Wrong Solution , assuming the opposite doesn't imply that R(a)=0 ..

As deadline is over so we can now post solution. My solution : Suppose $a^{n+1}+a-1=m\prod (a+a_i-1)$ So $a-1|m\prod a_i$ and also let $a=n\prod a_i$. Now it's easy to see all of $a_i>1$ So finally $a^{n+1}+a-1\geq m (a+1)^n$ which implies $m<a$ Now it's easy to see $a-1|m-n$ but as both of $m,n<a$ so clearly $m=n$ But then we've $m=n|1$ so $m=n=1$ Then we've $\prod a_i=a|\prod (a_i-1)+1$ .So now we must have $\prod(a_i-1)+1=\prod a_i$ As all of $a_i>1$ so clearly solution exist if $n=1$. A contra !!.

subham1729 wrote: So $a-1|m\prod a_i$ How you get that conclusion? The problem was proposed by me to RMM 2012 but it was used in Romania TST. Not good move for Romanian guys. Anyway I post the solution as it originally was proposed. Let $a_1, a_2, \cdots, a_n, c$, $c>2$ be positive integers such that the number $\dfrac{c}{a_1a_2\cdots a_n}$ is an integer. Prove that the number \[\frac{c^{n+1}+c-1}{(c+a_1-1)(c+a_2-1)\cdots(c+a_n-1)}\] is not an integer. Solution: Denote by $t$ the number $\dfrac{c}{a_1a_2\cdots a_n}$. Suppose that the number \[\frac{c^{n+1}+c-1}{(c+a_1-1)(c+a_2-1)\cdots(c+a_n-1)} = m\] is an integer. Obviously $1 \leq m \leq c$ and the number $c^{n+1}+c-1$ is coprime with the numbers $c$ and $c-1$ hence, $1 \leq m < c-1$, $1 \leq t < c$ $(1).$ More over \[m=\frac{c^{n+1}+c-1}{(c+a_1-1)(c+a_2-1)\cdots(c+a_n-1)} \equiv \frac{1}{a_1a_2\cdots a_n} \equiv t \pmod{c-1} \; (2).\] From $(1)$ and $(2)$ we get that $m=t$. And because the numbers $m$ and $c$ are coprime we get that $a_1a_2\cdots a_n=c$, $m=t=1$. Lemma 1. $(c+x-1)(c+y-1) \leq c(c+xy-1)$ for all positive integers $x,y$. Proof. $(c+x-1)(c+y-1) \leq c(c+xy-1) \Longleftrightarrow (c-1)(x-1)(y-1) \geq 0$. Now consequently applying Lemma 1, we get that $(c+a_1-1)(c+a_2-1)\cdots(c+a_n-1) \leq c^{n-1}(2c-1)<2c^n<c^{n+1}+c-1$ which contradicts with the condition $m=1$ and we are done.

Suppose the opposite is true and let $P(x)=x^{n+1}+x-1-k\prod_{i=1}^n(x+a_i-1)$ where $k\in \mathbb{Z}$ and $P(a)=0$. Let $P(x)=(x-a)\cdot Q(x)$.$P(1)\equiv 1\pmod{\prod_{i=1}^n a_i}$ and hence as $1-a\equiv 1\pmod{\prod_{i=1}^n a_i}$ $\implies$ $Q(1)\equiv 1\pmod{\prod_{i=1}^n a_i}$.$|Q(1)|= \frac{\mid k\prod_{i=1}^n a_i-1\mid}{\mid a-1\mid} $.Note that if all $a_1,a_2,...a_n$ are one than we have that $a\mid 1$ contradiction as $a\not =1$ $\implies$ $$0=a^{n+1}+a-1-k\prod_{i=1}^n (a+a_i-1)\leq a^{n+1}+a-1-k \cdot a^{n-1}(a+1)$$And so if $k\geq a$ we would have that $-1\geq 0$,impossible and so $k\leq a-1$ $\implies$ $|Q(1)|\geq \prod_{i=1}^n a_i$ and hence as $Q(1)\equiv 1\pmod{\prod_{i=1}^n a_i}$ $\implies$ $Q(1)=1$ $\implies$ $P(1)=1-a$ and hence $a=k\prod_{i=1}^n a_i$ but as $k\mid a$ and $P(a)=0$ $\implies$ $k\mid -1$ $\implies$ $k=1$ (everything is positive after all).However $a\mid \prod_{i=1}^{n}(a_i-1)+1$ by rational root theorem but the left side is larger than the right and hence a contradiction.$\blacksquare$

