Given a monic polynomial $P(x) = x^n + a_1x^{n-1}+\ldots a_n$, with integer coefficients, I need to show that it has no rational roots (in this case integer) using the following facts
1) $ n>1,$
2) $ a_n=17, $
3) $1+a_1+\ldots+a_n \neq 0$, $1-a_1+a_2 \ldots +(-1)^na_n \neq 0$,
4) $|a_m| \leq 15$, $ \forall m <n$.
Using the rational root theorem and the first 3 conditions, I was able to rule out $\pm 1$ as roots among the four possible roots $\pm 1, \pm17$. It is clear that I have to use the last condition to rule out $\pm 17$, but I am unable to do that. I may be missing something simple here, but any hints are welcome.
Best Answer
Since $P$ is monic and $a_n=17$, we have $$ P(17) = 17^n + a_1 17^{n-1}+a_2 17^{n-2}+\ldots +a_{n-1}17+17 $$Now use the fact that $|a_m|\leq 15$ for $1\leq m\leq n-1$: $$ |a_1 17^{n-1}+a_2 17^{n-2}+\ldots +a_{n-1}17+17| $$ $$\leq |a_1| 17^{n-1}+|a_2| 17^{n-2}+\ldots +|a_{n-1}|17+17 $$ $$ \leq 15(17^{n-1}+17^{n-2}+\ldots +17)+17 $$ $$ = \frac{1}{16} \left(17+15\cdot 17^n\right) < 17^n $$In other words, $|P(17)|$ is strictly greater than $0$, and a similar argument holds for $|P(-17)|$.