1. Statement of the problem: Prove 6 is a factor of $n(n^2+5)$ for every integer $n$. 2. We will show the expression is divisible by 2 and by 3, and since 2 and 3 are coprime, it follows that 6 divides the expression. 3. Divisible by 2: If $n$ is even then $n(n^2+5)$ is divisible by 2. If $n$ is odd, write $n=2k+1$ so $n^2=4k^2+4k+1=2(2k^2+2k)+1$ and thus $n^2+5=2(2k^2+2k+3)$ which is even, hence divisible by 2. 4. Divisible by 3: Consider residues mod 3. If $n\equiv 0 \pmod{3}$ then $n(n^2+5)$ is divisible by 3. If $n\equiv 1 \pmod{3}$ then $n^2\equiv 1$ so $n^2+5\equiv 6\equiv 0 \pmod{3}$ and thus divisible by 3. If $n\equiv 2 \pmod{3}$ then $n^2\equiv 4\equiv 1$ so again $n^2+5\equiv 6\equiv 0 \pmod{3}$. 5. Conclusion: Because the expression is divisible by both 2 and 3 and $\gcd(2,3)=1$, we conclude that 6 divides $n(n^2+5)$ for every integer $n$. Final answer: 6 is a factor of $n(n^2+5)$.