Note that $x_k-1=\frac{m-n}{n+k}$ must be an integer for all $1\leq k\leq n+1$, so $m-n$ is divisible by $N=\text{lcm}(n+1,n+2,\ldots,2n+1)$. Let $m-n=Ni$. We have that $P=x_1x_2\ldots x_{n+1}-1=\left(\frac{N}{n+1}\cdot i+1\right)\left(\frac{N}{n+2}\cdot i+1\right)\ldots\left(\frac{N}{2n+1}\cdot i+1\right)$. Suppose that $P$ is not divisible by any odd prime, so it is $2^k$ for some nonnegative integer $k$. Then, note that there is exactly one power of 2 among $n+1,n+2,\ldots,2n+1$, say $2^j$. Since $2^j\mid\mid N$, we have that $\frac{N}{a}$ is odd for $a=2^j$ and even for all other $n+1\leq a\leq 2n+1$. Let $v_2(i)=z$, and note that we clearly have $i<\frac{N}{2n+1}\cdot i+1<2^k$, so $z<k$. But then we have $\frac{N}{a}\cdot i\equiv 2^z\pmod{2^{z+1}}$ for $a=2^j$ and $\frac{N}{a}\cdot i\equiv 0\pmod{2^{z+1}}$ for all other $n+1\leq a\leq 2n+1$. Hence, $0\equiv P\equiv\Pi\left(\frac{N}{a}\cdot i+1\right)-1\equiv 1\cdot1\cdot\ldots\cdot1\cdot(2^i+1)-1\equiv 2^i\pmod{2^{z+1}}$, a contradiction.

Since all elements of the sequence are integers we must have $m \equiv n \pmod{n+k}; \forall k=1,2,...,n+1$. Let $S = \{n+1,n+2,...,2n+1\}$ and $LCM = lcm[S]$ then by Chinese Remainder Theorem we have that $m = t \cdot LCM + n$. Now it's fairly easy to prove that there is exactly one element in $S$ equal to $2^s$, s.t none of the elements is divisible by $2^{s+1}$. Now let's consider the two cases: t is odd Then we have that $v_2(LCM) = s \implies v_2(t\cdot LCM) = s$. Now as mentioned there exist some number in $S$ equal to $2^s$, so we have: $$x_k = \frac{m+k}{n+k} = \frac{t \cdot LCM + n + k}{n+k} = \frac{t \cdot LCM}{2^s} + 1$$ But now $\frac{t \cdot LCM}{2^s}$ is odd, hence $x_k$ is even and therefore $x_1x_2...x_{n+1} - 1$ is odd and have a odd prime divisor. t is even Let $v_2(t) = a$ and let $n+k \in S$ s.t. $n+k \not = 2^s$. Then $x_k = \frac{t \cdot LCM}{n+k} + 1$ and $v_2\left(\frac{t \cdot LCM}{n+k}\right) = v_2(t \cdot LCM) - v_2(n+k) \ge a+1 \implies x_k \equiv 1 \pmod{2^{a+1}}$. Now when $n+k = 2^s$ we have that $x_k = \frac{t \cdot LCM}{2^s} + 1$ and $v_2\left(\frac{t \cdot LCM}{2^s}\right) = v_2(t \cdot LCM) - v_2(2^s) \ge a \implies x_k \equiv 2^a \pmod{2^{a+1}} \implies x_k \not \equiv 1 \pmod{2^{a+1}}$ Obviously we have $x_k>2^a$ for all $k$. and we have that $2^{a+1}$ doesn't divide $x_1x_2...x_k - 1$, so as $x_1x_2...x_k-1>2^{a}$ it means that the product can't be a power of $2$, hence there is a odd prime divisor.

