ii) contradicts i) for $r=q$...

Assume that $ p$ divides $ c$ but $ p^{2}$ does not, and such that $ p^{2}$ divides $ f(n)$ for some $ n$. Then, $ p^{2}$ divides $ (n+b)^{2}-c$, or $ p$ divides $ (n+b)^{2}$. Thus, $ p$ divides $ n+b$, or $ p^{2}$ divides $ c=(n+b)^{2}-f(n)$, contradiction. Second part should read: "Let $ q\neq2$ such that $ q$ DOES NOT divide $ c$. If $ q$ divides $ f(n)$ for some integer $ n$, then show that for any positive integer $ r$, integer $ n'$ exists such that $ q^{r}$ divides $ f(n')$".

It is a fundamental of quadric reside $ x^{2}\equiv c(\mod p)$ If $ p$ odd then There exist $ x^{2}\equiv c(\mod p^{n})$ More general $ f(x_{1})\equiv 0(\mod p)$ $ \gcd(f'(x_{1}),p) = 1$ Then for each n there exist $ x_{n}$ such that $ f(x_{n})\equiv 0(\mod p^{n})$ (Use Helsen Lemma) An apply Prove that for all m then exist $ (a,b,k)\in Z$ such that $ 2m = a^{19}+b^{99}+k2^{1999}$ Where $ (a,b)$ odd.