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#### Flaming_Arrow

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##### Recreational Problems (MM level)
« on: June 12, 2008, 09:33:08 pm »
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btw im in yr 11 so this is question from unit 1 and 2 book

ill start off

$if\; x + \frac {1}{x} = 9$

$then\; x^2 + \frac {1}{x^2} = ?$
« Last Edit: June 12, 2008, 10:55:49 pm by chathuranj »
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##### Recreational Problems (MM level)
« Reply #1 on: June 12, 2008, 09:46:08 pm »
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Stickied!
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#### AppleXY

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##### Recreational Problems (MM level)
« Reply #2 on: June 12, 2008, 09:59:15 pm »
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uhm? Ok.... it's not hard lol.

solve for x and sub in.

$x^2 + 1 = 9x$

$x^2 - 9x + 1 = 0$

$\mbox{find solutions for x}$

$x =\pm \frac{\sqrt{77} + 9}{2}$

sub in to $x^2 + \frac{1}{x^2}$

$(\frac{\sqrt{77} + 9}{2})^2) + \left(\frac{1}{\frac{(\sqrt{77} + 9}{2})^2}}\right)$

etc etc,

x = 79 for +ve and -ve

MOD EDIT: prav you need to use "\frac" and "\sqrt", and +- is "\pm"
« Last Edit: June 12, 2008, 10:10:52 pm by AppleXY »
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#### unknown id

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##### Recreational Problems (MM level)
« Reply #3 on: June 12, 2008, 10:10:48 pm »
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$\left(x + \frac{1}{x} \right)^2 = 9^2$

$x^2 + 2 \cdot x \cdot \frac{1}{x} + \frac{1}{x^2} = 81$

$x^2 + \frac{1}{x^2} = 81 - 2$

$x^2 + \frac{1}{x^2} = 79$
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#### Flaming_Arrow

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##### Recreational Problems (MM level)
« Reply #4 on: June 12, 2008, 10:56:19 pm »
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yep solved by applexy and unknown id
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#### Flaming_Arrow

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##### Recreational Problems (MM level)
« Reply #5 on: June 14, 2008, 11:36:34 am »
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come on guys post some questions
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##### Recreational Problems (MM level)
« Reply #6 on: June 14, 2008, 12:32:52 pm »
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Show that $n^4 + 2n^3 + 2n^2 + 2n + 1$ for $n = 1,2,3,\ldots$ is never the square of an integer.
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#### Mao

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##### Recreational Problems (MM level)
« Reply #7 on: June 14, 2008, 12:48:54 pm »
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Show that $n^4 + 2n^3 + 2n^2 + 2n + 1$ for $n = 1,2,3,\ldots$ is never the square of an integer.

[IGNORE ALL THIS, COMPLETELY WRONG ]

in order for this to be a square, this function need to be able to be represented as $[f(n)]^2$ for a particular value of n

initial factorising gives

\begin{align*}
n^4 + 2n^3 + 2n^2 + 2n + 1 &= n^3(n+2)+n^2+n(n+2)+1 \\
&= (n^3+n)(n+2)+(n^2+1)\\
&= (n^2+1)(n^2+2n+1)\\
\end{align*}

$n^2+1$ cannot be further factorised [without going into imaginary, hence for that expression to represent a square of an integer, $(n+1)(n+2)=n^2+1$

$\implies n^2+2n+1=n^2+1 \implies 2n=0 \implies n=0$, NOT a positive integer.

hence, for $n=1,2,3...$, $n^4 + 2n^3 + 2n^2 + 2n + 1$ is never the square of an integer
« Last Edit: June 14, 2008, 07:00:11 pm by Mao »
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##### Recreational Problems (MM level)
« Reply #8 on: June 14, 2008, 12:56:13 pm »
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Hmm, Mao, $x^2 - 2x + 4$ cannot be expressed as $[f(x)]^2$ (without leaving the integers) yet it is a square when $x=2$.

If you prove that an expression cannot be expressed as a square, it means it can not be a square for every x. But it's still possible that it's a square for a particular value of x.

I also thought I'd add 1 more problem,

Let $f_n$ be the familiar fibonacci numbers (with $f_0 = 0,\; f_1 = 1$).

