n^3|(2^{n^2}+1)

Svejk

n^3|(2^{n^2}+1)

**Posted:** Sun Mar 25, 2007 11:32 am

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Radu Titiu

RDeepMath91

In IMO 1990 no. 3, the solution for n where $n | 2^{n}+1$ is $n = 1$ or $n = 3$ . (Will be proved if necesarry)
Thus, if $n^{2}| n^{3}| 2^{n^{2}}+1$ , we obtain $n = 1 \Rightarrow m = 3$
Hence, since $27 | 2^{9}+1$ , it follows that $m^{3}| 2^{m^{2}}+1$ . Mr. Green

**Posted:** Sun Mar 25, 2007 5:17 pm

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Svejk

#3

I don't think that $n|2^{n}+1$ implies only $n=1$ or $n=3$

**Posted:** Sun Mar 25, 2007 6:41 pm

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Radu Titiu

zscool · Riemann Hypothesis Offline Joined: 23 Jun 2003 Posts: 482

#4

the actual IMO question is if $n^{2}|2^{n}+1$ where $n$ is a positive integer then $n=1 or 3$ , so it doesn't quite apply.

**Posted:** Sun Mar 25, 2007 8:03 pm

Rust

#5

There are infinetely many solutions, suth that $n^{3}|2^{n^{2}}+1$ .
For example n=3 give $m=57$ , n=57 give $m=\frac{2^{3249}+1}{3249}$ e t.c.
Proofe. If $2^{n^{2}}+1=kn^{3}$ , then $m=kn$ , k,n,m - are odd.
Therefore $2^{m^{2}}+1=(2^{n^{2}}+1)(2^{(k^{2}-1)n^{2}}-2^{(k^{2}-2)n^{2}}+...+1)$ . We have $2^{n^{2}}+1=m*n^{2},$ therefore $2^{an^{2}}=(m*n^{2}-1)^{a}$ and
$\sum_{a=0}^{k^{2}-1}(-1)^{a}2^{an^{2}}=\sum_{a=0}^{k^{2}-1}(1-mn^{2})^{a}=k^{2}-\frac{k^{2}(k^{2}-1)}{2}mn^{2}+O(m^{2}).$
It mean
$2^{m^{2}}+1=mn^{2}k^{2}(1-\frac{k^{2}-1}{2}mn^{2})+O(m^{3})=m^{3}(1-\frac{k^{2}-1}{2}mn^{2}})+O(m^{3})=O(m^{3})$ .

**Posted:** Mon Mar 26, 2007 8:22 am

me@home

#6

$\sum_{a=0}^{k^{2}-1}(-1)^{a}2^{an^{2}}=\sum_{a=0}^{k^{2}-1}(1-mn^{2})^{a}=k^{2}-\frac{k^{2}(k^{2}-1)}{2}mn^{2}+O(m^{2}).$
In the final expression, what if there are binomial coefficients which divide $m^{2}$ ?

**Posted:** Tue Mar 27, 2007 11:55 am

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Rust

#7

Let $k=\frac{2^{n^{2}}+1}{n^{3}},m=kn.$
Then
$2^{m^{2}}+1=(2^{n^{2}})^{k^{2}}+1=1+(mn^{2}-1)^{k^{2}}=\sum_{i=1}^{k^{2}}C_{k^{2}}^{i}(-1)^{i-1}m^{i}n^{2i}=k^{2}n^{2}m-k^{2}...$

**Posted:** Tue Mar 27, 2007 12:14 pm