<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] The equation is quadratic in $c$.
If $b, c$ are positive integers, we have
$$c = \frac{18b + \sqrt{3[(2b^2+1)^3 - 54]}}{6b^2 + 3}.$$
So, $(2b^2+1)^3 - 54 = 3m^2$ for some positive integer $m$.
Since $3 | 54$ and $3 | 3m^2$, we know that $3 | (2b^2 + 1)$. Let $x = \frac{2b^2+1}{3}$ and $y = \frac{m}{3}$.
We have $x^3 - 2 = y^2$. Since $x$ is a positive integer, we know that $y$ is also a positive integer.
The equation $x^3 - 2 = y^2$ is well-known.
The only positive integer solutions to $x^3 - 2 = y^2$ are $(x, y) = (3, 5)$.
So, $(b, c) = (2, 3)$ is the only positive integer solution.
See:
[1] Theorem 3.4 (page 7), https://kconrad.math.uconn.edu/blurbs/gradnumthy/mordelleqn1.pdf
[2] Solving the diophantine equation $y^{2}=x^{3}-2$
[3] https://en.wikipedia.org/wiki/Mordell_curve
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