First we investigate when does the sequence $x_1, x_2, \cdots x_{n+1}$ are all integers. If we let $y_k=x_k-1=\frac{m-n}{n+k}$, then $x_k\in \mathbb{Z} \iff y_k\in \mathbb{Z}\iff n+k\mid m-n$ for all $k=1,2,\cdots n+1$. The last condition is clearly true if and only if $\text{lcm}(n+1,n+2,\cdots,2n+1)\mid m-n$. So for convience denote $A=\text{lcm}(n+1,n+2, \cdots, 2n+1)$, and we may write $m-n=A\ell$ for some $\ell\in \mathbb{N}$. From now onwards, we will only consider the expression $(y_1+1)(y_2+1)\cdots (y_{n+1}+1)-1=x_1x_2\cdots x_{n+1}-1$. We will prove that it is never a power of two, which will solve the problem. Observe that the numbers $\frac{y_1}{\ell},\frac{y_2}{\ell},\cdots \frac{y_{n+1}}{\ell}$ has greatest common divisor $1$, otherwise suppose $d>1$ is a common divisor for $\frac{y_k}{\ell}=\frac{A}{n+k}$ for all $k=1,2,\cdots, n+1$, then this implies $\frac{A}{d}<A$ is a common multiple of $n+1, n+2, \cdots 2n+1$, contradiction. Now let us consider $2$ cases. Case 1: There exists an odd prime $p$ that divides $\ell$, or $\ell=1$. If we have $\ell=1$, then since $y_1, y_2\cdots, y_{n+1}$ has no common divisor, it is impossible for all of them to be even, so there is at least one odd number among them, say $y_j$. So $(y_j+1)$ is even, implying $(y_1+1)\cdots(y_j+1)\cdots(y_{n+1}+1)-1$ is odd, thus not a power of two. Suppose now $\ell>1$, then let $p$ is an odd prime that divides $\ell$, then $p\mid \ell\mid y_i \Rightarrow y_i+1\equiv 1 \pmod p$ for all $i=1, 2, \cdots n+1$. So $(y_1+1)(y_2+1)\cdots(y_{n+1}+1)-1\equiv 1-1\equiv 0\pmod p$. So $(y_1+1)(y_2+1)\cdots(y_{n+1}+1)-1$ has an odd prime factor $p$, thus not a power of two. Case 2: $\ell=2^{\alpha}$ is a power of two. Consider $2^{\beta}$ to be the largest power of two that is no greater than $2n+1$. If $2^{\beta}\le n$, then $2^{\beta+1}<2n+1$, so $2^{\beta+1}$ is a larger power of two no greater than $2n+1$, contradiction. So $2^{\beta}\ge n+1$. So suppose that $n+j=2^{\beta}$ for some $j=1, 2, \cdots, n+1$. Then if $v_2(A)>\beta$, then by the maximality of $\beta$, $$v_2(\frac{y_k}{\ell})=v_2(\frac{A}{n+k})\ge v_2(\frac{A}{2^{\beta}}) \ge 1$$for all $k=1, 2, \cdots n+1$, so each of are even, a contradiction since these numbers has greatest common divisor equal to one. So $v_2(A)\le \beta$. But we also have $v_2(A)\ge v_2(n+j)=\beta$, we must have $v_2(A)=\beta$. So $v_2(\frac{y_j}{\ell})=0$, and thus we have $v_2(y_j)=v_2(\frac{y_j}{\ell})+v_2(\ell)=\alpha$. We claim that $v_2(y_i)>\alpha$ for all $i\neq j$, and $i=1,2,\cdots ,n+1$. Suppose such $i\neq j$ exists, then $v_2(\frac{y_i}{\ell})=0$, so $\frac{A}{n+i}$ is odd, implying $v_2(n+i)=\beta$. So $n+i$ is an odd multiple of $2^{\beta}=n+j$, so $2n+1\ge n+i\ge 3\cdot 2^{\beta}= 3(n+j)>2(n+1)$, a contradiction. So we have $2^{\alpha+1}\mid y_i$ for all $i\neq j$, and $i=1, 2, \cdots, n+1$, and $2^{\alpha} \mid\mid y_j$. So take modulo $2^{\alpha+1}$, we obtain $(y_1+1)\cdots (y_j+1)\cdots(y_{n+1}+1)-1\equiv (y_j+1)-1\equiv 2^{\alpha}\pmod {2^{\alpha+1}}$. So if $(y_1+1)(y_2+1)\cdots(y_{n+1}+1)+1$ is a power of $2$, then it must be equal to $2^{\alpha}=\ell$, but then we obtain $(y_1+1)(y_2+1)\cdots(y_{n+1}+1)-1>(y_1+1)>y_1=\frac{A}{n+1}\cdot \ell>\ell$, a contradiction. So in either case, $(y_1+1)(y_2+1)\cdots (y_{n+1}+1)-1=x_1x_2\cdots x_{n+1}-1$ can never be a power of two, and we are done.

Since $n+k|m+k$ for all $k = 1, 2, \ldots, n+1,$ we have $m = n \pmod{n+k}.$ Define $M = [n+1,n+2,\ldots,2n+1]$, and by CRT, $m = n \pmod{M}$, so we can write $m = n + qM$ for some $q$. Thus, $x_k$ is: $$x_k = \frac{m+k}{n+k} = \frac{n+qM+k}{n+k} = \frac{qM}{n+k} + 1$$ and the product $x_1x_2\cdots x_{n+1} - 1$ is: $$\left( \frac{qM}{n+1} + 1 \right) \left( \frac{qM}{n+2} + 1 \right) \cdots \left( \frac{qM}{2n+1} + 1 \right) - 1$$ and then setting $A \subset \{1, 2, \ldots, n+1\},$ we see that this is equal to: $$q\sum^{n+1}_{|A| = 1} \prod_{B \in A} \frac{M}{n+B}$$ and then we note that the summation of the product is odd because it's a summation of all even terms except for one term.

Let $m=n+h$, from $n+k \mid m+k$ we follow $n+k \mid h$ for all $k=1,2, \cdots , n+1$. Hence, $h=\text{lcm}(n+1, \cdots ,2n+1) \cdot K$ so $m=n+ \text{lcm} (n+1, \cdots, 2n+1) \cdot K$ for some $K \in \mathbb{N}$. Note that for $n \ge 5$ then from $n+1$ to $2n+1$, there are at least $2$ numbers that are both divisible by $3$, which means $3 \mid \frac{\text{lcm}(n+1, \cdots , 2n+1) \cdot K}{n+i}$ for all $1 \le i \le n+1$. This follows $$3 \mid \prod_{i=1}^{n+1} \left(1+ \frac{\text{lcm}(n+1, \cdots , (2n+1) \cdot K}{n+i} \right)-1= x_1x_2 \cdots x_{n+1}-1.$$Hence, it suffices to check for the case $1 \le n \le 4$. For $n=4$ then we also have two numbers among $[n+1,2n+1]$ that are both divisible by $3$. For $n=3$ then $2 \nmid \frac{\text{lcm}(4,5, \cdots ,7)}{4}$, which means $x_1x_2 \cdots x_{n+1}=K \cdot (2l+1)$ for $l \ge 1$. The case $n=1,2$ is similar to $n=3$. Thus, $x_1x_2 \cdots x_{n+1}-1$ is divisible by an odd prime, as desired.

