Finding perfect squares in a quadratic equation

Question

Well, I have a number $\text{p}$ that is given by:

$$\text{p}=9+108x^2(1+x)\tag1$$

I want to find $x\in\mathbb{Z}$ such that $\text{p}$ is a perfect square.

I found no solutions. Is there a way to prove that there aren't any solutions? Except for the obvious ones: $x=0$ and $x=\pm1$.

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My work

Since $\text{p}$ must be a perfect square we can write:

$$\text{p}^2=9+108x^2(1+x)\tag2$$

Now, because $\text{p}^2\ge0$ we need:

$$9+108x^2(1+x)\ge0\space\Longleftrightarrow\space x\ge-\frac{1}{3}\left(1+2^\frac{2}{3}+2^{-\frac{2}{3}}\right)\approx-1.07245\tag3$$

Answer 1

Dividing by $9$, you want to find $x \in \mathbb{Z}$ such that $12x^3 +12x^2 +1$ is a square number.

$x=0$ is one obvious solution. Even if you restrict $x$ to strictly positive integers there is another obvious solution.

(Note that this was posted before Jan added “Except for $x=0$ and $x=1$” to his question)

Answer 2

Dividing by $9$, you'll get something similar to $$p^2 - 1=12 x^2(1+x)$$

First, let's ssume $x$ is odd. Then either $x^2\mid p-1$ or $x^2\mid p+1$. Suppose $x^2\mid p-1$, so that $p=1+kx^2$, then $ k^2x^4+2kx^2=12 x^2(1+x)\Rightarrow k^2x^2-12x-12+2k=0\Rightarrow k\in \{\pm 1, \pm 2\}.$ You should get something similar for the other case.

Now suppose $x$ is even, say $x=2^qy$ with $y$ odd. Then the equation becomes $$p^2 - 1=3\times 4^{q+1} y^2(1+2^qy).$$ Then as before $y^2 \mid p\pm 1$, so that $p=\pm 1+ky^2$. This time, after simplification, we get $$k^2y^2 \pm 2k = 3\times 4^{q+1} (1+2^qy)\Rightarrow k^2y^2 - 3\times 2^{3q+2} y\pm 2k-3\times 4^{q+1}=0.$$ It follows $k^2 \mid 3\times 2^{3q+2} y \Rightarrow k=\pm 2^t,\, $ for some $t=0,1,\ldots, \left[\frac{3q+2}{2}\right].$ But you can easily see that $t\leq 1$.