I'm trying to solve this task:
$P(x)$ is a polynomial with integer coefficients. For some natural number $c$, c does not divide $P(1), …, P(c)$. Prove that $P(X)$ does not have integer roots.
My observations:
if $P(c) = a_1c^n+a_2c^{n-1}+…+a_{n-1}c+a_n$ is not divisible by $c$ so must be $a_n$. Also $P(0) = a_n$ so we can say that c does not divide $P(0)$.
None of values $P(0), P(1), …, P(c)$ can equal to zero since c will divide zero.
Then $P(c) = a_1c^n+a_2c^{n-1}+…+a_{n-1}c+a_n = a_1c^n+a_2c^{n-1}+…+a_{n-1}c+P(0)$
and we know that $c$ does not divide $P(c)$ and also $P(0)$ so this sum
$a_1c^n+a_2c^{n-1}+…+a_{n-1}c $
should not be divisible by $c$ but it is. So there seems to be a contradiction.
Is this approach correct? I am not sure how it shows that $P(x)$ does not have integer roots.
Could somebody please give a hint? Thanks in advance.
Best Answer
You are using the argument that if $a\nmid (b+c)$ and $a\nmid c$ then $a\nmid b$ which is evidently not true. Because $2\nmid (2+3)$ and $2\nmid 3$. But $2$ divides 2.
Proper argument might go like this. I am avoiding the use of congruences which I feel you might be unaware of:
Suppose for some integer $m$, we have $p(m)=0$. Then, as you noted $c$ divides $p(m)$. Now, consider the integers $m-1, m-2, \dots, m-c$. One of these numbers must be divisible by $c$, say it is $m-r$, $1\leq r \leq c$.
It can now be shown that $c$ divides $p(r)$. This is easy to see. Simply consider: $$ p(m)-p(r) = a_n(m^n-r^n)+a_{n-1}(m^{n-1}-r^{m-1})+\dots+ a_1(m-r),$$ where $p(x) = a_nx^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0$.
Clearly $$p(m)-p(r) = (m-r) C,$$ for some integer $C$ (This is because $m^k-r^k = (m-r)\sum\limits_{i=0}^{k-1} m^ir^{k-1-i}$. Hence we can take $(m-r)$ common throughout).
Since $c$ divides $m-r$, $ c$ must divide $p(m)-p(r)$ and hence it must divide $p(r)$ as it divides $p(m)$. This is a contradiction as $1\leq r \leq c$.