define $(a_n)_{n \in \mathbb{N}}$ such that $a_1=2$ and $$a_{n+1}=\left(1+\frac{1}{n}\right)^n \times a_{n}$$Prove that there exists infinite number of $n$ such that $\frac{a_1a_2 \ldots a_n}{n+1}$ is a square of an integer.
Problem
Source: Iran 2nd round 2022 P5
Tags: number theory, Sequence
Tintarn
10.05.2022 10:52
The problem is really a bit of a red herring, as it is more about being an integer than being a square.
Indeed, a direct computation shows that
\[a_1a_2\dots a_n=(n+1)^n \cdot n^{n-2} \cdot (n-1)^{n-4} \cdots 3^{4-n} \cdot 2^{2-n}.\]Now for even $n$, this is always a square and so if $n+1$ is an odd square, then our number is a rational square.
But of course we need to check whether it is integral.
Indeed, we claim that
\[(n+1)^{n-1} \cdot n^{n-2} \cdot (n-1)^{n-1} \cdots 3^{4-n} \cdot 2^{2-n}\]is always an integer when $n \ge 2$, this will clearly show that all $n=(2k+1)^2-1$ do the job.
We check each prime at a time: So if $p$ is a prime, we need to check that
\[(n-1)v_p(n+1)+\sum_{k=1}^n (2k-n-2)v_p(k) \ge 0.\]Separating moreover the contributions from all powers of $p$, it suffices to check that for all $q=p^k$ we have
\[(n-1)\chi_{q \mid n+1}+\sum_{k \le n, q \mid k} (2k-n-2) \ge 0.\]Computing the LHS, we get the equivalent
\[(n-1)\chi_{q \mid n+1}+q \cdot \left\lfloor \frac{n}{q}\right\rfloor \cdot \left(\left\lfloor \frac{n}{q}\right\rfloor+1\right) \ge (n+2) \cdot \left\lfloor \frac{n}{q}\right\rfloor\]Writing $n=aq+r$ with $0 \le r \le q-1$, this is equivalent to
\[(n-1)\chi_{r=q-1}+q \cdot a(a+1) \ge (aq+r+2) \cdot a\]or equivalently
\[(n-1)\chi_{r=q-1} \ge a(r+2-q).\]If $r \le q-2$, we have $LHS=0$ and $RHS \le 0$ and if $r=q-1$ we need to check that $n-1 \ge a$ which is obvious.
Pmshw
10.05.2022 14:51
$a_n=\frac{2n^n}{n!}$ so we can see $a_1a_2...a_n=2^n\binom{n}{0}\binom{n}{1}...\binom{n}{n-1}$ so if we put $n=2^{2k-1}-1$ it would be sufficient , so we can say it is an square of an integer because $\binom{n}{k}=\binom{n}{n-k}$
Tintarn
10.05.2022 18:11
Pmshw wrote: we can see $a_1a_2...a_n=2^n\binom{n}{0}\binom{n}{1}...\binom{n}{n-1}$ Are you sure? I think $a_2=\frac{9}{2}$ so $a_1a_2=9$ but your result would be $a_1a_2=8$.
Pmshw
10.05.2022 19:36
Tintarn wrote: Pmshw wrote: we can see $a_1a_2...a_n=2^n\binom{n}{0}\binom{n}{1}...\binom{n}{n-1}$ Are you sure? I think $a_2=\frac{9}{2}$ so $a_1a_2=9$ but your result would be $a_1a_2=8$. I think I wrote the problem wrong... my bad... it should know be fixed. I think your solution now needs editing ...
math-helli
11.05.2022 09:00
This solution is the same given by Dr. Mohsen Jamaali!!
Hopeooooo
17.05.2022 23:10
Another olution with a little bash$;$ We know that $:a_n=2 \times (\frac{3}{2})^2 \times (\frac{4}{3})^3 \times \cdots \times (\frac{n}{n-1})^n$ Let $n=2k+1.$ We know that $;\prod_{i=1}^n a_i =2^n \times 2^{n-1} \times (\frac{3}{2})^{2(n-2)} \times (\frac{4}{3})^{3(n-3)} \times \cdots \times (\frac{n}{n-1})^n $ We define $f$ and $g$ the sequence of power in $\prod_{i=1}^n a_i.$ So $f$ is the power of first $k$ terms in $\prod_{i=1}^n a_i$ and $g$ is the last $k+1$ terms in $\prod_{i=1}^n a_i.$ We claim that $f$ is strictly increasing and analogously $g$ is strictly decreasing$.$ $Proof:$ We just need to prove that $\forall t<k:k(k+1) \ge t(2k+1-t).$ $k^2+k-2tk-t+t^2=(k-t)^2+k-t \ge 0 $ so we are done. Because $n$ is odd so all of the power is even. It is east to see that $A=\prod_{i=k+1}^n a_i=x^2$ and also $B=\prod_{i=1}^k a_i=\frac{2r^2}{s^2}.$ So we claim that $AB \in N.$ Let $a_i$ the first $k$ terms we have $a_j$ in the last $k+1$ terms such that $a_i.a_j \in N.$ Let $a_i=(\frac{a}{a-1})^{a-1}$ we know that we have a multiple of $a-1$ in the last $k+1$ terms now take $a_j$ the greatest multiple of $a_i$ so we have $a_j=(\frac{at}{(a-1)t})^{(a-1)t}$ now we just need to prove that $(n-a+1)(a-1)-(n-a+2)(a-2)+t(a-1)(n-t(a-1)-t(a-2)(n-t(a+2)) \ge$ and this is absolutely correct. Now $\forall \,\, s^2 | AB,\,\,$ take$ \,\,\,\,\, n+1=2^{2s+1}\times s^2.$ So we are done.$\boxed {}$