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.

Essentially, you know $x^2(n+1)=y^2(n+15)$, let $m=n+1$ for convenience, then you know that since $x$ and $y$ are relatively prime, $y^2 \mid m$
and $x^2\mid m+14$. Since $\frac{m+14}{m}=\frac{x^2}{y^2}$, we then have
that $m=y^2a$ for some $a$ and $m+14=x^2a$ for that same $a$. Then $y^2a+14=x^2a$. Then $14=(x^2-y^2)a$. Now there are only a few possibilities. $x^2-y^2=1,2,7,14$. But we factor, so that we have $(x-y)(x+y) =1,2,7,14$. Then if $(x-y)(x+y)=2k$ for an odd number $k$, we have that since $2$ divides exactly one of the factors, $(x-y)+(x+y)$ is odd, but $x-y+x+y=2x$, which is even, contradiction. Thus $x^2-y^2=7$ or $x^2-y^2=1$. In the first case no matter how we assign $x+y$ and $x-y$ to $1$ and $7$ or $-1$ and $-7$, their difference is always $\pm 6$, so that $y=\pm 3$. But then $m=3^2\cdot 2=18$, which is fine, so now we just need to check the other case, $x^2-y^2=1$. The difference of the factors will always be $0$, since if $(x+y)(x-y)=1$, $x+y=x-y$. Thus $y=0$. But then $m=0$, which is impossible.
Hence $m$ is always 18, or $n$ is always 17.