cjquines0 wrote: $$q\sum^{n+1}_{|A| = 1} \prod_{B \in A} \frac{M}{n+B}$$ and then we note that the summation of the product is odd because it's a summation of all even terms except for one term. I don't think this is true, because $\frac{M}{n+B}$ is always even for large enough $n$ (all $n \ge 5$ for example).

shinichiman wrote: I don't think this is true, because $\frac{M}{n+B}$ is always even for large enough $n$ (all $n \ge 5$ for example). You’re right. I have to think about this more, thank you! On a side note, the above solution is an abbreviated one of one I submitted to one of our TSTs. At the time, I couldn’t see any errors, but looking back on my solution from two months ago, I think I addressed the issue of $\frac{M}{n+B}$ being always even… but I can’t remember exactly. I will look into this, thank you.

As $n+k|m-n$ for any $1\le k\le n+1$, we infer $m=n+\beta [n+1,n+2,...,2n+1]=n+t$. Suppose $x_1...x_{n+1}-1=2^{\alpha}$. This rewrites as $$\left ( \dfrac{t}{n+1}+1 \right ).... \left ( \dfrac{t}{2n+1}+1 \right )=2^{\alpha}+1\ \ (*)$$As $[n+1,...,2n+1]$ is divisible by $n+k$ for any $1\le k\le n+1$, we infer that the LHS is congruent with $1$ modulo $\beta$, so $b$ divides $2^{\alpha}$. Let $b=2^a$. Let $n+i$ be the number that maximizes $v_2(n+k)$ for $1\le k\le n+1$. Note that its $v_2$ is unique and $v_2(n+k)<v_2(n+i)$ for any $k\ne i$. Therefore, $\dfrac{t}{n+k}+1\equiv 1(\mathrm{mod}\ 2^{a+1})$ for any $k\ne i$ and $v_2 \left ( \dfrac{t}{n+i} \right )=2^a$. This implies that the LHS of $(*)$ is $2^as+1$ with odd $s$, so $a=\alpha$. But then any paranthesis from the LHS of $(*)$ is greater than $2^{\alpha}+1$, a contradiction.

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So there is a unique power of $2$ in $[n+1,2n+1]$, so let that be $n+t$. Clearly $m-n=u\text{lcm}(n+1,n+2, \cdots 2n+1)=uM$. Assume the contrary and let $x_1x_2\cdots x_{n+1}-1=2^N$. Now, plugging everything in, we have $\prod_{k=1}^{n+1} (1+\frac{uM}{n+k}) = 2^N+1$. Denote $u=2^lu'$, where $(u',2)=1$. Let us look at $\frac{uM}{n+k}$ now. If $k=t$, we have $1+\frac{uM}{n+k} \equiv 1+2^l \cdot (\text{odd number}) \equiv 1 \pmod{2^l}$. Of course, in this case, $1+\frac{uM}{n+k} \not\equiv 1 \pmod{2^{l+1}}$ In every other case, $1+\frac{uM}{n+k} \equiv 1 + 2^l \cdot (\text{even number}) \equiv 1 \pmod{2^{l+1}}$. Therefore, the L.H.S is $1$ in mod $2^l$, but not $1$ in mod $2^{l+1}$. This forces that $N=l$, but clearly $1+\frac{uM}{n+k} \ge 1+u \ge 1+2^N$, so as $n \ge 1$, we have a contradiction.

Shoutout to the first ever N3 that I have solved. Clearly if $m$ did equal $n$ then all $x_i$'s would be integers. But since $m > n$ we need $m = n + z$ where $z$ is divisible by all of $n + 1, n + 2, ..., 2n + 1$. Indeed we have $z = kl$, where $k$ is a positive integer and $$l = \text{lcm}(n + 1, n + 2, ... , 2n + 1).$$Call $a$ the quantity $x_1x_2 \cdots x_{n + 1} - 1$. We have $a = \prod_{j=1}^{n+1} \left(\frac{kl + n + j}{n + j}\right) - 1.$ Expanding out the product gives $a = \sum_{j = 1}^{n + 1}a_j(kl)^j$ where $a_j$ is the sum of products of distinct integers in $\{n + 1, n + 2, ..., 2n + 1\}$ (call this set $A$), taken $n + 1 - j$ at a time, divided by $(n + 1)\cdots (2n + 1)$. Our ultimate goal is to show that this sum is not a power of $2$ since this immediately implies that $a$ would be divisible by an odd prime. Factor out $k$ from the sum. We now have $a = k(a_1l + a_2kl^2 + ... + a_{n+1}(k^nl^{n+1}))$. All we need to do is verify that the right factor is odd to solve the problem. To this end, we show that $a_jl^j$ is indeed an integer for all $j$, even for all $j > 1$, and odd for $j = 1$. We use the fact that $a_j$ equals the sum of products of distinct members of the set $A$ divided by $(n + 1)\cdots (2n + 1)$, with each product consisting of $n + 1 - j$ members. Call $t$ the maximum power of a prime $p$ that divides some element of $A$. Call $u_{p, j}$ the smallest power of $p$ that we need to multiply a summand of $a_j$ by to form an integer. The maximum of $u_{p, j}$ across all summands is $jt$. But since the highest power of $p$ that divides $l$ is $t$, we have that $p^{jt}$ divides $l^j$, so $a_jl^j$ must be an integer for all $j$.

