If $\sqrt{\frac{n+15}{n+1}}\in\mathbb Q$, then $n=17$ How can one show that : 
If $\sqrt{\frac{n+15}{n+1}}\in\mathbb Q$ so $n=17$
I tried using the fact that any number $a\in\mathbb Q$ so $a=\frac{x}{y}$ such that $\gcd(x,y)=1$ 
So $\frac{n+15}{n+1}=\frac{x^2}{y^2}$
But here I'm stuck.

If we let $m=n+1$, the equation
$$
\frac{m+14}{m}=\frac{x^2}{y^2}
$$
is equivalent to
$$
m=\frac{14y^2}{(x-y)(x+y)}
$$
we can assume that $x\ge0$ and $y\gt0$ and $(x,y)=1$. Thus,
$$
(x-y,x+y)\mid2\quad\text{and}\quad(x-y,y)=1\quad\text{and}\quad(x+y,y)=1
$$
If $(x-y,x+y)=2$, then, since $(x-y,y)=1$, $y$ must be odd and the numerator only has one factor of $2$. Since the denominator has two factors of $2$, $m\not\in\mathbb{Z}$. Therefore,
$$
(x-y,x+y)=1
$$
and since either both $x-y$ and $x+y$ are even or both are odd, both must be odd. Because $y\gt0$, we cannot have $x-y=x+y$.
Therefore, since $(x-y)(x+y)\mid7$, we have either
$$
x-y=1\quad\text{and}\quad x+y=7\quad\implies\quad x=4,y=3,\color{#C00000}{m=18}
$$
or
$$
x-y=-1\quad\text{and}\quad x+y=7\quad\implies\quad x=3,y=4,\color{#00A000}{m=-32}
$$
or
$$
x-y=-1\quad\text{and}\quad x+y=1\quad\implies\quad x=0,y=1,\color{#0000F0}{m=-14}
$$
Thus,
$$
\begin{array}{c}
\color{#C00000}{n=17}&\text{or}&\color{#00A000}{n=-33}&\text{or}&\color{#0000F0}{n=-15}\\
\color{#C00000}{\Downarrow}&&\color{#00A000}{\Downarrow}&&\color{#0000F0}{\Downarrow}\\
\color{#C00000}{\sqrt{\frac{n+15}{n+1}}=\frac43}&&\color{#00A000}{\sqrt{\frac{n+15}{n+1}}=\frac34}&&\color{#0000F0}{\sqrt{\frac{n+15}{n+1}}=0}
\end{array}
$$
