Integral values only

Number Theory Level 4

What is the sum of all integer values of $n$ for which $\frac{n^{2}-9}{n-1}$ is also an integer?

1 solution

Rajen Kapur
Aug 30, 2015

The given expression can be re-written as $(n + 1) - \dfrac{8}{(n - 1)}$ , which is integer for n = -7, -3, -1, 0, 2, 3, 5, and 9. Thus answer is 8.

how you came upon this form of writing the expression ?

Kirti Sharma - 5 years, 6 months ago

Let $n-1=x$ ,

$\frac{n^2-9}{n-1} = \frac{(n-3)(n+3)}{n-1} \\ \frac{(x-2)(x+4)}{x} = \frac{x^2+2x-8}{x} \\ x+2-\frac{8}{x}$

Replacing $n-1=x$ ,

$\therefore n+1-\frac{8}{n-1}$

Akshat Sharda - 5 years, 5 months ago

Same way. But I got it wrong cause I forgot -7

Shreyash Rai - 5 years, 5 months ago

The result can be achieved using polynomial division .

You can also use remainder theorem , which is very easy to use in this example.

Arulx Z - 5 years, 5 months ago

The quantity which is integer for every integer value of n is separated out, in this case (n + 1), as first step to arrive at the solution..

Rajen Kapur - 5 years, 5 months ago