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?

Question

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?

Answer 1

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.

Answer 2

$6b^2c^2 + 3c^2 - 36bc - 4b^4 - 4b^2 + 53=0$

In order to quadratic equation for $c$ has rational solution, the discriminant must be square number.

$v^2=96b^6+144b^4+72b^2-636$

So, above quartic equation must have rational solutions.
This quartic equation can be transformed to below elliptic curve.

$Y^2=X^3+36X^2+432X-91584$ with $b^2=X/24.$

Using online Magma calculator as follows.
$E$:=EllipticCurve([$0, 36, 0, 432, -91584$])
IntegralPoints($E$)

It says that all integral point is $ [ (96 : -1080 : 1) ]$.
Hence all integral point is $(b,c)=(2,3).$

Answer 3

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

Answer 4

Hint

$$6b^2c^2+3c^2-36bc-4b^4-4b^2+53=(6b^2c^2-36bc+54)-(4b^4+4b^2+1)+3c^2=6(bc-3)^2-(2b^2+1)^2+3c^2=0\\\implies 6(bc-3)^2+3c^2=(2b^2+1)^2$$ Then $3\mid 2b^2+1$. Now proceed like this by checking modulo primes like $2,3$ and so on.