When does a certain number is a perfect square

Question

I've the following number:

$$12\left(n-2\right)^2x^3+36\left(n-2\right)x^2-12\left(n-5\right)\left(n-2\right)x+9\left(n-4\right)^2\tag1$$

Now I know that $n\in\mathbb{N}^+$ and $n\ge3$ (and $n$ has a given value) besides that $x\in\mathbb{N}^+$ and $x\ge2$.

I want to check if the number is a perfect square.

Yesterday, this question was answered using the software SageMathCell. And the code that was used is the following:

E = EllipticCurve([0, β, 0, γ, δ])
P = E.integral_points()
for p in P:
    if p[0] % α == 0:
        print(p[0] // α, p[1] // α)

Using $(1)$ I found that:

• $$\alpha=12(n-2)^2\tag2$$
• $$\beta=36(n-2)\tag3$$
• $$\gamma=-144(n-5)(n-2)^3\tag4$$
• $$\delta=1296(n-4)^2(n-2)^4\tag5$$

Now, when I tried $n=71$ it should have found that $x=1585$ is a solution but it gave me nothing in return.

What mistake have I made?

Answer 1

You must first multiply all coefficients by $144(n-2)^4$ in order to bring your equation to elliptic curve form ( the coefficient of $x^3$ and $y^2$ has to be $1$ in particular) this is why in the end you will also mod out and divide by $a$.

To clarify use instead the following in the code: $\beta'=\beta,\gamma'=12(n-2)^2\gamma,\delta'=144(n-2)^4\delta$ and $a'=12(n-2)^2=a$.

In general you have $y^2=ax^3+bx^2+cx+d$, multiplying by $a^2$ then gives $(ay)^2=(ax)^3+b(ax)^2+ac(ax)+a^2d$. So $b'=b,c'=ac,d'=a^2d$.