Difficult implications of a detail in an(other) approach to the simultaneous Pell equations $24a^2+1=t^2$ and $48a^2+1=u^2$ In reviewing & improving my attempt to solve the question on simultaneous Pell-equations (see here in MSE)
I came across a detail which has surely wider implications and which I cannot encompass (correct term?) so far.
The original question is about:
How many solutions (and which) have the simultaneous Pell equations
$$ 24a^2+1=t^2 \\ 48a^2+1=u^2 \tag 1
$$

My ansatz, first part: I approached this problem the following way. First look at the difference equation:
$$ 24a^2=u^2-t^2 \tag 2 $$
and use the method of parametrization of the pythagorean triples for $b^2 = c^2-a^2$ [which comes out to be $(a,b,c)=(n^2-m^2,2nm, n^2+m^2)$ (wikipedia)] such that I arrive at
$$ u=2m^2+3n^2 \\ t=2m^2-3n^2 \\ a= nm \tag 3
$$
Expansion shows that it is indeed a solution for (2)
$$ 24m^2n^2 \overset{?}= (2m^2+3n^2)^2-(2m^2-3n^2)^2 = 12m^2n^2- (-12m^2n^2) = 24m^2n^2 \tag 4
$$
But while this shows solutions for the difference-eq (2) it is easy to see, that this parametric solution for $(a,b,c)$ cannot at the same time be a solution for the first equation
$$   24 a^2+1 = t^2 \tag {5a}
$$
which leads to
$$ 24(n^2m^2)  =(2m^2-3n^2)^2 -1 \tag {5b}$$
This has no solution which can be seen when taken $\pmod 5$.
 $\qquad \qquad \qquad $update2: this is corrected from previous version after helpful comment of user servaes, see revision history 
So my conclusion on my scribble notes was first: no solutions, simply :-)) ...
But actually there are known solutions for the simultaneous system, namely,
$(a,t,u)=(0,\pm1,\pm1)$ and $(\pm1,\pm5,\pm7)$ and thus it looks, as if my approach is useless here and should be discarded.

My ansatz, second part: In a second look on it, after some coffee,  I see that this ansatz can be revived, if we allow non-integer values for $(m,n)$. For instance

*

*$(m,n)=(\sqrt 3,\frac1{\sqrt 3})$ gives $(a,t,u)=(\frac {\sqrt 3}{\sqrt 3},2\cdot 3-\frac33, 2\cdot 3+ \frac33 )=(1,5,7)$ .

*Analoguously $(m,n)=(\frac1{\sqrt 2},\sqrt 2,)$ gives $(a,t,u)=(\frac {\sqrt 2}{\sqrt 2},\frac22-3\cdot2, \frac22+ 3 \cdot2)=(1,-5,7)$ .

In this example, the values $\sqrt3$ and $\sqrt 2$ are quite obvious when eq's (3) are considered. But to allow irrational numbers in that well known and such commonly used parametrization of the pythagorean triples shall likely have many more implications- but which I do not see here, even if we uset the restriction that the irrational values have to be selected in a way that $(a,t,u)$ are still all integer.
This is surely of importance, if I want to use the full logic of my small approach for other, but comparable, cases like $Aa^2+1=t^2 ; Ba^2+1=u^2$ .

Q1: can the introduction of that irrational values be understood as meaningful generalization of the Pythagorean-triple parametrization?


Q2: under the assumption of quadratic irrational values for $(n,m)$ - can we still prove, that the set of solutions is that set of the two which are already known?

 A: I'm an amateur and I didn't understand parts of the question but the solutions for these simultaneous equations is not difficult so it is logical to begin there.
$$\text{Given }\quad 24a^2+1=t^2\quad\text{and}\quad  48a^2+1=u^2$$
We first eliminate $a$ and then find pair values for the other two variables.
$$24a^2+1=t^2\implies 48a^2+2- 2t^2=0\\
48a^2+1=u^2\implies 48a^2+1-u^2=0
$$
$$(48a^2+2-2t^2)-\space (48a^2+1-u^2)
=(u^2+1-2t^2)=0
$$
$$
(u^2+1-2t^2)=0\\
 \implies  u=\pm\sqrt{2t^2-1}\quad\text{ for }\quad   
t\le -\dfrac{1}{\sqrt{2}},\space
t\ge \dfrac{1}{\sqrt{2}}\quad\\
\quad u=\pm\sqrt{2t^2-1}
\implies u=0\iff t=\pm\dfrac{1}{\sqrt{2}}\\
\quad u=\pm\sqrt{2t^2-1}
\implies u=\pm1\iff t=\pm1\\
\quad u=\pm\sqrt{2t^2-1}
\implies u=\pm 7\iff t=\pm 5
$$
By inspection we can see these easy solutions are valid and that there are "probably" infinite solutions that are not so easy. Still it is an easy exercise to substitute these  values into the original equations to find some of the $a$-values.
\begin{align*}
 24a^2+1-t^2=0& ,\space t=\pm\dfrac{1}{\sqrt{2}}
\implies  a = ± \frac{i}{4 \sqrt{3}}\\
24a^2+1-t^2=0& ,\space t=\pm1\qquad\implies a = 0\\
24a^2+1-t^2=0& ,\space t=\pm5\qquad\implies a = \pm1
\end{align*}
\begin{align*}
48a^2+1-u^2=0& ,\space u=0 \implies
a = ± \frac{i}{4 \sqrt{3}}\\
48a^2+1-u^2=0& ,\space u=\pm1\implies a = 0\\
48a^2+1-u^2=0& ,\space u=\pm7\implies a =  \pm 1
\end{align*}
So these easy solutions represent the following triples.
$$(\pm\frac{i}{4 \sqrt{3}},  \pm\dfrac{1}{\sqrt{2}}, 0)
\qquad(0, \pm1, \pm1)\qquad
(\pm1, \pm5, \pm 7)$$
A few other triples found in Excel are
$$ (\pm10,\pm49,\pm\sqrt{4801})\\
   (\pm 14, \pm\sqrt{4705}, \pm 97)\\
   (\pm\sqrt{35}, \pm 29,\pm 41)\\
   (\pm\sqrt{1190}, \pm 169, \pm 239)\\
   (\pm\sqrt{40426},\pm 985,\pm 1393)
$$
A1: The introduction of irrational  parameters is not meaningful for  Pythagorean triples because all such triples are integers.
A2: the introduction of irrationals  provides strong evidence but cannot prove that the only integers solutions are those  two already found.
