With $x = \sqrt{ \dfrac{97n+2}{2n-1} } $, find $n\in\mathbb{N}$ such that $x \in \mathbb{N}$.

Question

I have the number:

$$x = \sqrt{ \dfrac{97n+2}{2n-1} } $$

and I have to find $n \in \mathbb{N}$ such that the number $x$ is natural. I know that in order for this to happen, we must have:

1. $\dfrac{97n+2}{2n-1}$ is a perfect square.
2. $(2n-1) | (97n+2)$

but I don't know how to use this knowledge to find the values of $n \in \mathbb{N}$ for which $ x\in \mathbb{N}$.

Answer 1

The only value of n possible here is n=51 when x=7. The solution goes like

Rearrange to get $n=\frac{x^2+2}{2x^2-97}$

$n=\frac{1}{2}\frac{2x^2-97+101}{2x^2-97}$

$n=\frac{1}{2}+\frac{1}{2}\frac{101}{2x^2-97}$

Now,$2x^2-97>0$ and $2x^2-97< 101$ so possible values of x are 7,8,9. Put the value of x in the equation and select all the values for which n is a natural number. The only value of n possible here is n=51 when x=7

Answer 2

So you need $\frac {97n +2}{2n-1} = k^2$

We have $97n + 2 = \frac {97}2(2n-1) + \frac {97}2 + 2 = \frac {97}2(2n-1)+\frac{101}2$.

So $\frac {97n+2}{2n-1} = \frac {97}2 + \frac {101}{2(2n-1)}$ must be an integer which means $97 + \frac {101}{2n-1}$ must be an even integer which means $\frac {101}{2n-1}$ must be an odd integer.

But $101$ is prime so that means $2n-1 = 101$ and $n =51$ or that $2n-1=1$ and $n =1$. So $\frac {97n+2}{2n-1}=\frac {97*51+2}{101}=\frac {4949}{101}= 49$ which is a square, or $\frac {97n+2}{2n-1}=\frac {97*1+2}1=99$ where is not a square.

So our only option is $n=51$ and $k = \pm 7$.

So $n = 51$ is the only solution.