Proving that a list of perfect square numbers is complete

Question

Well, I have a number $n$ that is given by:

$$n=1+12x^2\left(1+x\right)\tag1$$

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

I found the following solutions:

$$\left(x,n\right)=\left\{\left(-1,1^2\right),\left(0,1^2\right),\left(1,5^2\right),\left(4,31^2\right),\left(6,55^2\right)\right\}\tag2$$

Is there a way to prove that this a complete set of solutions? So I mean that the solutions given in formula $(2)$ are the only ones?

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My work:

• We know that: $$ 1 + 12x^2 \left(1+x \right) \ge 0 \space \Longleftrightarrow \space x \ge -\frac{1+2^{-2/3}+2^{2/3}}{3} \approx -1.07245 \tag3 $$ So we know that for $x<-1$ there are definitely no solutions.

Answer 1

$y^2=1+12x^2(1+x) \implies (12 y)^2 = (12 x)^3 + 12 (12 x)^2 + 144$

Magma code for positive $y$ only:

S:= IntegralPoints(EllipticCurve([0,12,0,0,144]));
for s in S do
  x:= s[1]/12;
  if x eq Floor(x) then
    print "(",x,", ",Abs(s[2]/12),")";
  end if;
end for;

Output:

( -1 ,  1 )
( 0 ,  1 )
( 1 ,  5 )
( 4 ,  31 )
( 6 ,  55 )

Answer 2

COMMENT:

n is odd, let $n=(2m+1)^2$ ; m∈Z, then:

$4m(m+1)=12x^2(1+x)$ ⇒ $m(m+1)=3 x^2 (1+x)$

Suppose $m=3t$, then:

$t(3t+1)=x^2(1+x)$

⇒ $n=(2m+1)^2=(6t+1)^2$

For example $n=31^2= (6\times 5 +1)^2$ or $n=55^2=(6\times 9+1)^2$

Therefore there may be more solutions. Now let $m+1=3t$ then:

$(3t-1)3t=3x^2(1+x)$ ⇒ $(3t-1) t=x^2(1+x)$ ⇒ $n=(2m+1)^2= (6t-1)^2$

For example $n=5^2=( 6\times 1 -1)^2$

In this case there may also be some solutions.

Moreover n can be the square of negative numbers, so it can have general forms $n=[±(6t+1)]^2$ or $n=[±(6t-1)]^2$.We have to show whether equation $x^3+x^2=3t^2±t$ can have infinite solutions or limited number of solutions in Z.