The largest prime factor always increases. It suffices to show that the largest prime factor of $2^n-1$ with odd exponent is greater than the largest prime factor of $n$. Let $p$ be the largest prime factor dividing $n$, and let $q$ be some prime dividing $2^p-1$ such that $\nu_q(2^p-1)$ is odd, which exists since $2^p-1\equiv3\pmod 4$ is not a perfect square. Then, the order of $2$ modulo $q$ is $p$, so $p\mid q-1$ implies $q>p$. By Lifting the Exponent, \begin{align*} \nu_q\left(2^n-1\right)&=\nu_q\left((2^p)^{\frac np}-1\right)\\ &=\nu_q(2^p-1)+\nu_q\left(\frac np\right)\\ &=\nu_q(2^p-1). \end{align*}Since this is odd, this means that if $n$ is one term, then $q$ must divide the next term, so the largest prime factor is strictly increasing. Therefore, no two terms can be equal.

Does this work?? Define $g(n)$ as the squarefree part of $n$. Choose a prime $p_1\mid g(a_1)$. If this doesn't exist, shift the sequence by one term, noting that $2^\bullet-1$ can't be a square. Then let $p_{k+1}$ be the smallest prime factor of $g(2^{p_k}-1)$. It is well known that $p_k<p_{k+1}$. Now we claim that if $p_i\mid a_j$, then there is some $k>0$ such that $p_{i+k}\mid a_{j+1}$. Indeed, note that \[\nu_{p_{i+1}}(2^a-1)=\nu_{p_{i+1}}(2^{p_i}-1)+\nu_{p_{i+1}}(a).\]So if $p_{i+1}\nmid a$, then we're good. Otherwise, repeat the same argument but with $p_{i+2}$ and $p_{i+1}$. Eventually this has to work. So the largest $i$ such that $p_i\mid a_n$ is strictly increasing. $\blacksquare$

I believe the following is true: if $a>1$ and $k>0$ are integers such that $k(2^a-1)$ is a perfect square, then $k>a$. I will post my proof of this later.

Let $s(n)$ denote the squarefree part of $n$. We claim that the largest prime factor of $s(a_n)$ always increases. Suppose $a_n=r^2p_1p_2\cdots p_k$, with $p_1<p_2<\dots<p_k$. Consider a prime $q\mid 2^{p_k}-1$. Due to orders, $q>p_k$ since the order of $2$ mod $q$ must be $p_k$. Furthermore, $2^{p_k}-1$ is not a perfect square, so it has some prime divisor $q>p_k$ with odd exponent. However, $$v_q(2^{r^2p_1p_2\cdots p_k}-1)=v_q((2^{p_k})^{r^2p_1\dots p_{k-1}}-1^{r^2p_1\dots p_{k-1}})$$$$=v_q(2^{p_k}-1)+v_q(p_1\dots p_{k-1}r^2)=v_q(2^{p_k}-1)+2v_q(r).$$ Thus if $v_q(2^{p_k}-1)$ is odd, $v_q(2^{a_n}-1)$ is also odd, so $a_{n+1}$ has an odd exponent of $q$. This means that the largest prime with an odd exponent strictly increases and we are done.

We claim that the largest prime factor of $a_n$ increases. Lemma. If $p$ is prime, there exists $q > p$ such that \[ \nu_q(2^p -1) \equiv 1 \pmod{2}. \] Proof. Observe a $q$ exists with the second condition as $2^p - 1$ is 3 mod 4. Hence the order of 2 mod $q$ is $p$, hence $p \mid q-1$ and $q > p$. Now let $M$ be the largest prime divisor of $a_n$. We see there exists a primitive $q > M$ dividing $2^M-1$, and hence \[ \nu_q(2^{a_n}-1) = \nu_q(2^M - 1) + \nu_q(a_n/M). \]As $q > M$, the second term is 0, hence the total is odd, and $q \mid a_{n+1}.$ Notably this implies our claim, so no such sequence exists.

a_507_bc wrote: Consider an infinite sequence of positive integers $a_1, a_2, a_3, \dots$ such that $a_1 > 1$ and $(2^{a_n} - 1)a_{n+1}$ is a square for all positive integers $n$. Is it possible for two terms of such a sequence to be equal? What an exceptionally silly problem! The idea of looking at maximal/minimal prime divisors is very well-known. Additionally, this IMOC 2024 problem is pretty much identical. How is this problem 2 at the RMM? What is the committee doing?

oVlad wrote: a_507_bc wrote: Consider an infinite sequence of positive integers $a_1, a_2, a_3, \dots$ such that $a_1 > 1$ and $(2^{a_n} - 1)a_{n+1}$ is a square for all positive integers $n$. Is it possible for two terms of such a sequence to be equal? What an exceptionally silly problem! The idea of looking at maximal/minimal prime divisors is very well-known. Additionally, this IMOC 2024 problem is pretty much identical. How is this problem 2 at the RMM? What is the committee doing? Last year, RMM 2024/4 was pretty much the same as Zhautykov 2021/1, I doubt it was intentional. Accidents happen, and as long as no leader has noticed, nothing can be done. I hope no contestant has had strong advantage. Though to be fair, solving the problem for $k=1$ makes you try to type "$n(2^n-1)$" in the AoPS search bar to see what context is known, and you indeed get the IMOC problem