Since the next higher power of $2$ is greater than $2n + 1$ then we must have that the maximum of all $u_{2, j}$ must equal $t$ if $j = 1$ and must be less than $jt$ if $j \ge 2$. So each summand of $a_jl^j$ is even if $j \ge 2$ while only one summand of $a_1l$ is odd, namely the one with the product of members of $A$ excluding $2^t$. Thus $a_1l$ is odd and we are done.

$\dfrac{m+k}{n+k}=\dfrac{m-n}{n+k}+1\in \mathbb{Z} \implies \text{lcm}(n+1, \ldots , 2n+1) | m-n\implies m = n+qL$ where $q\in\mathbb{Z}^+$ and $L=\text{lcm}(n+1, \ldots , 2n+1)$. Let $a_i=\dfrac{L}{n+i}$ and $s_k = \displaystyle\sum_{\text{sym}}\prod_{i=1}^ka_i$. Assuming the contrary, we have $$2^y+1=\prod_{i=1}^{n+1}x_i = \prod_{i=1}^{n+1}\left(1+\dfrac{qL}{n+1}\right) = \prod_{i=1}^{n+1}(1+qa_i)=1+ \sum_{i=1}^{n+1}q^is_i$$which means $q=2^z$. Let $\ell$ satisfy $v_2(n+\ell) = \max\limits_{1\le i\le n=1}\{v_2(n+j)\}$; it is clear there does not exist $ \ell'$ s.t. $v_2(n+\ell ') = v_2(n+\ell)$ or else $\exists \ell''$ between $\ell$ and $\ell'$ s.t $v_2\left(n+\ell''\right) > v_2(n+\ell)$ contradiction. Thus, $v_2(L)=v_2(n+\ell)\implies a_i$ odd iff $i=\ell$. So $s_1$ is odd. But $2^{y-z} = s_1+\sum_{i=1}^nq^is_{i+1}\implies q=1$ else the RHS is odd, contradiction. But then $1+qa_{\ell}$ is even, so $\prod\limits_{i=1}^{n+1}\left(1+qa_i\right)=2^y+1$ is even, contradiction. Thus, our assumption is wrong and $x_1\cdots x_{n+1}-1$ has an odd prime factor.

This is also India TST 2016 Day 2 Problem 2.

I guess I really shouldn't be scared to try anything above N/G/A/C 1. This wasn't that bad! And I kind of like it. Also see my blog for some kind of quickly written motivation. As $x_1,x_2,\dots,x_{n+1}$ are all integers, we get \begin{align*} m &\equiv n \pmod{n+1} \\ m &\equiv n \pmod{n+2} \\ &\ \vdots \\ m &\equiv n \pmod{2n+1} \end{align*}Then $m \equiv n \pmod{ \text{lcm}(n+1,n+2,\dots,2n+1)}$. Thus, we let \[ m = \text{lcm}(n+1,n+2,\dots,2n+1) \cdot e +n. \]Since $m>n$, we get $e>0$. Using that, we arrive at \[ x_1x_2\dots x_{n+1}-1 = \left(\prod_{k=1}^{n+1} \frac{\text{lcm}(n+1,n+2,\dots,2n+1) \cdot e}{n+k}+1 \right) -1 \]Now it is easy to show that there is exactly one power of two in the interval $[n+1,2n+1]$. Hence, let $n+i = 2^{\ell}$ with $1 \leq i \leq n+1$. Then \[ v_2 \left(\frac{\text{lcm}(n+1,n+2,\dots,2n+1) }{n+i} \right)< v_2 \left(\frac{\text{lcm}(n+1,n+2,\dots,2n+1) }{n+j} \right) \]for any $j \neq i, 1 \leq j \leq n+1$. But then \[ \frac{\text{lcm}(n+1,n+2,\dots,2n+1) }{n+j} \cdot e +1 \equiv 1 \pmod {2^{v_2 \left(\frac{\text{lcm}(n+1,n+2,\dots,2n+1) }{n+i} \cdot e \right)+1}} \]for all $j \neq i$. Assuming $x_1x_2 \dots x_{n+1}-1$ was a power of two after all for the sake of contradiction, we arrive at \[ \frac{\text{lcm}(n+1,n+2,\dots,2n+1) }{n+i} \cdot e+1 \equiv 1 \pmod {2^{v_2 \left(\frac{\text{lcm}(n+1,n+2,\dots,2n+1) }{n+i} \cdot e \right)+1}}. \]That's cleary a contradiction for $e>0$. So we are done. $\hfill \square$

Assume for the sake of contradiction $x_1x_2 \cdots x_{n+1}=2^k+1$. Note $x_i -1 = \frac{m-n}{n+i}$. Let $m-n=2^a b$ where $b$ is odd. Here is the key idea: Amongst the numbers $\{n+1, n+2,\dots ,2n+1\}$ there exists exactly one power of $2$ say $2^t=n+j$. Then $v_2( x_j -1)=a$. and $\forall i \neq j \ \ v_2(x_i-1)>a$. Examining every term in the expansion $\prod_{i=1}^{n+1} \bigg( \frac{m-n}{n+i}+1 \bigg)$ we infer, $$x_1x_2 \dots x_{n+1} \not\equiv 0 \mod 2^{a+1}$$And $$x_1x_2 \dots x_{n+1} \equiv 0 \mod 2^{a}$$So, $a=k$. But clearly $x_1x_2\dots x_{n+1} \geq 2^{a+1}$, since for any $i \neq j, \ \ 2^{a+1} \mid x_i-1 $. Contradiction. We are done.

