Let $a$ be an integer. Prove that the following equation cannot have more than one integer $x^4+7x^3+(a+2)x^2-11x+a=0$ Let $a$ is a interger. Prove that the following equation cannot have more than one integer $$x^4+7x^3+(a+2)x^2-11x+a=0$$

$$x^4+7x^3+(a+2)x^2-11x+a=0$$
Or $$-\frac{x^4+7x^3+2x^2-11x}{x^2+1}=x^2+7x+1-\frac{18x+1}{x^2+1}=a$$
When $a\in \mathbb{Z}$ that means $$(x^2+1)\mid (x^4+7x^3+2x^2-11x)$$
Or $$(x^2+1)\mid (18x+1)\rightarrow 18x+1\ge x^2+1$$
Or $$0\le x\le 18$$
Now i tried all value and got only $x=0$ and $x=18$ are roots that satified.
$$x=0\rightarrow a=0 \text{   or   } x=18\rightarrow a=-450 $$
Now i tried to solve this equation egain with $a=0$ and $a=-450$ and each value of $a$ get only one integer root of $x$ and it is done.
I don't know if my solution is correct.Help me check it and give me some solution,ty..

Using Vieta's formulas helps.
Suppose that the equation has an odd root $x$. Then, LHS of the equation$$\underbrace{x^4+7x^3+2x^2-11x}_{\text{odd}}+a(\underbrace{x^2+1}_{\text{even}})$$
is odd, which contradicts that $x$ is a root.
So, we see that if the equation has an integer root, then it is even.
Now, suppose that the equation has four roots $2b,2c,p+qi,r-qi$ where $b,c\in\mathbb Z$ and $p,q,r\in\mathbb R$.
Then, by Vieta's formulas, we get
$$2b+2c+p+r=-7\tag1$$
$$4bc+(2b+2c)(p+r)+(p+qi)(r-qi)=a+2\tag2$$
$$2(b+c)(p+qi)(r-qi)+4bc(p+r)=11\tag3$$
It follows from $(1)$ that $p+r$ is an integer.
Also, it follows from $(2)$ that $(p+qi)(r-qi)$ is an integer.
Then, LHS of $(3)$ is even, and RHS of $(3)$ is odd, which is a contradiction.
