Find all integer numbers $n$ such that $\frac{11n-5}{n+4}$ is a perfect square.

Question

Find all integer numbers $n$, such that, $$\sqrt{\frac{11n-5}{n+4}}\in \mathbb{N}$$

I really tried but I couldn't guys, help please.

Answer 1

If our square root is to be an integer, we need to have $\frac{11n-5}{n+4}$ a non-negative integer. Note that $$\frac{11n-5}{n+4}=11-\frac{49}{n+4}.$$ So $n+4$ must divide $49$. But $49$ has very few divisors, so there are very few possile integer values of $\frac{49}{n+4}$. Try them all, including the negative ones. For each candidate, check whether the number $11-\frac{49}{n+4}$ is a perfect square.

Answer 2

hint: $\dfrac{11n-5}{n+4} = k^2\to 11n-5=nk^2+4k^2\to n(11-k^2)=5+4k^2= 49+4(k^2-11)\to n = \dfrac{-49}{k^2-11}-4\to k^2-11 = \pm1,\pm7,\pm49.$ From this you can find all possible values of $k$, and then find $n$.