A System of Simultaneous Pell Equations 
Are $0$ and $\pm1$ the only integer solutions for which both $\sqrt{24n^2+1}$ and $\sqrt{48n^2+1}$ are simultaneously integers ?


Whilst pondering upon the Biblical concept of the Jubilee year, I couldn't help but notice that a time period of $7^2=49$ years, apart from approximating half of a decimal century of $10^2=100$ years, also lies conspicuously close to a third of a duodecimal “century” of $12^2=144$ years. In other words, I was left with solving the system of Diophantine equations $$\dfrac{x^2}3+1~=~y^2~=~\dfrac{z^2}2-1,$$ which, after a bit of basic number theory, boiled down to the system of Pell equations described above. Then, a Mathematica search of depth up to $10^4$ into the $($ periodic $)$ continued fraction expansion of $~\lim\limits_{n\to\infty}\sqrt{\dfrac{48n^2+1}{24n^2+1}}~=~\sqrt2~$ failed to reveal any other solutions, save for the ones mentioned earlier, implying that any other possible values of n possess at least $3,828$ digits. 



*

*A $2007$ paper by Mihai Cipu and Maurice Mignotte shows that there might be at most one such positive solution, for $n>1.$ 

*In the same year, the $34^{th}$ volume of the Annales Mathematicae et Informaticae, published by the Eszterhazy Karoly University of Applied Sciences, contains a relevant paper by Laszlo Szalay on the resolution of simultaneous Pell equations.
 A: Here a general solution of this problem  using Fermat's method of infinite descent (the simplest version of it currently very sophisticated).
From
$$24n^2+1=x^2\\48n^2+1=y^2$$ we have, since $x$ and $y$ are odd and $x\lt y$
$$(x+2h)^2=y^2\Rightarrow h^2+hx-6n^2=0\Rightarrow h= \frac{-x\pm\sqrt{x^2+24n^2}}{2}=$$ 
Because of $24n^2=x^2-1$ the radicand becomes $2x^2-1$ and must be equal to $t^2$ so we  have $$2x^2-1=t^2\iff1+t^2=2x^2$$
From  identity $$(r^2-s^2+2rs)^2+(r^2-s^2-2rs)^2=2(r^2+s^2)^2$$  comes the general solution of the equation $$X^2+Y^2=2Z^2..........(*) $$ it follows, because $(1,t,x)$ is a solution of $(*)$ 
$$\begin{cases}r^2-s^2+2rs=t \\r^2-s^2-2rs=1\iff2r^2=1+(r+s)^2 \\r^2+s^2=x\Rightarrow r\text{ and }s\lt x \end{cases}$$
 This shows another solution $(1,r+s,r)$ of $(*)$ with $r\text{ and }s\lt x$ so we have $$\begin{cases}r_1^2-s_1^2+2r_1s_1=r+s \\r_1^2-s_1^2-2r_1s_1=1\iff2r_1^2=1+(r_1+s_1)^2 \\r_1^2+s_1^2=r\Rightarrow r_1\text{ and } s_1\lt r\end{cases} $$ Again another solution of $(*)$ with $r_1\text{ and }s_1\lt r$ and the procedure can be iterated how many times you want.
By descent the proof is done.
A: A wild guess: 
Write the equivalent system 
$$x^2= 24 n^2 + 1 \\
y^2 = 48 n^2 + 1$$ so 
$$x^2 + 24 n^2 = y^2$$
This equation has some ( families ) of solutions that are parametrized by some pairs of integers $(a,b)$. Express $x$, $y$ in terms of $a$, $b$ (quadratically) and plug in into the first equation. We get an equation in $a$, $b$ of degree $4$, a Thue equation. There are some effective bounds for the solutions of such an equation, although they may be too large to check up tot them. Perhas this is the method in the last paper you quoted. 
Added: From the first equation we conclude $(x^2, 24 n^2)=1$, and so $(x^2, z^2)=1$. From the third equation we get 
$$(y-x)(y+x)= 24 n^2$$. $y$ and $x$ are both odd. We conclude that $(y-x, y+x)=2$. Therefore, we have several possibilities:
$$\begin{array}{c:c:c:c:c} 
\frac{y-x}{2} & 6 u^2&  3 u^2 & 2 u^2&  u^2 \\
\frac{y+x}{2} & v^2 & 2v^2 & 3 v^2 & 6 v^2 \\
n & u v&  u v&  u v &  u v \end{array} $$
