1. ## Congruence/Residue Class

Q:
Let $f(n) = a_0x^n + ... + a_n$ be a polynomial over Z
Show that if r consecutive vales of f (i.e. values for consecutive integers) are all divisible by r, then r|f(m) for all $m \in$ Z

2. the remainder of the r consecutive integer must be 0,1,...,r-1.
since $f(x)\mod r=f(x\mod r)$, $f(0)\mod r=f(1)\mod r=...=f(r-1)\mod r=0$, $\forall m,f(m)\mod r=f(m\mod r)=0$since $m \mod r\in\{0,1,...,r-1\}$

3. Cool, thx for the response

A few questions:
1. I believe I came up with something similar to the proof you gave but I was wondering if it is required to prove that m mod r $\in$ {0, 1, ... ,r-1} or is it obvious?

What I am thinking is that that since we are working with mod r and all numbers should be represented by another number mod r then it is safe to assume m mod r $\in$ {0, 1, ... ,r-1}. Is that correct?

2. In order for this to be true, what does r have to equal? A second part to this question says that this is true with r > 1 and $(a_0, ... , a_n) = 1$. The problem I see with this is the $a_n$ component of the function

4. Originally Posted by dlbsd
Cool, thx for the response

A few questions:
1. I believe I came up with something similar to the proof you gave but I was wondering if it is required to prove that m mod r $\in$ {0, 1, ... ,r-1} or is it obvious?

What I am thinking is that that since we are working with mod r and all numbers should be represented by another number mod r then it is safe to assume m mod r $\in$ {0, 1, ... ,r-1}. Is that correct?

2. In order for this to be true, what does r have to equal? A second part to this question says that this is true with r > 1 and $(a_0, ... , a_n) = 1$. The problem I see with this is the $a_n$ component of the function
let $p=x\mod r$, then $p=\min\{y|y\geq 0,x=y+kr,k\in Z\}$,then obviously, $p