Note that $x_i$ being an integer implies that $n+i|m-n$, so we can write $m=Lk+n$, where $L=\text{lcm}(n+1,n+2,\ldots,2n+1)$. We wish to show that $$\prod_{i=1}^{n+1}\left(k\frac{L}{n+i}+1\right)-1$$is not a power of 2. First, if $k$ is odd, then we can choose $i$ such that $v_2(n+i)$ is maximal. Then, $k\frac{L}{n+i}$ is odd, so we have an even term in the product. Therefore, this expression evaluates to an odd number (of course, larger than 1), and we win. On the other hand, if $k$ is even, then we expand the entire product, and regroup as a polynomial in $k$. If we look at the linear coefficient, it is $\sum_{i=1}^{n+1}\frac{L}{n+i}$, which is odd, since exactly one summand is an odd number. So, the linear term as a whole has $v_2$ equal to $v_2(k)$. Now, the rest of the terms are divisible by at least $k^2$, so the $v_2$ of the entire expression is just $v_2(k)$. However, this product is much larger than $k$, so it can't be a power of 2.

probably posted before Let $l=\text{lcm}(n+1,n+2,\ldots, 2n+1)$. Then we have that $n+k\mid m+k$, which implies that $n+k\mid m-n$ for all $k=1,2,...,n+1$. Therefore, we can let $m=n+cl$ for some positive integer $c$. Suppose for the sake of contradiction that $$-1+\prod_{k=1}^{n+1}\left(1+\frac{cl}{n+k}\right)=2^x$$for a nonnegative integer $x$. Note that this can always be factored into $$c\left(\,\text{some terms divisible by }c+\sum_{k=1}^{n+1}\frac{l}{n+k}\right).$$ Therefore, both terms must be a power of 2. However, there is exactly one integer $y$ such that $n+1\le 2^y\le 2n+1$, so $\nu_2(l)=y$. Since this is a unique maximum for $\nu_2(n+k)$, there is only one odd term in $\sum_{k=1}^{n+1}\frac{l}{n+k}$, so it must be odd. This is a contradiction since $c$ is even, so the factor inside the parentheses must be odd.

Let $m-n=x$; we have that for $1\le i\le n+1$, $x_i=\frac{x}{n+i}+1$ is an integer, meaning $lcm(n+1, ..., 2n+1)|x$. Now let $S=x_1...x_{n+1}-1$; note we can rewrite $S$ as: $S=(1+\frac{x}{n+1})(1+\frac{x}{n+2})...(1+\frac{x}{2n+1})-1=\sum \frac{x}{n+i_1}+\sum_{i_1<i_2} \frac{x^2}{(n+i_1)(n+i_2)}+...+\sum_{i_1<...<i_{n+1}} \frac{x^{n+1}}{(n+i_1)...(n+i_{n+1})}$, where for each sum, $1\le i_j\le n$ for all $j$. Now note that since $2n+1<2(n+1)$, there exists a power of 2 in $[n+1, 2n+1]$, which we will call $z=2^j$, and also let $\nu_2(x)=y$. We see that out of $\nu_2(\frac{x}{n+1}), ..., \nu_2(\frac{x}{2n+1})$, $\nu_2(\frac{x}{z})$ is the unique minimum; thus, $\nu_2(\sum \frac{x}{n+i_1})=\nu_2(\frac{x}{z})$. But now note for $k>1$, $\nu_2(\frac{x^{k}}{(n+i_1)...(n+i_{k})})>\nu_2(\frac{x}{z})$ because $i_1, ..., i_k$ are distinct and thus $n+i_1, ..., n+i_k$ can't all be $z$. Hence, we have $\nu_2(S)=j$. So suppose FTSOC that $S=2^j$; then, we have $S\le x$ since $\nu_2(x)\ge j$, meaning $\frac{x^{n+1}}{(n+1)....(2n+1)}\le x$. This means that $(n+1)...(2n+1)\ge x^n\ge lcm(n+1, ..., 2n+1)^n$, which is clearly absurd. Hence, we've arrived at a contradiction, and we are done.

FTSOC, lets assume that, $\displaystyle \prod_{k=1}^{n+1} x_k = 2^a+1$ for some $a\in \mathbb N$. Here, $x_k$ is a integer therefore, $n+k\mid m+k \implies n+k \mid m-n$. Doing this for all $k=1,2,\ldots,n+1$ gives, $\operatorname{lcm}(n+1, n+2, \dots, 2n+1)\mid m-n$. Let, $\operatorname{lcm}(n+1, n+2, \dots, 2n+1) = d$ then, $m = xd + n$ for some $x\in \mathbb N$. Now, $x_k = \dfrac{m+k}{n+k} = \dfrac{xd+n+k}{n+k} = \left(x\cdot\dfrac{d}{n+k} + 1\right)$. Therefore, $\displaystyle 2^a+1 = \prod_{k=1}^{n+1} x_k = \prod_{k=1}^{n+1} \left(x\cdot\dfrac{d}{n+k} + 1\right) \qquad(\spadesuit)$ Let, $b = v_2(x)$ then, $v_2\left(x\cdot\dfrac{d}{n+k}\right) \geq b$. Now, we claim that there is exactly one $k$ for which $v_2\left(\dfrac{d}{n+k}\right) = 0$. Clearly there will be at least one because $d$ is the lcm so, $\displaystyle v_2(d) = \max_{i=1}^{n+1}(v_2(x_i))$. Now, if there are two $k_1$, $k_2$ then we take $n+ \frac{k_1+k_2}2$ and it will have $v_2$ more than $n+k_1$ and $n+k_2$ giving us a contradiction with the lcm. Now, from $(\spadesuit)$ clearly, $a > b$ because there are at least two elements in the product. Now, by our claim all except exactly one $\left(x\cdot\dfrac{d}{n+k} + 1\right)$ are $1$ in modulo $2^{b+1}$. and that one is $2^b+1 \pmod{2^{b+1}}$ So, \begin{align*} 1 \equiv 2^a+1 &\equiv \prod_{k=1}^{n+1} x_k \pmod {2^{b+1}+1}\\ & \equiv \prod_{k=1}^{n+1} \left(x\cdot\dfrac{d}{n+k} + 1\right) \pmod {2^{b+1}+1}\\ & \equiv 2^b+1 \pmod {2^{b+1}+1} \end{align*}Which is clearly a contradiction. $\blacksquare$

