Finding integer values of $n$ for which the equation $$x^3+(n+1)x^2-(2n-1)x-(2n^2+n+4)=0$$ has at least one integer solution.
Try: let $\alpha,\beta,\gamma$ be the roots of the equation. Then
$\alpha+\beta+\gamma=-(n+1)$ and $\alpha\beta+\beta\gamma+\gamma\alpha=1-2n$ and $\alpha\beta\gamma=2n^2+n+4$.
Now i did not understand how to solve it. Could some help me to solve it , Thanks
Suppose $x,n$ are integers such that $$x^3+(n+1)x^2-(2n-1)x-(2n^2+n+4)=0\tag{eq1}$$
Suppose first that $n$ is odd.
Then, reducing mod $2$, we can replace $n$ by $1$, so \begin{align*} &x^3+(n+1)x^2-(2n-1)x-(2n^2+n+4)=0\\[4pt] \implies\;&x^3+(n+1)x^2-(2n-1)x-(2n^2+n+4)=0\;(\text{mod}\;2)\\[4pt] \implies\;&x^3-2x^2-x -7\equiv 0\;(\text{mod}\;2)\\[4pt] \implies\;&x^3-x\equiv 1\;(\text{mod}\;2)\\[4pt] \implies\;&0\equiv 1\;(\text{mod}\;2)\\[4pt] \end{align*} contradiction.
Hence $n$ must be even.
Then, reducing mod $2$, we can replace $n$ by $0$, so \begin{align*} &x^3+(n+1)x^2-(2n-1)x-(2n^2+n+4)=0\\[4pt] \implies\;&x^3+x^2+x-4=0\;(\text{mod}\;2)\\[4pt] \implies\;&x^3+x^2+x\equiv 0\;(\text{mod}\;2)\\[4pt] \implies\;&x\equiv 0\;(\text{mod}\;2)\\[4pt] \end{align*} so $x$ must be even.
Writing $x=2w$, and $n=2m$, $(\text{eq}1)$ reduces to $$4m^2 - (4w^2-4w-1)m -(4w^3+2w^2+w-2) = 0\tag{eq2}$$ Then, regarding $(\text{eq}2)$ as a quadratic equation in $m$, it follows that the discriminant $$D = 16w^4+32w^3+40w^2+24w-31$$ is a perfect square.
Noting that $D$ is odd, $D$ must be an odd square
Identically, we have $$ \begin{cases} D = (4w^2+4w+3)^2 - 40\\[4pt] D = (4w^2+4w+1)^2 + 16(w+2)(w-1)\\ \end{cases} $$ hence we must have $-2 \le w \le 1$, else $D$ would be trapped strictly between $(4w^2+4w+1)^2$ and $(4w^2+4w+3)^2$, which are consecutive odd squares.
If $w=-1$ or $w=0$, then $D < 0$, contradiction.
For $w=-2$, $(\text{eq}2)$ reduces to $(4m-7)(m-4)=0$, which yields the solution $(w,m)=(-2,4)$ for $(\text{eq}2)$, and the corresponding solution $(x,n)=(-4,8)$ for $(\text{eq}1)$.
For $w=1$, $(\text{eq}2)$ reduces to $(4m+5)(m-1)=0$, which yields the solution $(w,m)=(1,1)$ for $(\text{eq}2)$, and the corresponding solution $(x,n)=(2,2)$ for $(\text{eq}1)$.
Thus, $n=2$ and $n=8$ are the only values of $n$ for which $(\text{eq}1)$ has at least one integer solution for $x$.