Prove that for any integers $a,b,c,$ there exists a positive integer $n$ such that the number $n^3+an^2+bn+c$ is not a perfect square. 
Question: Prove that for any integers $a,b,c,$ there exists a positive integer $n$ such that the number $n^3+an^2+bn+c$ is not a perfect square. 

Solution: Let $f:\mathbb{N}\to\mathbb{Z}$ be such that $$f(n)=n^3+an^2+bn+c, \forall n\in\mathbb{N}.$$ 
Also assume for the sake of contradiction that $f(n)$ is a perfect square $\forall n\in\mathbb{N}$.
We have $f(1)=1+a+b+c, f(2)=8+4a+2b+c, f(3)=27+9a+3b+c$ and $f(4)=64+16a+4b+c$. 
Now since $f(4)$ is a perfect square $\implies f(4)\equiv 0,1\pmod 4\implies c\equiv 0,1\pmod 4.$
First let that $c\equiv 0 \pmod 4$. Then $f(2)\equiv 0\pmod 4\implies 2b\equiv 0\pmod 4\implies b\equiv 0,2\pmod 4.$ 
Also we have $f(1)\equiv 0 \pmod 4$. Now we have $b+c\equiv 0,2 \pmod 4\implies 1+b+c\equiv -1,1\pmod 4.$ Thus we have $a\equiv -1,1\pmod 4$. 
Also $f(3)\equiv 0\pmod 4$.
Now we have $f(3)-2f(2)+f(1)\equiv 0 \pmod 4\implies 12+2a\equiv 0\pmod 4\implies 2a \equiv 0\pmod 4 \implies a\equiv 0,2\pmod 4.$ But we have $a\equiv -1,1\pmod 4$, which is a contradiction. Thus it is not true that $f(n)$ is a perfect square $\forall n\in\mathbb{N}$ when $c\equiv 0 \pmod 4$. 
A similar analysis for $c\equiv 1\pmod 4$ will lead to a contradiction. Thus it is not true that $f(n)$ is a perfect square $\forall n\in\mathbb{N}$ when $c\equiv 1\pmod 4$.
Hence it is not true that $f(n)$ is a perfect square $\forall n\in\mathbb{N}$ in any case, i.e., $\exists n\in\mathbb{N}$ such that $f(n)$ is not a perfect square. 
Is there any better way to solve this problem? 
 A: Let $a,b,c \in \mathbb Z$, and let $f(n)=n^3+an^2+bn+c$, $n \in \mathbb N$. We show that at least one of $f(1)$, $f(2)$, $f(3)$, $f(4)$ is not a perfect square. We use the fact that $m^2 \equiv 0\:\text{or}\:1\pmod{4}$ for $m \in \mathbb Z$.
Suppose $f(n)$ is a perfect square, $n \in \{1,2,3,4\}$. We note that
$$ \begin{eqnarray*} f(1) \equiv a+b+c+1\pmod{4}, \\
f(2) \equiv 2b+c \pmod{4}, \\ 
f(3) \equiv a+3b+c+3 \pmod{4}, \\
f(4) \equiv c \pmod{4}. \end{eqnarray*} $$
Since $f(3)-f(1)$, $f(4)-f(2)$ are both even, each must be divisible by $4$. But then $4$ must divide both $2b$ and $2(b+1)$. This is impossible.
Therefore, at least one of $f(1)$, $f(2)$, $f(3)$, $f(4)$ must be a non-square, as claimed. $\blacksquare$
A: Let's put:
$$f(n)=n^3+an^2+bn+c$$
Suppose the opposite: that for some $a,b,c$ function $f(n)$ is always a perfect square for every $n$.
It means that, for example:
$$f(n-1)=(n-1)^3+a(n-1)^2+b(n-1)+c=p^2\tag{1}$$
$$f(n+1)=(n+1)^3+a(n+1)^2+b(n+1)+c=q^2\tag{2}$$
...with $p,q$ being integers. Now subtract (1) from (2) and you get:
$$f(n+1)-f(n-1)=6n^2+2+4an+2b=q^2-p^2$$
$$2(3n^2+1+2an+b)=(q-p)(q+p)$$
Obviously, $p,q$  must be either both odd or both even. In both cases the RHS is divisible by 4. It follows that:
$$2\mid3n^2+b+(2an+1)\tag{3}$$
It is obvious that $2an+1$ is always odd so from (3) it follows that for all $n$:
$$2\nmid 3n^2+b\tag{4}$$
But this is impossible: for odd $b$ take any odd value of $n$ and (4) does not hold. If $b$ is even take any even value of $n$ and (4) does not hold. So for some values of $n$ (4) and consecutively (3) cannot be true and our assumptions that $f(n-1)$ and $f(n+1)$ are both perfect squares does not hold.