I like this. The core of this problem is to think of the problem as a "polynomial" - how to vanish terms etc. Suppose otherwise. Let $m \in \mathbb{N}$ such that \[ a^{n + 1} + a - 1 = m \prod_{i = 1}^n (a + a_i - 1) \]Take modulo $a - 1$ both sides, we get \[ 1 \equiv m \prod_{i = 1}^n a_i \ (\text{mod} \ a - 1 ) \]Thus, we have $a - 1 \mid m \prod_{i = 1}^n a_i - 1$. Let $a = k \prod_{i = 1}^n a_i$ for some $k \in \mathbb{N}$, then we have \[ k \prod_{i = 1}^n a_i - 1\mid k - m \]First, we observe that if there exists $i$ such that $a_i = 1$, then $a \mid RHS = a^{n + 1} + a - 1$, forcing $a = 1$, a contradiction. Therefore, \[ a^{n + 1} + a - 1 = m \prod_{i = 1}^n (a + a_i - 1) \ge m(a+1)^n \rightarrow m < a \]Second, we notice that \[ a = k \prod_{i = 1}^n a_i \ge 2k \rightarrow k < a \]Therefore, we have $1 \le m,k < a$, forcing $|m - k| < a - 1$. This forces $m = k$. Let's get back to the first equation. Consider the equation modulo $a$, we get \[ -1 \equiv m \prod_{i = 1}^n (a_i - 1) \ (\text{mod} \ a) \]This, gives us \[ \left( m \prod_{i = 1}^n a_i \right) - 1 \mid m \prod_{i = 1}^n (a_i - 1) \]As $m$ relatively prime to $LHS$, we get \[ \left( m \prod_{i = 1}^n a_i \right) - 1 \mid m \prod_{i = 1}^n (a_i - 1) \]forcing $m = 1$, or otherwise we fail by size reasons. Thus, we must have \[ \left( \prod_{i = 1}^n a_i \right) - 1 = \prod_{i = 1}^n (a_i - 1) \]which works if and only if $n = 1$ and $a_i = 2$ (easy inequality and bounding argument). However, this gives us $a = 2$ and $a_i = 2$. But, it is easy to check that this doesn't work. Therefore, the original statement is false.

For IndoMathXdZ's solution, I understand why \[ 1 \equiv m \prod_{i = 1}^n a_i \ (\text{mod} \ a - 1 ) \]but it is confusing to me why $a - 1 \mid m \prod_{i = 1}^n a_i$. If the product $ma_1a_2...a_n$ was congruent to $1 (\text{mod } a-1)$, I believe this means $a-1| m \prod_{i = 1}^n a_i -1$. So, where did that 1 go?

Solved with Narwhal234, who just gave me the crux step (look $\pmod{a-1}$) and then randomly disappeared. We assume that there exists $a_1,a_2,\dots,a_n,a$ for which $a^{n+1}+a-1$ is divisible by the product $(a+a_1-1)(a+a_2-1)\ldots(a+a_n-1)$ and work towards a contradiction. We start off with the following key claim. Claim : Let $a=ka_1a_2\dots a_n$ and $a^{n+1}+a-1 = m \prod_{i=1}^n (a+a_i-1)$ for positive integers $k$ and $m$. Then, $k=m=1$ if all the conditions assumed above are satisfied. Proof : We look at our relations $\pmod{a-1}$. Note that, \[1\equiv a =ka_1a_2\dots a_n \pmod{a-1}\]and, \[1\equiv (1)^{n+1} + 0 \equiv a^{n+1}a-1 = m\prod_{i=1}^n (a+a_i-1) \equiv m\prod_{i=1}^n a_i \pmod{a-1}\]Now, further note that if $k>a-1$, \[a=ka_1a_2\dots a_n \ge (a)(2) = 2a > a\]for all $a>1$, which is a very clear contradiction. Thus, $k\le a-1$. Similarly if $m>a-1$, since $a_i \ge 1$ for all $1\le i \le n$ and $a>1$ (so there exists some $j$ for which $a_j >1$), \[ a^{n+1}a-1 = m\prod_{i=1}^n (a+a_i-1) \ge a(a^{n-1})(a+1)=a^{n+1}+a^n > a^{n+1}+a-1 \]for all $a>1$. Thus, $m\le a-1$ as well. But now, we observed above that \[ka_1a_2\dots a_n \equiv 1 \equiv ma_1a_2\dots a_n \pmod{a-1}\]since $\gcd(a-1,a_1\dots a_n)=1$ this implies that $k\equiv m \pmod{a-1}$ and since $0<k,m\le a-1$ this implies that in particular $k=m$. Next, if $a=ka_1a_2\dots a_n$ and \[(ka_1a_2\dots a_n)^{n+1} + ka_1a_2\dots a_n -1 = a^{n+1}+a-1 = k \prod_{i=1}^n (a+a_i-1)\]the right hand side is divisible by $k$ but not the left hand side for all $k>1$. This allows us to conclude that $k=1$ proving the claim. Now we simply resort to a size argument. First of all note that if $a>1$ and there exists some $a_i=$ then, \[a \mid (a+a_1-1)(a+a_2-1)\dots (a+a_n-1) \mid a^{n+1}+a-1\]which is a clear contradiction. Thus, $a_1,a_2,\dots,a_n \ge 2$. Next, observe that if $a=a_1a_2\dots a_n$ and $a^{n+1}+a-1 = (a+a_1-1)(a+a_2-1)\ldots(a+a_n-1)$ then since $a_1-1 < a_1 < a$ we have \[a^{n+1}+a-1 = (a+a_1-1)(a+a_2-1)\ldots(a+a_n-1) < (2a)^n = 2^na^n \]But, \[a^{n+1}+a-1 > a^{n+1}= a^n(a_1a_2\dots a_n) \ge 2^na^n\]which contradicts our previous inequality. Thus, there cannot exist such $a_1,a_2,\dots , a_n,a$ finishing the proof of the problem.

We then have this goat who did all the cleanup I never could have finished. (Also he admitted orz before I left)