Question

The number of pairs of positive integers $(p,q)$ such that $GCD(p,q)=1$ and $\frac{p}{q}+\frac{14q}{9p}$ is an integer are
(correct answer + 2, wrong answer - 0.50)

• A. $12$
• B. infinite
• C. $4$
• D. $9$

Answer 1

The correct option is C $4$

Let $x=\frac{p}{q}$. The problem becomes equivalent to finding all the positive rational numbers $x$ such that $x+\frac{14}{9x}=n$ for some integer $n$

This equation can be rewritten into the quadratic equation
$9x^2-9xn+14=0,$ whose discriminant must be a square number in order for the root $x$ to be a rational number.
i.e. $△=(-9n{)}^2-4\times 9\times 14=k^2$
$\Rightarrow 9n^2-4\times 14=m^2(\text{Let}{m}^2=\frac{k^2}{9})$
$\Rightarrow 9n^2-m^2=2^3\times 7$
$\Rightarrow (3n-m)(3n+m)=2^3\times 7$

We know that $3n-m$ and $3n+m$ must both either be even or odd and since, their product is even, both should be even.
$\begin{array}{cc}3n-m& 3n+m\\ 2& 2^2\times 7\\ 2^2& 2\times 7\end{array}$

This gives us two pairs of $n$ and $m$: $(5,13)$ and $(3,5).$
Plugging them into the original quadratic equation,
$9x^2-9xn+14=0$ and solving for $x$, gives us
$\Rightarrow 9x^2-9xn+14=0$
For $n=5$
$9x^2-45x+14=0$
$\Rightarrow x=\frac{14}{3}$ or $x=\frac{1}{3}$

Now, for $n=3$
Original quadratic equation becomes
$9x^2-27x+14=0$
$\Rightarrow x=\frac{7}{3}$ or $x=\frac{2}{3}$
Therefore, there are four pairs $(a,b)$ that satisfy the given conditions, namely $(1,3),(2,3),(7,3)$ and $(14,3)$.