TITLE: Does this system of simultaneous Pell-like equations have any non-trivial positive integer solutions?
QUESTION [21 upvotes]: Let $a,b,c$ be positive integers satisfying
\begin{align}
2a^2-1 &= b^2,  \\
2a^2+1 &= 3c^2.
\end{align}
The trivial solution is $(a,b,c)=(1,1,1)$. Are there others?

REPLY [10 votes]: ${\bf Summary:}$ We prove that for any positive integer solution $a$, $b$ and $c$ to the given equations there exist $x,y\in \Bbb{Z}$ such that
$$
4a=x^2+3y^2,\quad b=\frac{|x^2-3y^2|}{2}\quad\text{and}\quad c=|xy|.
$$
We insert these values into the second equation and obtain
$(3(x^2-y^2))^2=8(x^4-1)$ which implies $x^2=y^2=1$, hence $a=b=c=1$. 
${\bf Edit:}$ Thanks to a hint of Zhoraster to the rational slope method, the first part has been simplified.
Assume $a$, $b$ and $c$ are an integer solution to the given equations. Adding the two equations we obtain:
$$
4a^2=b^2+3c^2\quad\Rightarrow\quad 1=\left(\frac{b}{2a}\right)^2+3\left(\frac{c}{2a}\right)^2.
$$
So $\left(\frac{b}{2a},\frac c{2a}\right)$ is a rational point of the ellipse
$$
1=u^2+3v^2.
$$
If $u,v\in\Bbb{Q}$, then the slope of the line through $(u,v)$ and $(1,0)$ is rational, i.e., there exists $t\in\Bbb{Q}$ with $t(u-1)=v$. 
But then, if $(u,v)$ is on the ellipse, 
$$
1=u^2+3(t(u-1))^2\quad \Rightarrow\quad (u-1)(3t^2(u-1)+u+1)=0.
$$
If $u\ne 1$, we obtain 
$$
u=\frac{3t^2-1}{3t^2+1}\quad\text{and}\quad v=t(u-1)=-\frac{2t}{3t^2+1},
$$ 
so all rational points of the ellipse different from $(1,0)$ are parametrized by $t\in \Bbb{Q}$. Write $t=\frac yx$ for some coprime integers $x,y$,
then 
$$
u=\frac{3y^2-x^2}{3y^2+x^2}\quad\text{and}\quad v=t(u-1)=-\frac{2xy}{3y^2+x^2}.
$$
In particular, for $(u,v)=\left(\frac{b}{2a},\frac c{2a}\right)\ne (1,0)$ there exist $x,y$ coprime integers, such that
$$
\frac{b}{2a}=\frac{3y^2-x^2}{3y^2+x^2}\quad\text{and}\quad \frac{c}{2a}=-\frac{2xy}{3y^2+x^2}.
$$
Now $x,y$ cannot be both even, since they are coprime, and if one is even and the other is odd, then $3y^2-x^2$ and $3y^2+x^2$ are both odd, which
contradicts $\frac{b}{2a}=\frac{3y^2-x^2}{3y^2+x^2}$, since $2\not|b$. So $x$ and $y$ are both odd. Moreover, if $3|x$, then we can set
$x_1=y$, $y_1=\frac x3$, and then $x_1, y_1$ are coprime, $3\not| x_1$ and
$$
\frac{b}{2a}=\frac{x_1^2-3y_1^2}{3y_1^2+x_1^2}\quad\text{and}\quad \frac{c}{2a}=-\frac{2x_1y_1}{3y_1^2+x_1^2}.
$$
So, we have achieved the following result: If $a$, $b$ and $c$ are integer solutions to the given equations, there exists $x,y$ coprime odd integers,
such that $3\not| x$ and such that
$$
\frac{b}{2a}=\pm\frac{3y^2-x^2}{3y^2+x^2}\quad\text{and}\quad \frac{c}{2a}=-\frac{2xy}{3y^2+x^2}.
$$
Since $gcd((3y^2+x^2)/2,xy)=1=gcd(c,2a)$, we have 
$$
c=\pm xy, \quad \pm 4a=3y^2+x^2\quad\text{and}\quad b=\pm\frac{3y^2-x^2}2.
$$
Now we insert these values into the second equation $2a^2+1=3c^2$ (or in the equivalent equation $16a^2+8=24 c^2$) and obtain
$$
(x^2+3y^2)^2+8=24x^2y^2\quad \Leftrightarrow \quad (3(x^2-y^2))^2=8(x^4-1).
$$
By the Lemma below, in $\Bbb{Z}$ the solutions of this equation satisfy $x^2=1$ and $x^2=y^2$, 
hence $(a,b,c)=(1,1,1)$ is the only positive integer solution.
${\bf Lemma:}$ The only two integer solutions of $R^2=8(x^4-1)$ satisfy $R=0$ and $x^2=1$.
${\bf Proof:}$ Evidently $R=0$ and $x^2=1$ give solutions. Assume by contradiction that $x\ne \pm 1$. We also can assume that $x>0$. We have
$$
R^2=2^3(x^2+1)(x+1)(x-1),
$$
and clearly $x$ is odd (else $2^3$ would be a square). 
Since 
$$
gcd(x^2+1,x+1)=gcd(x^2+1,x-1)=gcd(x+1,x-1)=2,
$$
any prime $p>2$ which divides $x^2+1$ cannot divide $x+1$ or $x-1$, hence $x^2+1=2^i r^2$ for some $i,r\ge 1$, with $r$ odd.
Similarly $x+1=2^js^2$ and $x-1=2^k t^2$ for some $j,k,s,t\ge 1$, with $s,t$ odd.
Moreover, since $2^{i+j+k+3}$ is a square, necessarily $i+j+k$ is odd. But, if $i$ is even, then $x^2+1=(2^{i/2}r)^2$, which is impossible,
since $x\ne 0$. So $i$ is odd and $j+k$ is even.
We have
$$
2=x+1-(x-1)=2^j s^2-2^k t^2\quad \Rightarrow\quad 1=2^{j-1} s^2-2^{k-1} t^2,
$$
and so $j=1$ or $k=1$, hence both $j$ and $k$ are odd. 
But then $1=(2^{(j-1)/2}s)^2-(2^{(k-1)/2}t)^2$, which is impossible, since $t>0$. 
This contradiction concludes the proof. $\Box$