Prove that there exists a positive integer n, such that for all integers $k$ the number $k^2+k+n$ has no prime divisors less than $2008$.
Here is the solution my book gave
I don't understand why such $r$ exists? Could anyone explain to help me understand? Thanks in advance
The example stated means that as already $2$ values of $k$( $k \equiv 0 \mod{p}$ and $k \equiv p-1 \mod{p}$) give the same remainder($k^2+k \equiv 0 \mod{p}$), then there are $p-2$(congruent to $\pmod{p}$) more values of $k$ which have to attain $p-1$ remainders. This can not happen and atleast a remainder will exist which is never attained by $k^2+k$.