<s>[INST] Is there a quick (hopefully elementary) way to prove that $6b^2c^2 + 3c^2 - 36bc - 4b^4 - 4b^2 + 53=0$ has only one solution? I have the Diophantine equation $$6b^2c^2 + 3c^2 - 36bc - 4b^4 - 4b^2 + 53=0.$$ Numerical calculations suggest this has only one positive integer solution, namely $(b,c)=(2,3)$. Is there a quick way to prove or disprove that?
 [/INST] Solving for $c$, we get
$$ c=-{\frac {-18\,b \pm \sqrt {24\,{b}^{6}+36\,{b}^{4}+18\,{b}^{2}-159}
}{3 (2\,{b}^{2}+1)}}$$
If there is to be an integer solution, we need $24\,{b}^{6}+36\,{b}^{4}+18\,{b}^{2}-159$ to be a square.  Taking $s = b^2$, let's look for integer solutions of
$t^2 = 24 s^3 + 36 s^2 + 18 s - 159$, which is an elliptic curve.
Taking $s = x/24 - 1/2$ and $t = y/24$, this becomes $y^2 = x^3 - 93312$: note that if $s$ and $t$ are integers, then $x = 24 s + 12$ and $y = 24 t$ are also integers.
Now $y^2 = x^3 - 93312$ is an elliptic curve in Weierstrass form.  According to Sage, its only integer solutions are $x = 108, y = \pm 1080$.  This corresponds to $s=4$ and $t = \pm 45$.  And indeed $s=4$ gives us $b = \pm 2$; $b=2$ gives us an integer solution with $c=3$, and $b=-2$ gives us an integer solution with $c=-3$.  Those are the only integer solutions.
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