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. 



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*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.

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.
