2
$\begingroup$

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?

$\endgroup$

2 Answers 2

1
$\begingroup$

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,

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

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

$\endgroup$
7
  • 1
    $\begingroup$ How do you know the first has no solutions? $\endgroup$
    – Puzzled417
    Aug 7, 2016 at 21:56
  • $\begingroup$ @Puzzled417: By inspection of the parametric solution to the Pell equation $x^2-48r^2=1$. Some users may give a short and rigorous proof for the non-existence of solutions for the first case. $\endgroup$ Aug 7, 2016 at 22:04
  • $\begingroup$ That is, the Pell equation, since even links to things you can look up yourself are useful. $\endgroup$ Aug 7, 2016 at 22:06
  • $\begingroup$ How did you determine solutions to the second one? $\endgroup$
    – Puzzled417
    Aug 7, 2016 at 22:09
  • $\begingroup$ @Puzzled417: Also by inspection and modular arithmetic. Sometimes the existence of solutions is easier to establish than its non-existence. $\endgroup$ Aug 7, 2016 at 22:13
0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ How do we have $r^2 = 9km$? $\endgroup$
    – Puzzled417
    Aug 7, 2016 at 22:48
  • $\begingroup$ @Puzzled417: We don't, that was a brainfart, sorry. See my edit, please. $\endgroup$ Aug 8, 2016 at 2:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.