Easy problem

Tomekk

Hello,

find all positive integers $n$ such that

$\frac{2010}{n^{2}-1}$ is also a positive integer.

**Posted:** Wed Nov 04, 2009 6:02 am

ernie

We have to let $n^2 - 1$ be a factor of $2010$ . So $n^2$ has to be a factor of $2011$ . The only two factors of $2011$ are $\{1,2011\}$ . However, $n^2 \neq 1$ , because then $\frac{2010}{0}$ would be undefined. Thus, there are no answers.

**Posted:** Wed Nov 04, 2009 6:56 am

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If pro is the opposite of con, then what's the opposite of progress? Razz

mavropnevma

**Posted:** Wed Nov 04, 2009 7:49 am

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Listen to REMBETIKA for decoding the handle.

ernie

Oh wow, that was a fail.

**Posted:** Wed Nov 04, 2009 1:19 pm

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If pro is the opposite of con, then what's the opposite of progress? Razz

dragon96

**Posted:** Wed Nov 04, 2009 11:15 pm

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$\mathfrak{Dragon96}$

grn_trtle

**Posted:** Thu Nov 05, 2009 7:14 am

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$\arctan\alpha + \arctan\beta = \text{arg}\{(1+\alpha i)(1+\beta i)\} = \arctan\left( \frac{\alpha+\beta}{1-\alpha\beta} \righ...$

Tomekk

Try $n=4$ also Smile

.

**Posted:** Fri Nov 06, 2009 1:05 pm

cryax

2010=2*3*5*67. in order to 2010/(n^2-1) is a positive integer, n^2-1 must be estimated number of 2010 with n is a integer so that n=4,2(n^2-1=15=3*5;n^2-1=3) is the right answer.

**Posted:** Sat Nov 07, 2009 8:06 am

cenx

I think $2010/(n^2-1)$ can be factored as $1005/(n-1) - 1005/(n+1)$ and the problem turns to sequentive odd factor hunt, at least we have less to worry about ^^

**Posted:** Sat Nov 07, 2009 10:46 am

mathkid95

The thing is, if you factor out the bottom to $(n + 1)(n - 1)$ you get n cannot be 1.
Because 0 is neither negative nor positive, it cannot be counted. The statement,

**Posted:** Sun Nov 08, 2009 5:58 am

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Rotfl

$\text{M^{A}T H_{R^{U_{L_{E{S}}}}}}$

Tomekk

I am not sure if I understood what you said, but even if you dont factor $n^2 -1$ you still get that $n$ cannot be 1.

In this problem, $n$ chasing isnt hard at all, since you dont have much options anyway. Solutions are $n=4$ and $n=2$ .

**Posted:** Sun Nov 08, 2009 6:17 am