Find $\frac{f_1}{10} + \frac{f_2}{10^2} + \frac{f_3}{10^3} + \ldots$

Back to chemistry study for me
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#### Neobeo

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##### Recreational Problems (MM level)
« Reply #9 on: June 14, 2008, 01:15:14 pm »
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Let $f_n$ be the familiar fibonacci numbers (with $f_0 = 0,\; f_1 = 1$).

Find $\frac{f_1}{10} + \frac{f_2}{10^2} + \frac{f_3}{10^3} + \ldots$

Let $n=\frac{f_1}{10} + \frac{f_2}{10^2} + \frac{f_3}{10^3} + \ldots$

Then
\begin{align*}
10n &= f_1 + \frac{f_2}{10} + \frac{f_3}{10^2} + \frac{f_4}{10^3} + \ldots \\
&= 1 + \frac{f_2}{10} + \frac{f_3}{10^2} + \frac{f_4}{10^3} + \ldots
\end{align*}

Giving us
\begin{align*}
n + 10n &= 1 + (\frac{f_1}{10} + \frac{f_2}{10}) + (\frac{f_2}{10^2} + \frac{f_3}{10^2}) + (\frac{f_3}{10^3} + \frac{f_4}{10^3}) + \ldots \\
&= 1 + \frac{f_3}{10} + \frac{f_4}{10^2} + \frac{f_5}{10^3} + \ldots
\end{align*}

Also
\begin{align*}
100n &= 10f_1 + f_2 + \frac{f_3}{10} + \frac{f_4}{10^2} + \frac{f_5}{10^3} + \ldots \\
&= 10 + 1 + \frac{f_3}{10} + \frac{f_4}{10^2} + \frac{f_5}{10^3} + \ldots \\
&= 10 + 11n \mbox{ (from above)}
\end{align*}

$\therefore n = \frac{10}{89}$

I should come up with a new problem shortly. But study comes first!
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#### bigtick

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##### Recreational Problems (MM level)
« Reply #10 on: June 14, 2008, 05:39:36 pm »
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=(n^2+an+1)^2, no rational a will satisfy
« Last Edit: June 14, 2008, 05:42:48 pm by bigtick »

#### Mao

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##### Recreational Problems (MM level)
« Reply #11 on: June 14, 2008, 06:20:15 pm »
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Hmm, Mao, $x^2 - 2x + 4$ cannot be expressed as $[f(x)]^2$ (without leaving the integers) yet it is a square when $x=2$.

If you prove that an expression cannot be expressed as a square, it means it can not be a square for every x. But it's still possible that it's a square for a particular value of x.

but i did try to prove for a particular value of n!

[my basic principle is, if it is a perfect square for a particular value of n, then at some point it must be able to be expressed as a perfect square]

that is, at some point, $n^2+1=n(n+1)+1$, which is false for integer values of n.
« Last Edit: June 14, 2008, 06:23:48 pm by Mao »
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#### Neobeo

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##### Recreational Problems (MM level)
« Reply #12 on: June 14, 2008, 06:24:37 pm »
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$\implies n^2+2n+1=n^2+1 \implies 2n=0 \implies n=0$, NOT a positive integer.

You don't necessarily have to multiply two equal numbers to get a square.
$\mbox{e.g. }8\times18=12^2$
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#### Mao

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##### Recreational Problems (MM level)
« Reply #13 on: June 14, 2008, 07:07:34 pm »
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IN HINDSIGHT (thanks to dcc):

$\implies (n+1)^2(n^2+1)$

let this be a perfect square:

$(n+1)^2(n^2+1)=k^2$

$\implies (n+1)\sqrt{n^2+1}=\pm k$ where k is an integer

hence, we must require $\sqrt{n^2+1}$ to be rational.

but if we look at $n=1,2,3,4,5.....$, $n^2=1,4,9,16,25.....$, with the difference between the terms increasing. hence $n^2+1$ will not be a perfect square for the designated set of n, and $\sqrt{n^2+1}$ will be irrational for $n=1,2,3,4,5.....$

hence, it cannot be a perfect square

[i think that's what dcc meant ]
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#### Mao

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##### Recreational Problems (MM level)
« Reply #14 on: June 14, 2008, 07:19:42 pm »
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$\begin{array}{ccccc}
(1-\lambda)x_1 & + & 2x_2 & = & 0\\
x_1 & - & \lambda x_2 & = & 0\\
\end{array}$

find $\lambda$ such that the above system has non-zero solutions for $(x_1,x_2)$
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