When the polynomial $16(n+1)^2(p(x)-1)^3p(x)+1$ is a perfect square

Question

Let $n$ be a positive integer. Let $p(x)$ be a positive polynomial with positive integer coefficients. I am asking when the polynomial $16(n+1)^2(p(x)-1)^3p(x)+1$ is a perfect square, i.e., there exist a positive integer $m$ such that $$ 16(n+1)^2(p(x)-1)^3p(x)+1=m^2 $$

for some values of $x$.

Answer 1

COMMENT:

we consider a certain case. Let $p(x)=t$ , we may write:

$$16(m+1)^2[t-1)^2[(t-1)t]=(m-1)(m+1)$$

Suppose:

$m-1=[4(n+1)(t-1)]^2$

$m+1=t^2-t$

⇒ $t=\frac{1±\sqrt{4m+5}}{2}$

We get :

$(m, t, n)=[(1, 2, -1), (1, -1, -1)], [(5, 3, -3/4), (5, -2, 5/4) $

It can be seen in series:

$m=1, 5, 11, 19, . . . $

only $1-1=0$ and $5-1=4$ are perfect square but for $m=1$, n is not positive integer and for $m=5$, n is a rational number.

Similar result comes out if we let:

$m+1=[4(n+1)(t-1)]^2$

$m-1=t^2-t$

$t=\frac{1±\sqrt{4m-3}}{2}$

For series:

$m=1, 3, 7, 13, 21, . . .$

Corresponding values of t are:

$t=(0, 1), (2, -1), (3, -2), . . .$

Where discriminant is positive, only $3+1=4$ is perfect square which gives $t=(2, -1)$ and $n=(1/2, 3/2), (-3/4, -5/4)$ which is not acceptable.Hence only $p(x)=t=0$ can give solutions if we let $n∈ \mathbb R$.In this case all zeroes of $p(x)$ can provide the condition. We can consider other cases, may be find a suitable t and m.