$p(x)=ax^2+bx+c$ sends all integers to perfect squares, whether the polynomial is perfect square? $p(x)=ax^2+bx+c$ sends all integers to perfect squares. I wonder whether $p(x)$ is prefect square?
I only know that $a$ & $b$ should be half integers. But how to use perfect square numbers here? 
 A: First, let's establish that $a,b,c$ are integers.  As observed, we have $2a,2b,c\in\mathbb{Z}$ from the fact that $p$ takes integer values.
Since $p(4)=16a+4b+c$ is a square, it is $0$ or $1$ modulo $4$.  But $c$ is a square, and therefore also $0$ or $1$ modulo $4$, and $4b$ is an even integer, therefore $4b\equiv 0 \pmod{4}$ and $b\in\mathbb{Z}$.
Similarly, $p(2)=4a+2b+c$ is a square, and $2b+c$ is $0$ or $1$ modulo $4$, so $a\in\mathbb{Z}$.
Now we can follow the argument here: choose a prime $q$ such that $q \mid p(n)$ for some $n\in\mathbb{Z}$, large enough so that $q$ does not divide the resultant $-a(b^2-4ac)$ of $p(x)$ and $p'(x)$.
Then $q^2 \mid p(n)$, because $p(n)$ is a square.  But $p(n+q)\equiv p(n)+q p'(n) \pmod{q^2}$.  So $q\mid p(n+q)$, therefore $q^2\mid p(n+q)$, and $q\mid p'(n)$, so that $p$ and $p'$ have a common root modulo $q$, contradicting the assumption that $q$ does not divide the resultant.
The only way this could have happened is if the resultant were zero, which happens precisely when $p(x)$ has a double root, i.e. is the square of a linear polynomial.

Apparently it is an open problem in algebraic geometry to determine how many consecutive square values a quadratic may take.  But here is a quadratic that takes $8$ consecutive square values, at $x=-1,0,1,2,3,4,5,6$:
$$p(x) = -420x^2 + 2100x + 2809$$
