Is $A$ a perfect square? Consider
$$A=a^2+2ab^2+b^4-4bc-4b^3,$$
where $a,b,c\in\mathbb{Z}$ and $b\neq0$ such that $b|a$ and $b|c$, so $b|A$.
Now I want to know that Is $A$ a perfect square?
 A: In general, A is obviously not a perfect square. A counterexample is when a = 3, b=1 and c=1 in which case A = 8. However, there are many values of a, b, and c where A is a perfect square. One special case is when b = 1. If b = 1, and you let a + 1 be the hypotenuse of a right triangle whose sides are natural numbers and let $c = \frac{E^{2}}{4}-1$, where E is an even leg of the right triangle, then you can find infinitely many solutions to meet the conditions of your equation. For example, take the Pythagorean triple 5, 4, 3.
a + 1 = 5 which means a = 4
b = 1
$c = \frac{4^{2}}{4}-1=3$
You can view a list of Pythagorean triples here.
A: No, $A$ is not a perfect square for such $a,b,c$ values. As a counterexample, take $a=b=c=1$, for which $A=-4$.

EDIT: The answer below corresponds to the original question "Can $A$ be a perfect square?".
Yes, it can be 0. Let $a=bn$ and $c=bm$. Then:
$$\begin{align}
A&=a^2+2ab^2+b^4-4bc-4b^3\\
&=b^2n^2+2b^3n+b^4-4b^2m-4b^3\\
&=b^2 \big((n+b)^2-4(m+b)\big)
\end{align}$$
So any values that make $(n+b)^2-4(m+b)$ a square are valid ones. So, for instance, any $n=-b$ and $m=-b$ are enough, but you can find many others.
