So Many Years!

It is given that:

$s - \dfrac {sx} {2013} - 1 = 0$

$\Rightarrow 2013s -sx = 2013 \quad \Rightarrow s(2013 - x) = 2013$

For integer solutions, both $s$ and $(2013-x)$ are integers and since $2013 = 3 \times 11 \times 61$ , there are $8$ possible unordered pairs:

• $(1, 2013) \quad (-1, -2013)$
• $(3, 671) \quad (-3, -671)$
• $(11, 183) \quad (-11, -183)$
• $(33, 61) \quad (-33,-61)$

For each of these $8$ unordered pair there are $2$ solutions for $s$ and $x$ and therefore are $\boxed {16}$ .

The equation can be rewritten as:

$\frac { s-1 }{ s } =\frac { x }{ 2013 }$

So the ratio between $x$ and $2013$ should be equal to the ratio between $s-1$ and $s$ . Since $s$ must be an integer, it would be a factor of $2013$ or a multiplication of factors of $2013$ . Since $x$ is merely

$(s-1)($ factor or multiplication of the factors of 2013 $)$

and that the "factor or multiplication of the factors of 2013" is definitely an integer, $x$ would always be an integer if $s$ is an integer.

Listing the factors of 2013:

$2013=3*11*61$

The number of possible ways of pairing the $3$ factors $3*11*61$ is $7$ ways:

$3, 11, 61, 3*11, 3*61, 11*61, 11*61*3$

It also has to be noted that the value of $s$ can also be $1$ as $2013$ can be divisible by $1$ .

It also has to be noted that the value of $s$ can be negative.

Therefore, the total number of solutions is $2(7+1)=\boxed{16}$

Using more (simple) algebra, one can proof that $s$ cannot be more than $2013$

For those who have more general solutions, please post them too.

s= 2013 /2013-x; s will be integer if and only if 2013-x is some integer factor of 2013 that integer factor may be positive or negative 2013 factors are 1, 3, 11, 33, 61, 183, 671, 2013 -1, -3, -11, -33, -61, -183, -671, -2013(considering negative also because in the question he mentioned it is integer i.e can be negative also ) therefore total factors are [16] therefore there are 16 ordered pairs possible for the given problem