Suppose that $n + k \mid m + k$ for $k = 1, 2, \dots n + 1$. Note that if $\nu_2\left(\frac{m-n}{n+k}\right) = a$, then $\frac{m+k}{n+k} \equiv 1 \pmod{2^a}$. As such, we want to show that $x_1x_2\dots x_{n+1} - 1$ which is \[ \prod_{k=1}^{n+1} \left(\frac{m-n}{n+k}+1\right) - 1 \]is not a power of two. Note that $\nu_p\left(\frac{m-n}{n+k}\right)$ is minimized at $\nu_2(m-n) - \left\lfloor \log_2(2n+1) \right\rfloor$ with the unique power of two between $n+1$ and $2n+1$ inclusive, which by expanding is also the $\nu_2$ of the product, which is a contradiction if it is a power of two.

Note that $x_k-1=\tfrac{m-n}{n+k}$. Out of the numbers $n+1,n+2,\dots,2n+1$, one is a power of two and thus has a larger $2$-adic valuation than any of the other numbers. Let that number be $n+k_0=2^r$. Since the numerators are fixed, and $n+k_0$ has the largest $2$-adic valuation, we have $x_{k_0}-1$ has the smallest $2$-adic valuation. Let $x_k-1=2^{a_k}b_k$ for all $k$. Then, in the expansion of \[\prod_{k=1}^{n+1}{(2^{a_k}b_k+1)}-1\]every term will be divisible by $2^{a_{k_0}+1}$, except for one of them, so $2^{a_{k_0}+1}\nmid x_1x_2\ldots x_{n+1}-1$, but then everything else is odd, so we're done.

nice problem Note that for all $k \in {1, 2, \dots, n+1}$ we have \[ n + k \mid m + k \implies n + k \mid m - n; \]hence by CRT we have $N = \text{lcm}(n+1, n+2, n+3, \dots, 2n+1) \mid m - n$; set $m - n = iN$ for some $i$. Now \[x_1x_2\ldots x_{n+1}-1 \prod_{k=1}^{n+1} \left(\frac{m+k}{n+k}\right) -1= \prod_{k=1}^{n+1} \left(\frac{iN}{n+k}+1\right) -1.\]Assume FTSOC that this value is a power of $2$, say $2^a$. Taking $\mod{i}$ tells us that $i \mid 2^a$, so set $i = 2^b$. Now expand \[\prod_{k=1}^{n+1} \left(\frac{iN}{n+k}+1\right) - 1\]and consider the $j$ such that $v_2(n+j)$ is maximal; this $j$ is clearly unique. Now expanding our product, the minimal possible $v_2$ among every term is precisely $v_2\left(\frac{iN}{n+i}\right)$, which is attained at the term $\frac{iN}{n+i}$; hence $v_2(x_1x_2\ldots x_{n+1}-1) = b$, but this fails due to size and we're done. $\square$

Assume the contrary $x_{1}x_{2}\dots x_{n + 1} - 1$ be power of 2. Since $x_{k} = \frac{m + k}{n + k}$ is integer, so $n + k \mid m - n$. Let $\alpha = \max(\nu_{2}(n + 1), \nu_{2}(n + 2), \dots, \nu_{2}(2n + 1))$. Then it's not hard to see that there are exist exactly 1 index $i$ such that $\nu_{2}(n + i) = \alpha$. Let $\beta = \nu_{2}(m - n)$. Thus $x_{1}x_{2}\dots x_{n + 1} \equiv 2^{\beta - \alpha} + 1 (2^{\beta - \alpha + 1})$. (Note that $\beta > \alpha$, since $x_{1}x_{2}\dots x_{n + 1}$ is odd) Thus we reach a contradiction. Thus we're done. $\blacksquare$

Feels a bit superficial. Note that there is a unique minimum, say $a$, among the quantities $\nu_2\left(\frac{m-n}{n+k}\right)$ for $1 \leq k \leq n+1$. This implies that $$\frac{m+k}{n+k} \equiv 1 \pmod {2^a}$$and in particular, $\nu_2(x_1x_2\cdots x_n - 1) = a$ by uniqueness of $k$. So assume that $x_1x_2\cdots x_n - 1 = 2^a$ precisely. But this is a clear size contradiction because $\frac{m+k_1}{n+k_1}$ for some $k_1 \neq k$ is $1$ mod $2^b$ for $b > a$ hence already larger than $2^a$.

