Given that $c$ is a positive integer that satisfies $$x - \frac{1147}{x} = c$$ where $x$ is a rational number, find all possible pairs $(x,c).$
I first factored $1147$ as $31 \cdot 37,$ so I got two pairs $(1147,1),(37,6).$ However, I didn't know if there were any more pairs.
Observe that it is equivalent to look for $(x,c)$ such that $x^2-cx-1147=0$. So, you are looking for RATIONAL solutions of $$X^2-cX-1147=0$$ Now, since $c$ is integer, the Rational Root Theorem says that any RATIONAL root of the above equation must be of the form $p/q$ where $p$ is an integer factor of the constant term and $q$ an integer factor of the leading coefficiente. In our case, $p=\pm1,\pm31,\pm37,\pm1147$ and $q=1$.
So the only possible pairs you need to check are those for which $x=\pm1,\pm31,\pm37,\pm1147$.
$\displaystyle x - \frac{1147}{x} = c \implies x^2 - cx - 1147 = 0, \ c \in \mathbb{Z+}$
If $a, b$ are roots of the quadratic, by Vieta's formula,
$a + b = c, \ ab = - 1147$
As both sum and product of $a$ and $b$ are integers, $a$ and $b$ must be integers.
So given $1147 = 37 \times 31$ or $1147 = 1147 \times 1$ and $c$ is a positive integer, the only possible values of $a$ and $b$ are
$(37, - 31) \text{ for } c = 6$
$(1147, -1) \text { for } c = 1146$.