CyclicISLscelesTrapezoid wrote: I believe the following is true: if $a>1$ and $k>0$ are integers such that $k(2^a-1)$ is a perfect square, then $k>a$. I will post my proof of this later. First, we establish two lemmas. Lemma 1 (special case of Mihăilescu): For an integer $a \ge 2$, $2^a-1$ is not a perfect square $2^a+1$ is a perfect square if and only if $a=3$. Proof: Since $2^a-1 \equiv 3 \pmod{4}$, it is not a perfect square. If $2^a+1=x^2$ for a positive integer $x$, then $2^a=(x-1)(x+1)$, which means $x-1$ and $x+1$ are both powers of $2$. However, at least one of the terms $x-1$ and $x+1$ is not divisible by $4$, so that term must be either $2$ or $1$. A quick inspection gives $x=3$ and $a=3$ as the only solution. $\square$ Lemma 2: Let $a \ge 1$ be an integer. If a prime $p$ divides $2^a-1$, then $p$ is larger than the smallest prime factor of $a$. Proof: Let $r=\operatorname{ord}_p(2)$. Then $r \mid p-1$ by Fermat and $r \mid a$, so $r \mid \gcd(p-1,a)$. If $p$ is at most the smallest prime factor of $a$, then all prime factors of $p-1$ are less than $p$ and all prime factors of $a$ are at least $p$, so $\gcd(p-1,a)=1$, a contradiction. $\square$ Assume WLOG that $k$ is square-free. Then, $2^a-1=kx^2$ for some positive integer $x$. We are ready to tackle the main problem by induction on the number of prime factors of $a$, with multiplicity. We have two base cases. $a$ is prime. Then, $2^a-1$ is not a perfect square by Lemma 1, so it must have a prime factor $p$ of odd multiplicity, ergo, $p \mid k$. By Lemma 2, we have $p>a$, so $k>a$, as desired. $a=6$. Then, $k=7>a$, as desired. We now resolve the inductive step. Let $p$ be the smallest prime divisor of $a$, and let $q=\tfrac{a}{p}$. Since $a$ has at least $2$ prime factors with multiplicity, we have $q \ge p$. Let \begin{align*} A &= 2^q-1 \\ B &= 2^{(p-1)q}+2^{(p-2)q}+\dots+2^q+1. \end{align*}We have $AB=2^a-1=kx^2$ and \begin{align*} \gcd(A,B) \mid B-(2^{(p-1)q}-1)-(2^{(p-2)q}-1)-\dots-(2^q-1)=p. \end{align*}Lemma 2 implies that $p \nmid A$, so $\gcd(A,B)=1$. Choose positive integers $k'$ and $l$ such that $k'l=k$, $k' \mid A$, and $l \mid B$. Then, $\tfrac{A}{k'}$ and $\tfrac{B}{l}$ are relatively prime and multiply to a perfect square, so we can write $A=k'x'^2$ and $B=ly^2$ for positive integers $x'$ and $y$. If $p=2$, then $B=2^q+1$ is not a perfect square by Lemma 2 unless $q=3$, but this gives $a=6$, which we have already resolved. Otherwise, $k'>q$ by the inductive hypothesis and $l \ge 3$, so $k=k'l>3q>a$, as desired. The rest of the solution is dedicated to resolving $p \ge 3$. Claim: $B$ is not a perfect square. Proof: The main idea is to bound $B$ between consecutive squares. For nonnegative integers $i$, let \begin{align*} c_i &= (-1)^i\binom{-1/2}{i} \\ &= (-1)^i\frac{(-\frac{1}{2})(-\frac{3}{2}) \cdots (-\frac{2i-1}{2})}{i!} \\ &= \frac{(2i)!}{i!(2i)!!} \\ &= \frac{\binom{2i}{i}}{4^i}. \end{align*}Consider the formal power series \[P(X)=(1-X)^{-1/2}=c_0X^0+c_1X^1+\cdots\]with the property that \[P(X)^2=(1-X)^{-1}=X^0+X^1+\cdots.\] Let $m=\tfrac{p-1}{2}$, and define \[f(x)=c_0x^m+c_1x^{m-1}+\dots+c_{m-1}x.\]Since \[(f(x)+c_m+c_{m+1}x^{-1}+\cdots)^2=x^{2m}+x^{2m-1}+\cdots,\]we have $f(2^q)^2<2^{2mq}+\dots+2^q+1=B$. We also have \[(f(2^q)+1)^2>2^{2mq}+2^{(2m-1)q}+\dots+x^{(m+1)q}+2x^{mq}>B.\] Finally, we check that $f(2^q)$ is an integer. We have $\nu_2(c_i) \ge -2i \ge -2m+1>-p$ for $i \le m-1$. Furthermore, each $c_i$ is multiplied by a non-unity power of $2^q$. We have shown that $q \ge p$, so $\nu_2(2^q)>p$. Thus, each term in the expansion of $f(2^q)$ is an integer, so $f(2^q)$ is an integer. We have bounded $B$ between two consecutive perfect squares, so it is not a perfect square, as desired. $\square$ Since $B \mid 2^a-1$, all prime divisors of $B$ are greater than $p$ by Lemma 2. Thus, we have $l>p$. We also have $k'>q$ by the inductive hypothesis, so $k=k'l>pq=a$. This completes the inductive step, so we are done. $\blacksquare$

