List of funny problems (3rd)

19/12/2011

List of funny problems (3rd)

Filed under: Mathematical matters — bboyjordan @ 7:31 PM

Problem25- Maybe Fermat primes are infinte?

Show that $\displaystyle \lim_{n\to +\infty}{\sigma_{-1}(2^{2^n}+1)}=1$ .
Solution. For every positive integer $\displaystyle m=\prod_{1\le i\le \omega(m)}{p_i^{\alpha_i}}$ we have by definition $\sigma_{-1}(m)=\displaystyle \sum_{d\mid m}{d^{-1}}=\prod_{1\le i\le \omega(m)}{\left(\sum_{0\le j \le \alpha_i}{p^{-j}}\right)}= \prod_{1\le i\le \omega(m)}{\frac{p-p^{-\alpha_i}}{p-1}}$ .

It means that for $m=2^{2^n}+1$ we have $\sigma_{-1}(2^{2^n}+1)=\displaystyle \prod_{p_i\mid 2^{2^n}+1}{\frac{p_i-p_i^{-\alpha_i}}{p_i-1}} < \prod_{p_i\mid 2^{2^n}+1}{\frac{p_i}{p_i-1}}= \prod_{p_i\mid 2^{2^n}+1}{\left(1+\frac{1}{p_i-1}\right)}$ .

Note now that if a prime $p_i \mid 2^{2^n}+1$ then $\text{ord}_{p_i}=2^{n+1}$ ; but since $\text{ord}_{p_i}(2)\mid \varphi(p_i)=p_i-1$ then there exists a positive integer $k_i>0$ such that $p_i=k_i2^{n+1}+1$ .

Also this trivial inequality holds: $2^{(n+1)2^n}>2^{2^n}+1=\displaystyle \prod_{p_i\mid 2^{2^n}+1}{p_i^{\upsilon_{p_i}(2^{2^n}+1)}} \ge \prod_{1\le i\le \omega(2^{2n}+1)}{p_i}> 2^{(n+1)\omega(2^{2^n}+1)}$ : so $\omega(2^{2^n}+1)<2^n$ .

Summing up, $\sigma_{-1}(2^{2^n}+1)<\displaystyle \prod_{1\le i\le \omega(2^{2^n}+1)}{\left(1+\frac{1}{p_i-1}\right)}$ $= \displaystyle \prod_{1\le i\le \omega(2^{2^n}+1)}{\left(1+\frac{1}{k_i2^{n+1}}\right)} \le \prod_{1\le i\le \omega(2^{2^n}+1)}{\left(1+\frac{1}{i2^{n+1}}\right)}$ .

Extendig this product, we have $\sigma_{-1}(2^{2^n}+1)<$ $\displaystyle \sum_{0\le i\le \omega(2^{2^n}+1)}\left(\frac{1}{2^{i(n+1)}}\left(\sum_{1\le x_1<x_2<\ldots<x_i\le \omega(2^{2^n}+1)}{\frac{1}{x_1x_2\ldots x_i}}\right) \right)$ $\displaystyle < \sum_{0\le i\le 2^n}\left(\frac{1}{2^{i(n+1)}}\left(\sum_{1\le x_1<x_2<\ldots<x_i\le 2^n}{\frac{1}{x_1x_2\ldots x_i}}\right) \right)$ $< \displaystyle \sum_{0\le i\le 2^n}\left(\frac{1}{2^{i(n+1)}}\left(\sum_{1\le x_1\le x_2\le \ldots\le x_i\le 2^n}{\frac{1}{x_1x_2\ldots x_i}} \right)\right)$
$\displaystyle = \displaystyle \sum_{0\le i\le 2^n}\left(\frac{1}{2^{i(n+1)}}\left(\sum_{1\le j\le 2^n}{j^{-1}} \right)^i\right)$ .

For every integer $y\ge 2$ it’s also true that $\displaystyle \sum_{1\le j\le y}{j^{-1}}<1+\int_1^y{x^{-1}dx}=1+\ln(y)$ .

Then we can say that there exists a positive constant $C>0$ such that $\displaystyle \sum_{1\le j\le 2^n}{j^{-1}}< Cn$ .

Turning back to our problem, we proved that $\displaystyle \sigma_{-1}(2^{2^n}+1)<\displaystyle \sum_{0\le i\le 2^n}\left(\frac{1}{2^{i(n+1)}}\left(\sum_{1\le j\le 2^n}{j^{-1}} \right)^i\right)$ $\displaystyle < \sum_{0\le i\le 2^n}{\left(\frac{Cn}{2^{n+1}}\right)^i}$ $\displaystyle< \sum_{0\le i\le \infty}{\left(\frac{Cn}{2^{n+1}}\right)^i}=$ $\displaystyle \frac{2^{n+1}}{2^{n+1}-Cn}$ .

Now since for every $\epsilon>0$ the following inequality $\displaystyle 2^{n+1}\left(1-\frac{1}{1+\epsilon} \right)>Cn$ holds definitively, then we finished, indeed:

$\displaystyle \lim_{n\to +\infty}\sigma_{-1}(2^{2^n}+1)< \displaystyle \frac{2^{n+1}}{2^{n+1}-C\ln(n)} < 1+\epsilon$ . []

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19/12/2011

List of funny problems (3rd)

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