Prove that the expression has infinitely many solutions

Question

Does $m = \dfrac{1}{72}\left(\sqrt{48r^2+1}+1\right)$ have infinitely many solutions in positive integers $(m,r)$?

Does $m = \dfrac{1}{18}\left(\sqrt{48r^2+1}-1\right)$ have infinitely many solutions in positive integers $(m,r)$ where $m \equiv 3 \pmod{4}$?

I wasn't sure how to prove that there are infinitely many solutions, but for both of them we need $48r^2$ to be one less than a perfect square. Thus we need $48r^2 = x^2-1$ and so $x^2-48r^2 = 1$, which has infinitely many solutions, but we also need $m$ to be an integer. How do we ensure that in both cases?

Answer 1

The first has no solutions with positive integer $m$. However, the second case,

$$\begin{aligned} m&=\dfrac{1}{18}\left(\sqrt{48r^2+1}-1\right)\\ &=\dfrac{1}{18}\left(x-1\right)\\ &\equiv {3\pmod{4}} \end{aligned}$$

has infinitely many given by,

$$x= \frac{ (7+\sqrt{48})^{6k+3}+(7-\sqrt{48})^{6k+3}}{2}=1351,\; 9863382151,\; 72010600134783751\dots$$

$$r = \frac{ (7+\sqrt{48})^{6k+3}-(7-\sqrt{48})^{6k+3}}{2\sqrt{48}}=195,\; 1423656585,\; 10393834843080975\dots$$

Answer 2

As for the first equation, rearranging and squaring both sides we get $$72^2m^2-144m=48r^2,$$ which simplifies to $$108m^2-3m=3m(36m-1)=r^2.$$ Then $r^2=9k^2$ and $m=3n$ for some $k,n$, which yields $$n(108n-1)=k^2.$$ Since the factors of the LHS are coprime, both must be squares: say $n=y^2, 108n-1=x^2;$ therefore, $$x^2-108y^2=-1.$$ But $108$ is divisible by $4$, and this excludes the existence of any solution $(x,y)$, hence the original equation itself has no solutions.