<s>[INST] Solving a pair of simultaneous Pell-like equations I'm trying to prove the following.

Conjecture: If $a,b,c$ are positive integers satisfying the system of equations
  \begin{align}
  a^2+3b^2 &= 4c^2,  \\
  a^2-3b^2 &= -2,
\end{align}
  then $(a,b,c)=(1,1,1)$.

Unfortunately, all my efforts have ended up going around in circles. I’m assuming there’s a relatively easy proof (likely by descent). Any suggestions/hints would be appreciated.
 [/INST] This is the same problem you once posted 3 years ago. Replacing variable names to avoid confusion, the old system is
\begin{align}
  2x^2−1 &= y^2, \\
  2x^2+1 &= 3z^2.
\end{align}
Now put $y=a$, $x=c$, and $z=b$, so that the system becomes
\begin{align}
  2c^2-1 &= a^2,  \\
  2c^2+1 &= 3b^2.
\end{align}
Substitute $2c^2=a^2+1$ (as given by the first equation) into the second to get $a^2+2=3b^2$ or $a^2−3b^2=−2$. Now multiply the first equation by $2$ and equate with $-2$ to get $2a^2-4c^2 = a^2-3b^2$, which is $a^2+3b^2=4c^2$.
</s>