Determine whether the equation $ 36x^3+36x^2+12x+1 = c^2$ has solutions for the nonzero integers $x$ and $c$ One idea was the following : $c^2$ is a perfect square so it can be written in the form $( x+a) (x+a)$ for some integer $a$ but since there is no way to write a cubic polynomial as a product of two polynomials of degree one there is no solution. Of course this reasoning is very wrong since the number $c$ can be written in many forms as an integer combination of powers of $x$ if $x$ is an integer. For example , we know there exist integers such that $ x+a = c = px^2+qx+r $ and so , $c^2$ can be expressed as an integer combination of powers of $x$ where there is a nonzero term of degree $3$.
Any suggestions?
Edit : I just graphed it on Desmos and it doesn't look like it has solutions at least for the small values I checked. Is there an analytical way to prove this? Please try to make your methods as elementary as possible as I'm still an undergraduate student.
 A: Just a starting idea, don't know if it helps.
Write $$36x^3 = c^2-(6x+1)^2 = (c-6x-1)(c+6x+1)$$ Say $p$ is a prime which divides $c-6x-1$ and $c+6x+1$ then it also divide $36x^3$ and $12x+2$. If $p\neq 2$ then $$6x\equiv_p -1 \;\;\;\wedge \;\;\;36x^3\equiv _p0$$
So: $$36x^2\equiv_p 1\implies 36x^3\equiv _p x \implies x\equiv_p 0 \implies 0\equiv_p -1$$  which is impossible. Clearly $c$ must be odd so $\gcd(c-6x-1,c+6x+1)=2$ and thus $$ c-6x-1 = 2u \;\;\;\wedge \;\;\; c+6x+1 = 2v$$ where $u,v$ are relatively prime, so $$9x^3 = uv \implies (u=9m^3, \;\;v=n^3,\;\;3\nmid n )  \;\;\;\vee \;\;\;(u=m^3,\;\;v=9n^3,\;\;3\nmid m )$$
where $m,n$ are such relatively prime numbers that $x=mn$. So we have two cases:


*

*First case: $$12x+2 = 2(v-u)\implies \boxed{6mn +1 = n^3-9m^3}$$
but I have no idea how to proceed...

A: There are exactly two rational solutions.
We analyze the zero locus of the the polynomial
$$p(u,v)=v^2-(36u^3+36u^2+12u+1)\text{.}$$
Consider the invertible substitution
$$\begin{align}
u&=\frac{x-3}{9}&v&=\frac{2y+1}{9} \\
x&=9u+3&y&=\frac{9v-1}{2}\text{.}
\end{align}$$
Then
$$81 p(u,v)=4q(x,y)$$
where
$$q(x,y)=y^2+y-x^3+7\text{.}$$
This is the minimal Weierstrass model of elliptic curve 27.a3.
Now consider the substitution
$$\begin{align}
x&=\frac{3}{a+b} & y&=\frac{4a-5b}{a+b}\\
a&=\frac{5+y}{3x} & b &=\frac{4-y}{3x}\text{,}
\end{align}$$
invertible on the zero locus of $q$. Then
$$\frac{q(x,y)}{x^3}=r(a,b)$$
where
$$r(a,b)=a^3+b^3-1\text{.}$$
But (infamously) the only rational zeros of this last equation  are
$(a,b)=(1,0)$ and $(a,b)=(0,1)$. Retracing our steps, the only rational zeros of $p$ are $(u,v)=(0,1)$ and $(u,v)=(0,-1)$.
