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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.

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    $\begingroup$ This is the same problem you once posted 3 years ago. $\endgroup$ – Hellen Jul 27 '17 at 19:53
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    $\begingroup$ @Hellen: Really? (Talk about going around in circles!!!) How can I show they’re equivalent? $\endgroup$ – Kieren MacMillan Jul 27 '17 at 19:58
  • $\begingroup$ The new $a$ is the old $b$, the new $b$ is the old $c$. $\endgroup$ – Hellen Jul 27 '17 at 19:59
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    $\begingroup$ The old system is, with variables names replaced to avoid confusion, \begin{align} 2x^2-1 &= y^2, \\ 2x^2+1 &= 3z^2. \end{align}Now put $y=a$, $x=c$, and $z=b$.The system becomes \begin{align} 2c^2-1 &= a^2, \\ 2c^2+1 &= 3b^2. \end{align} Replace $2c^2$ from the first equation into the second to get $a^2-2=3b^2$ or $a^2-3b^2=-2$. Now multiply by $2$ the first equation and equate $-2$ with $-2$ to get $2a^2-4c^2=a^2-3b^2$, which gives $a^2+3b^2=4c^2$. $\endgroup$ – Hellen Jul 27 '17 at 20:08
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    $\begingroup$ @Hellen: Please transfer your comment to an answer (including a link to the other question) and I'll accept it. $\endgroup$ – Kieren MacMillan Jul 28 '17 at 0:35
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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 $a^2+3b^2=4c^2$.

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