Integer Values of Fractions

Observe that $\frac { n^2-2013} {n+5} =n-5 - \frac {1988} {n+5}$ . Since $n$ is an integer, $n-5$ is an integer too. It follows that $\frac {1988}{n+5}$ should be an integer too. We find that $1988=2^2 \times 7 \times 71$ has $(2+1) \times (1+1) \times (1+1)=12$ positive and 12 negative factors. For each of the factors, we get a different value of $n+5$ . Hence we get 24 different values of $n$ for which $n-5+\frac {1988}{n+5}$ is an integer.

[Slight edits for clarity - Calvin] [Edited for -ve sign, as pointed out by Aaron]

(n²-2013)=n²-25-1988

(n²-25-1988)/(n+5)=(n²-25)/(n+5)-1988/(n+5)

but (n²-25)/(n+5)=n-5 ( always integer) so 1988/(n+5) must also be an integer

1988 has 12 divisors (24 considering both positive and negative values)

follows there are 24 values of (n+5) which makes (n²-2013)/(n+5) an integer -> there are 24 values of n which verifies the condition

factorize it as n-5-1988/(n+5), thus when 1988 can be divided by (n+5), it is an integer. as 1988=2^2 x 7 x 71, thus by Euler's theorem, 1988 has (2+1) x2x2=12 positive integer dividers. consider both negative and positive n, there are 12x2=24 in total.

dividing the numerator with the denominator we get n-5-(1998/n+5) which reduces the problem to find the number of divisors of 1998 which is 12.therefore no of solutions of n is also 12

Since the denominator cannot equal $0$ , we have $n \neq -5$ . Thus,

$\frac {n^2 - 2013} {n+5} = \frac {n^2-25} {n+5} - \frac { 1988} {n+5} = n - 5 - \frac {1988}{n+5} .$

For this expression to be an integer, $n+5$ must be a factor (either positive or negative) of $1988 = 2^2\times 7 \times 71$ . Since $1988$ has $3 \times 2 \times 2 = 12$ positive factors and 12 negative factors, there are $24$ possible values for $n$ .

Let ${n+5} = \ x$ : $\frac{(x-5)^2 -2013}{x} = \frac{x^2 - 10x -1988}{x} = (x-10) - \frac{1998}{x}$ Now denote the number of positive divisors of any non-zero integer ${a}$ as $\psi$ ( ${a}$ ) such that: $\psi(1988) = \psi(2^2)\psi(7)\psi(71) = 3 \times\ 2 \times \ 2 = 12$ Now we include the negative values to get an answer of $\boxed{24}$ possible integer values for ${n}$