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
10.05.2022 10:52
10.05.2022 14:51
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 $a_1a_2=8$.
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 ...
11.05.2022 09:00
This solution is the same given by Dr. Mohsen Jamaali!!
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.$\boxed {}$