Proving that for any three integers $a,b,c$ there exists a positive integer $n$ such that $\sqrt{n^3+an^2+bn+c}$ is not an integer Prove that for any three integers $a,b,c$ there exists a positive integer $n$ such that $\sqrt{n^3+an^2+bn+c}$ is not an integer.
In order to solve this problem I have tried looking at the expression under the radical modulo n. Thus I want to find n such that c is a quadratic non-residue modulo n. For example, if c = 2 (mod 3), since 2 is a non-residue mod 3 we may take n to be 3 and thus the expression is never a perfect square. I need a way to do this for arbitrary c, which I could not find on my own.
 A: I'm not sure offhand how to solve the problem using your particular approach. Instead, here is an alternate method. First, let
$$f(n) = n^3 + an^2 + bn + c \tag{1}\label{eq1A}$$
Note all perfect squares are congruent to either $0$ or $1$ modulo $4$. Thus, the difference of any $2$ perfect squares will be congruent to $-1$, $0$ or $1$ modulo $4$. In particular, it'll never be congruent to $2$ modulo $4$, i.e., have only one factor of $2$.
You don't mention the integers $a$, $b$ and $c$ need to be positive, so there may be some values of $n$ where $f(n)$ in \eqref{eq1A} is negative and, thus, its square root would not even be a real value. In any case, there will always be a positive integer $n_0$ such that for all $n \ge n_0$ we get $f(n) \ge 0$.
For any integers $n_1 \ge n_0$ and $d \gt 0$, we get
$$\begin{equation}\begin{aligned}
f(n_1 + d) - f(n_1) & = ((n_1 + d)^3 + a(n_1 + d)^2 + b(n_1 + d) + c) \\
& \; \; \; \; - (n_1^3 + an_1^2 + bn_1 + c) \\
& = (n_1^3 + 3n_1^2d + 3n_1d^2 + d^3 + an_1^2 + 2an_1d + ad^2 \\
& \; \; \; \; \; \; + bn_1 + bd + c) - (n_1^3 + an_1^2 + bn_1 + c) \\
& = 3n_1^2d + 3n_1d^2 + d^3 + 2an_1d + ad^2 + bd \\
& = d(3n_1^2 + 3n_1d + d^2 + 2an_1 + ad + b)
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Now, consider $d$ to be any even positive integer with just one factor of $2$, e.g., $d = 2$. All of the terms inside the brackets in \eqref{eq2A} would then be even except for $3n_1^2 + b$. If $b$ is even, choose an odd $n_1$, else if $b$ is odd, choose an even $n_1$. This makes the part inside the brackets odd, so the right side of \eqref{eq2A} has just one factor of $2$, which means it's congruent to $2$ modulo $4$. Thus, at least one of $f(n_1)$ and $f(n_1 + d)$ cannot be a perfect square, so its square root would not be an integer.
A: John Omielan has already provided a nice answer using $\text{mod}\ 4$.
Here is another approach using $\text{mod}\ 4$.
Let $f(n):=n^3+an^2+bn+c$.
Let us prove that at least one of $f(1),f(2),f(3),f(4)$ is not a square number.
Proof :
Let us consider in $\text{mod}\ 4$.
Suppose that $f(1),f(2),f(3),f(4)$ are square numbers.
Then, we have $f(i)\equiv 0,1$ for $i=1,2,3,4$.
Since $f(2)+f(4)\equiv 2b+2c$, we see that $f(2)+f(4)$ is even. So, we have $f(2)\equiv f(4)$.
Case 1 : $f(2)\equiv f(4)\equiv 0$
Then $2b+c\equiv c\equiv 0$ implies $b\equiv 0,2$ and $c\equiv 0$. If $b\equiv c\equiv 0$, then $f(1)\equiv 1+a\equiv 0,1$ implies $a\equiv 3,0$ for which $f(3)\equiv -1+a\equiv 2,3$, a contradiction. If $b\equiv 2$ and $c\equiv 0$, then $f(1)\equiv a-1\equiv 0,1$ implies $a\equiv 1,2$ for which $f(3)\equiv a+1\equiv 2,3$, a contradiction.
Case 2 : $f(2)\equiv f(4)\equiv 1$
Then $2b+c\equiv c\equiv 1$ implies $b\equiv 0,2$ and $c\equiv 1$. If $b\equiv 0$ and $c\equiv 1$, then $f(1)\equiv a+2\equiv 0,1$ implies $a\equiv 2,3$ for which $f(3)\equiv a\equiv 2,3$, a contradiction. If $b\equiv 2$ and $c\equiv 1$, then $f(1)\equiv a\equiv 0,1$ for which $f(3)\equiv a-2\equiv 2,3$, a contradiction.
So, we see that at least one of $f(1),f(2),f(3),f(4)$ is not a square number.
