#2 Measure Your Calibre

Suppose an integer solution existed, say $a$ . Note that for any integer $x$ , the given expression has the coefficients of $x^{2014} , x^{2013} , ... , x^{1}$ even. So $a^{2016} + (2016!+1!)a^{2015} + ... + (2!+ 2015!)a + (1!+ 2016!) \equiv a^{2016} + a^{2015} + 1 \pmod{2}$ . If $a$ were even, then

$0= a^{2016} + (2016!+1!)a^{2015} + ... + (2!+ 2015!)a + (1!+ 2016!) \equiv a^{2016} + a^{2015} + 1 \pmod{2} \equiv 0+0+1 \pmod{2} \equiv 1 \pmod{2}$ , which is not true.

Thus $a$ is odd. But then $0 = a^{2016} + (2016!+1!)a^{2015} + ... + (2!+ 2015!)a + (1!+ 2016!) \equiv a^{2016} + a^{2015} + 1 \pmod{2} \equiv 1+ 1+ 1 \pmod{2} \equiv 1 \pmod{2}$ , which is a contradiction.

Hence our assumption is incorrect, and no solution exists in integers.

Simply use RRT!!! A nice question btw @Md Zuhair Yeh bhi original hai kya??