Same idea as here. Let $p$ be the largest prime factor of $a_n$. Clearly $2^p-1\mid 2^{a_n}-1$ and for any $q\mid 2^p-1$ the order of $q$ modulo $2$ is $p$, hence $q>p$. Now from modulo $4$ reasons, $2^p-1$ can't be a perfect square and therefore there exists a prime $q\mid 2^p-1$ such that $\nu_q(2^p-1)$ is odd. In particular \[\nu_q(2^{a_n}-1)=\nu_q(2^p-1)+\nu_q(a_n/p)=\nu_q(2^p-1)\]Hence $q\mid a_{n+1}$ so $a_{n+1}$ has the largest prime factor greater than $p$, which suffices.

The answer is $\boxed{\text{No}}.$ Let $$\text{rad}(n) = \min_{a\in \mathbb{Z}, \sqrt{an}\in \mathbb{Z}}(a).$$ Claim:$\text{rad}(a_{n+1}) = \text{rad}(2^{a_{n}}-1).$ Proof: By the definition of the $\text{rad}$ function. Claim:$\text{rad}(2^{a}-1) > a.$ Proof: By Zsigmondy, we have that $2^{a}-1$ must have a prime factor greater than $a,$ since if all of the prime factors of $2^{a}-1$ are at most $a,$ then since $p \mid 2^{p-1}-1,$ we get a contradiction. Now, assume that these prime factors $p > a_{n}$ have $2 \mid \nu_{p}(2^{a}-1).$ Then, if $$x = \sqrt{\frac{2^{a}-1}{\text{rad}(2^{a}-1)}},$$we know that $a \cdot p$ will have the same $x$ by LTE, a contradiction. Thus, we have proven the claim. Now, $\text{rad}(a_{n+1}) = \text{rad}(2^{a_{n}}-1)>a \ge \text{rad}(a_{n}),$ and thus $\text{rad}(a_{n})$ can never be the same, and so $a_{m} \neq a_{n}$ for all $m,n,$ as desired.

oVlad wrote: a_507_bc wrote: Consider an infinite sequence of positive integers $a_1, a_2, a_3, \dots$ such that $a_1 > 1$ and $(2^{a_n} - 1)a_{n+1}$ is a square for all positive integers $n$. Is it possible for two terms of such a sequence to be equal? What an exceptionally silly problem! The idea of looking at maximal/minimal prime divisors is very well-known. Additionally, this IMOC 2024 problem is pretty much identical. How is this problem 2 at the RMM? What is the committee doing? It's 2018 IMO P3 again

Consider $q$ the largest prime divisor of $a_k$. Note that $2^q - 1 \mid 2^{a_k} - 1$. Consider a prime divisor $r$ of $2^q - 1$. We have that $\text{ord}_r(2) \mid q \implies \text{ord}_r(2) = q$. Note that $2^{r - 1} \equiv 1\mod{q} \implies q \mid r - 1$, so $q < r$. Hence, all prime divisors of $2^q - 1$ are greater than $r$. Since $2^q - 1 \equiv 3\mod{4}$, there exists a prime divisor $p$ of $2^q - 1$ such that $v_p(2^q - 1)$ is odd. Observe that $v_p(2^{a_k} - 1) = v_p(2^q - 1) + v_p(a_k)$. Note that $p \nmid a_k$ since $q$ is the largest prime divisor of $a_k$. So, $v_p(2^{a_k} - 1)$ is odd. This means that $p \mid a_{k + 1}$. Hence the largest prime divisor of each term increases. Suppose, FTSOC, there exist $x, y$ (where WLOG $x < y$) such that $(2^{a_x} - 1)a_{x + 1} = (2^{a_y} - 1)a_{y + 1}$. Consider $m$ the largest prime divisor of $a_y$. Note $2^m - 1 \mid 2^{a_y} - 1$, and all prime divisors of $2^m - 1$ are greater than $m$. Since $x + 1 \leq y$, $\gcd(2^m - 1, a_{x + 1}) = 1$. So, $2^m - 1 \mid 2^{a_x} - 1 \implies m \mid a_x$, a clear contradiction