$$f(x)=x^3+7x-14(1+n^2)$$Prove there are no integer solutions for any $n\in Z$
My Proof:
Let $A,B,C$ be the three roots of $f(x)$. Then
$$\begin{align}
A+B+C&=0\tag{1}\\
AB+BC+CA&=7\tag{2}\\
ABC&=14(1+n^2)\tag{3}\\
\text{By 1 and 2}, A^2+B^2+C^2&=-14\\
\implies B=\bar{C}&=b+ci\\
\text{Plugging these to (1)}\\
A&=-2b\\
\text{plugging to (2)}\\
c&=\sqrt{3}b\\
\text{plugging to (3)}\\
-4b^3&=7(1+n^2)
\end{align}$$
Which is not possible if $b$ and $n$ are both int due to nature of $mod\space4$
Is this proof valid? Also, Any other better way?
The proof is not correct, because it assumes that $b\in\mathbb Z$.
A possible proof is: you can appy Eisenstein's criterion, with $p=7$, in order to prove that $f(x)$ is irreducible in $\mathbb{Z}[x]$ and therefore that it has no integer roots. This would only fail if $7\mid n^2+1$, but this never happens: $n$ is congruent to $0$, $\pm1$, $\pm2$, or $\pm3$ modulo $7$ and therefore $n^2+1$ is congruent to $1$, $2$, $5$, or $4$ respectively.