Need help

I face problem while attempting questions like this one.

\[\huge \frac{n^{2}-9}{n-1}\]

Find sum of all values of $n$ as an integer for which $\frac{n^{2}-9}{n-1}$ is also a integer.

Options are :

$\huge 1)$ 0

$\huge 2)$ 7

$\huge 3)$ 8

$\huge 4)$ 9

Can anyone tell me how to do these type of questions with efficiency?

Thanks in advance.

#NumberTheory #Note #NMTC

Easy Math Editor

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`2 \times 3`	$2 \times 3$
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`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

Note first that $\dfrac{n^{2} - 9}{n - 1} = n + 1 - \dfrac{8}{n - 1}.$

Thus the given expression will be an integer for all $n$ such that $n - 1$ divides $8.$

Since there are $8$ integer divisors of $8,$ namely $\pm 1, \pm 2, \pm 4, \pm 8,$ there will be $8$ values of $n$ for which the given expression is also an integer.

(The $8$ values of $n$ are $-7, -3, -1, 0, 2, 3, 5, 9.$ )

Brian Charlesworth - 5 years, 10 months ago

Thank you again.:-)

Akshat Sharda - 5 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$