# 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.

• You want to use divisibility, right? So try clearing out your denominators Sep 22, 2015 at 22:01
• @DanielLittlewood you mean $X^2(n+1)=y^2(n+15)$ ? Sep 22, 2015 at 22:04
• first thing take $m=n+1$ and then you have to consider the cases $d=1,2,7,14$ where $d=\gcd(m,m+14)=\gcd(m,14)$ the problem will be very simplified (use Euclid's lemma if d=1), in other cases simplify and use it again Sep 22, 2015 at 22:07
• @Elaqqad thank you , and how did you know the cases that you have to consider ? Sep 22, 2015 at 22:12
• You do not state what $n$ is supposed to be. Do you want $n\in\mathbb{Z}$? Notice that $n=-33$ works, too.
– robjohn
Sep 22, 2015 at 22:56

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.

• Why can't $x^2-y^2=-1$ or $x^2-y^2=-7$? These both lead to integer values of $n$ for which $\sqrt{\frac{n+15}{n+1}}\in\mathbb{Q}$.
– robjohn
Sep 23, 2015 at 20:49

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, $$\begin{array}{c} \color{#C00000}{n=17}&\text{or}&\color{#00A000}{n=-33}&\text{or}&\color{#0000F0}{n=-15}\\ \color{#C00000}{\Downarrow}&&\color{#00A000}{\Downarrow}&&\color{#0000F0}{\Downarrow}\\ \color{#C00000}{\sqrt{\frac{n+15}{n+1}}=\frac43}&&\color{#00A000}{\sqrt{\frac{n+15}{n+1}}=\frac34}&&\color{#0000F0}{\sqrt{\frac{n+15}{n+1}}=0} \end{array}$$

• Thank you so much for answer ! But in the case of -33 the squart will be negative and this is false because thebsquart is always positive ! Sep 23, 2015 at 10:18
• How can the square root be negative if the square root is always positive? Just plug these values for $n$ into $\sqrt{\frac{n+15}{n+1}}$ and you get non-negative rational values.
– robjohn
Sep 23, 2015 at 12:00
• $\sqrt{\frac{-18}{-32}}=\frac34$
– robjohn
Sep 23, 2015 at 12:16