Find the number of integer solution of the equation: $$\displaystyle x^{2016}+(2016!+1!) x^{2015}+(2015!+2!) x^{2014}+ \cdots + (1!+2016!)=0$$ where $n!=n \cdot (n-1) \cdots 3 \cdot 2 \cdot 1$, for $n \geq1$
Please complete my proof:
Suppose $P(x)$ is any polynomial with integer coefficients and let it has a integer root, denoted by $a$, then $P(x)=(x-a)Q$. Now put $x=0$, then $P(0)=-aQ$ and put $x=1$, then $P(1)=(1-a)Q$. If we can prove $Q$ is even, then we can say that among the above two $(P(0)$ and $P(1))$ one is even and the other is odd. So let $$P(x)=\displaystyle x^{2016}+(2016!+1!) x^{2015}+(2015!+2!) x^{2014}+ \cdots + (1!+2016!)$$ clearly for $x=0,P(0)=2016!+1!$ which is an odd integer. now put $x=1, P(1)=1+2(2016!+1!)+2(2015!+2!)+\cdots+2(1009!+1008!)=1+2(2016!+2015!+\cdots+2!+1!)$ which is an odd integer,so it is a contradiction!! Thus,$$\displaystyle x^{2016}+(2016!+1!) x^{2015}+(2015!+2!) x^{2014}+ \cdots + (1!+2016!)=0$$ have no solutions.
But I can't prove $Q$ is not even, and point out where I'm lacking, please help for that.
Just recall that $x(x+1)$ is even for integer $x$, which implies $x^{2016}+x^{2015}=x^{2014}x(x+1)$ is even. And all $n!$, $n\ge 2$ is even, so is $2016!x^{2015}+(2015!+2!)x^{2014}+\ldots+(2!+2015!)x+2016!$ . We conclude by noting that $$P(x)=(x^{2016}+x^{2015})+\Big(2016!x^{2015}+(2015!+2!)x^{2014}+\ldots+(2!+2015!)x+2016!\Big)+1!\\ =\rm even + even + 1 = odd$$ hence is never $0$.