Let $M=\operatorname{lcm}(n+1,n+2,\dots,2n+1)$. By an easy Euclid argument, $M\mid m-n$, so let $m-n=yM$. Then the expression at the end looks like \[ \left(\frac{yM}{n+1}+1\right)\left(\frac{yM}{n+2}+1\right)\dots\left(\frac{yM}{2n+1}+1\right)-1. \]Clearly this is divisible by $y$, so if $y$ was not a power of $2$ we would already be done. Thus let $y=2^x$. Since there must be exactly one power of $2$ in $\{n+1,n+2,\dots,2n+1\}$, say $z$ is the power of $2$, then $\frac{M}{z}$ is odd but $\frac{M}{u}$ is even for any other $u$ in that set. Thus taking mod $2^{x+1}$ yields a contradiction(obviously the expression is not equal to $2^x$). $\blacksquare$

Note that $n + k \mid m - n$ for all $1 \le k \le n + 1$, hence $\text{lcm}(n+1, n+2, \dots, 2n + 1) \mid m - n$. Thus we can write $m = n + \text{lcm}(n+1, n+2, \dots, 2n + 1) \cdot s$ for some $s \ge 1$. Let $\alpha = \max(\nu_2(n+1), \nu_2(n+2), \dots , \nu_2(2n+1))$. Then note that $\alpha$ appears exactly once in the set $\{\nu_2(n+1), \nu_2(n+2), \dots, \nu_2(2n+1)\}$. Let $k$ be the integer in $1,2, \dots, n + 1$ such that $\nu_2(n+k) = \alpha$. Then taking modulo $2^{1 + \nu_2(s)}$, we get $x_1x_2 \cdots x_{n+1} = \prod_{i=1}^{n+1} (\frac{m-n}{n+i} + 1) \equiv (\frac{m-n}{n+k} + 1) \equiv (2^{\nu_2(s)} + 1) (2^{s+1})$. Hence $x_1x_2 \dots x_{n+1}$ cannot be a power of 2. $\blacksquare$

Let $L=\mathrm{lcm}(1,2,\dots,2n+1).$ Then we have $n+1,n+2,\dots,2n+1\mid m-n$ so $L\mid m-n$ and we can write $m=n+rL$ for some $r.$ Then we have \[x_1x_2\cdots x_{n+1}-1=\left(1+r\frac L{n+1}\right)\left(1+r\frac L{n+2}\right)\cdots\left(1+r\frac L{2n+1}\right)-1.\]Suppose this is a power of $2.$ Upon expansion we see that everything is a multiple of $r$ so $r$ is a power of $2.$ Now let $Z$ be the unique power of $2$ with $n+1\le Z\le 2n+1.$ If $r=1$ we have $\left(1+\frac LZ\right)\equiv 0\pmod 2,$ so $x_1x_2\cdots x_{n+1}$ is even and clearly $\ne 2,$ contradiction. If $r>1$ then upon expanding everything and dividing by $r$ we will have that $r$ now divides everything except possibly $\frac L{n+1}+\frac L{n+2}+\cdots+\frac L{2n+1},$ and if our assumption is true we must have $r$ divides this as well. However all terms of this are even except $\frac LZ,$ so the sum is odd, contradiction.

Suppose for contradiction that $x_1x_2\cdots x_{n+1}-1$ is a power of two, say $2^x$. Let $l=\text{lcm}(n+1,n+2,\ldots,2n+1)$. Note that for each $1\leq k\leq n+1$, we have $n+k\mid m-n$ as $x_k-1=\frac{m-n}{n+k}$ is an integer. Hence, $l\mid m-n$ so let $m-n=lq$. There always exists exactly one power of two in the set $\{n+1,\ldots,2n+1\}$, here say $2^z$. If $1\leq k\leq n+1$ is taken such that $n+k\neq 2^z$, then \[2^z\mid \mid \frac{l}{n+k}\mid \frac{m-n}{n+k}\Longrightarrow 2^z< x_k\Longrightarrow 2^z<x_1x_2\cdots x_{n+1}-1=2^x\Longrightarrow z<x.\]Now, let $j$ be such that $2^j\mid \mid q$. Then, for any $k$ with $n+k\neq2^z$, we have that $2^{j+1}\mid \frac{m-n}{n+k}$, which means $x_k\equiv 1\pmod{2^{j+1}}$, while for $x+k'=2^z$, we have $2^j\mid \mid \frac{m-n}{2^z}$, which means $x_{k'}\equiv 2^j+1\pmod{2^{j+1}}$. It follows that $2^x=x_1x_2\cdots x_{n+1}-1\equiv 1\cdot1\cdots(2^j+1)-1\equiv 2^j\pmod{2^{j+1}}$. This forces $x=j$. However, since $2^{j+1}\mid x_k-1$ for all $1\leq k\neq k'\leq n+1$, we have $2^{j+1}<x_1x_2\ldots x_k-1=2^x\Longrightarrow j+1<x$, which contradicts $x=j$.

