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.

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.
