Show there is no solution to this equation

Question

I have to show that $2x^4-20x+8$ cannot be divided by $16$ without remainder. The only thing comes to my mind is to write $16$ as $4^2$ which hasn't been of any help.

Could you give me some hints to prove this?

Answer 1

We have $$2(x^2+5x)-7=2(x+9)(x-4)+65$$

As $x+9-(x-4)=13$

If $13|(x+9)\iff13|(x-4)$

Now check

if $13\mid(x+9)$

and if $13\nmid(x+9)$

Answer 2

$\triangle' = 5^2 - 2(-7-169k) = 39 + 138k = 13(3+ 26k) \to 3+26k = 13n^2 \to 13|3$, contradiction !