Divisiblity of $n$ with $a,b,c$ is relative prime to p

Given an arbitrary prime $p > 2011$. Prove that there exist positive integers $a,b,c$ such that there exists some numbers from $a, b, c$ that are relatively prime to $p$, and for all positive integers $n$ such that $p|n^4 − 2n^2 + 9$, $p$ divides $24an^2 + 5bn + 2011c$.

I have no idea for this problem

Choose $a,b,c$ such that: $p| 24a-1 ; p| 5b+2; p|2011c-9$
• Hm, as $p>2011$, hence $gcd(24,p)=gcd(5,p)=gcd(2011,c)=1$. Then due to Chinese Remainder Theorem, there are such numbers . After that, due to $gcd(1,p)=gcd(2,p)=gcd(9,p)$; $gcd(a,p)=gcd(b,p)=gcd(c,p)=1$ – Toan Nguyen Dinh Jan 18 '15 at 5:09