Obviously each $x_k \geq 2$, so $S = x_1x_2\ldots x_{n+1}-1 > 1$ and hence it suffices to check that it is not a power of two. Let $d = m - n > 0$ and $y_k = x_k - 1 = \frac d{n+k}$ be positive integers. Write $n+1 \leq 2^r \leq 2n+1$ and $2^r = n + c$. Then $2^r$ is the unique element among $n+1, n+2, \dots, 2n+1$ with largest 2-adic valuation, so we have \[ t = \nu_2 \left(\frac d{2^r} \right) < \min_{\substack{1 \leq k \leq n + 1 \\ k \neq c}} \left\{ \nu_2 \left(\frac d{n+k} \right) \right\} \]and so \begin{align*} S &= (y_1+1) \dots (y_{c-1}+1) (y_{c}+1) (y_{c+1}+1) \dots (y_{n+1}+1) - 1 \\ &= (\text{even} \cdot 2^t+1) \dots (\text{even} \cdot 2^t+1) (\text{odd} \cdot 2^t+1) (\text{even} \cdot 2^t+1) \dots (\text{even} \cdot 2^t+1) -1\end{align*}taking mod $2^{t+1}$ gives \[ S \equiv (1) \dots (1)(2^t+1)(1) \dots (1) -1 \equiv 2^t \pmod {2^{t+1}} \]but one easily obtains $S > 2^t$, so $S$ is not a power of two.

For all $k \in \{1, 2, \ldots, n + 1\}$, we have $n + k \mid (m + k) - (n + k) = m - n$. We have $x_k = \frac{m + k}{n + k} = \frac{m-n}{n + k} + 1$. Suppose that $x_1 x_2 \cdots x_{n+1}$ was a power of $2$. Note there is unique number in $n+1, n + 2, \ldots, 2n + 1$ with maximal $\nu_2$ (the unique power of $2$). Suppose it is $2^x$. Let $t = \nu_2(m - n)$. We have $\nu_2 \left( \frac{m-n}{n + k} \right) = t - \nu_2(n + k) \ge t - x$. However, for $k \ne 2^x - n$, we have $\nu_2 \left( \frac{m-n}{n + k } \right) > t - x$. Thus, $\frac{m-n}{n + k} + 1\equiv 1\pmod{2^{t - x + 1} } $ for all $k\ne 2^x - 2n$. Hence $\prod_{i\ne 2^x - 2n} x_i \equiv 1\pmod{2^{t - x + 1}} $. However, $\nu_2 \left( \frac{m-n}{2^x} \right) = t - x$, so $x_{2^x - 2n} \not \equiv 1\pmod{2^{t - x + 1}}$ but $x_{2^x - 2n } \equiv 1\pmod{2^{t-x}}$. Thus, $\prod x_i - 1$ isn't divisible by $2^{t - x + 1}$ but is divisible by $2^{t - x}$, so it must equal $2^{t - x}$. However, $2^{t - x} = \frac{2^t}{2^x} \le \frac{m-n}{n + 1} $. Therefore, we have $x_1 x_2 \cdots x_{n + 1} - 1 = 2^{t-x} \le x_1 - 1$. However, $x_1 x_2 \cdots x_{n+1} - 1 \ge x_1 - 1$, with equality iff $x_i = 1$ for all $i \ne 1$ (and at most $n + 1$). However, $x_k = \frac{m - n}{n + k} + 1 > 1$ for any $k$. Now, since $n \ge 1$, $x_2$ is defined, so $x_2 > 1$ and therefore $x_1 x_2 \cdots x_{n+1} - 1 > x_1 - 1$, a contradiction. Hence $x_1 x_2 \cdots x_{n+1} - 1$ cannot be a power of $2$, and thus must be divisible by an odd prime.

We express the product \[P = x_1 \ldots x_{n+1} - 1 = \prod_{k=1}^{n+1} \left(\frac{m-n}{n+k}+1\right) - 1.\] Notice there is a unique $n+q$ which is a power of 2 in the interval $[n+1, 2n+1]$, which is the highest $v_2(n+q)$, which in turn gives the uniquely lowest $v_2\left(\frac{m-n}{n+k}\right)$. If we expand the above product $P$ and express it as a symmetric sum, we find \[v_2(P) = v_2\left(\frac{m-n}{n+q}\right), P > \frac{m-n}{n+q},\] and hence it is impossible for $P$ to be a power of 2. $\blacksquare$

Yay my first N3 Observe that $n+k \mid m+k \implies n+k \mid m-n$ for $k =1,2,…,n+1$, hence $lcm (n+1,n+2,…,2n+1)=N \mid m-n$ Let $m-n=tN, t \in \mathbb{N}$ Then, we have $x_{k}= \frac{tN+n+k}{n+k}=\frac{tN}{n+k}+1$ $$x_{1}x_{2}…x_{n+1} -1= (\frac{tN}{n+1}+1)… (\frac{tN}{2n+1}+1) -1 \equiv 0 (\mod t)$$If $t$ is not a power of $2$, then there exists an odd prime divisor of $t$, and we are done. Assume $t$ is a power of 2. Let $t=2^{\alpha}$ Clearly, $x_{1}x_{2}…x_{n+1} -1>2^{\alpha}$ Consider $l$ such that $v_{2}(N)=v_{2}(n+l)$ Since $n+1 \le n+l \le 2n+1 < 2(n+l)$, there is only one such $l$. Also, $\frac{N}{n+l}$ is odd, hence $\frac{2^{\alpha} N}{n+l} +1 \equiv 2^{\alpha+1} (\mod 2^{\alpha+1})$ For all $k \neq l$, $\frac{2^{\alpha}N}{n+l}+1 \equiv 1(\mod 2^{\alpha +1})$ Hence, $(\frac{2^{\alpha}N}{n+1}+1)… (\frac{2^{\alpha} N}{2n+1}+1) -1 \equiv 2^{\alpha} (\mod 2^{\alpha+1})$, so $x_{1}x_{2}…x_{n+1} -1$ cannot be a power of $2$, hence we are done.$\blacksquare$