)+v}-\beta = (p^{e_2}f+1)^{w_2} (p^{e_2}g+\beta) - \beta\] to be divisible by $p^{e_1+1}$, where $p \nmid g, h \in \mathbb Z$. Expanding, we get that this expression is \[p^{e_2}(w_2f\beta+g) \pmod{p^e} \implies w_2 \equiv -g(f\beta)^{-1} \pmod p\] gives a desired $w_2$. Hence we can always build a term with a larger $v_p$ given two terms with distinct $v_p$ values, making our $v_p$ unbounded, contradiction. $\blacksquare$

First, it is clear that \[ p \mid ca^{e} - db^{e} \iff \left( \frac{a}{b} \right)^e \equiv \frac{d}{c} \pmod p \]Now, let $e = \text{ord}_p\left( \frac ab \right)$. It follows that for all nonnegative integers $k$ and some positive integer $s$, \[ \left( \frac{a}{b} \right)^{s+ke} \equiv \frac{d}{c} \pmod p \]and by the minimality of orders, $\left( \frac ab \right)^{f} \equiv \frac{d}{c} \pmod p$ only if $f$ is of the form $s + ke$. So we are given that for all nonnegative integers $k$, \[ \nu_p \left( ca^{s+ke} - db^{s+ke} \right) \le M \]\[ \nu_p \left( \frac{a^{ke}}{b^{ke}} - \frac{db^s}{ca^s} \right) \le M \]Now, for brevity, write $x = \frac{a^e}{b^e}$ and $y = \frac{db^s}{ca^s}$, noting that both quantities are congruent to $1$ modulo $p$. Claim: Suppose there exists $k_1$ and $k_2$ with $\nu_p(x^{k_1} - y) = n$ and $\nu_p (x^{k_2} - y) = m$, where $m > n$. Then we can find an exponent $\ell$ such that $\nu_p (x^{\ell} - y) \ge m+1$. Proof. From the given condition, it follows that \[ n = \nu_p (x^{k_2} - x^{k_1}) = \nu_p (x^{k_2 - k_1} - 1) \]therefore, it follows by LTE (since $x \equiv y \equiv 1 \pmod p$), that \[ m = \nu_p \left( x^{(k_2-k_1)p^{m-n}} - 1 \right) \]Let $\left| (k_2 - k_1)p^{m-n} \right| = g$, so $m = \nu_p (x^g - 1)$. Now, consider \begin{align*} \nu_p &\left( x^{k_2} - y \right) \\ \nu_p &\left( x^{k_2 + g} - y \right) \\ \nu_p &\left( x^{k_2 + 2g} - y \right) \\ &\vdots \\ \nu_p &\left( x^{k_2 + (p-1)g} - y \right) \\ \end{align*}clearly, since $p^m \mid x^{k_2} - y$ and $p^m \mid x^g - 1$, all of these values are at least $m$. I claim that one of these values is at least $m+1$. It suffices to verify that the inner terms are all distinct resdiues modulo $p^{m+1}$. Assume not, so for $0\le i \neq j \le p-1$, then \begin{align*} x^{k_2 + ig} &\equiv x^{k_2 + jg} \pmod {p^{m+1}} \\ x^{g(i-j)} &\equiv 1 \pmod {p^{m+1}} \\ m+1 &\le \nu_p \left( x^{g(i-j)} - 1 \right) = \nu_p \left( x^g - 1 \right) + \nu_p (i-j) = m \end{align*}where we used LTE. This gives a contradiction. Thus, we have found a value $\ell$ for which $\nu_p (x^{\ell} - y) \ge m+1$. $\square$ Assume FTSOC that the sequence \[ \nu_p (x^0 - y), \nu_p (x - y), \nu_p(x^2 - y), \dots \]is nonconstant. Then consider the maximal and minimal value of the sequence, and apply the claim above which generates a new maximal value. Then we find that the sequence is unbounded, contradiction. Hence, the sequence must be constant, as needed. $\blacksquare$