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? 

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. $\blacksquare$
