Prove that n^3-n is divisible by 6 for every integer n. Is it induction to be used here?...
symplectic_manifold said:If 6 divides n^3-n then, since divisibility is transitive and 2 divides 6 , 2 must also divide n^3-n. Let n be even then n^3 is even and n^3-n is also even. Let n be odd then n^3 is odd and hence n^3-n is even. Since every even number is divisible by 2 it follows that 6 divides n^3-n.
n3-n= n(n2-1)= n(n-1)(n+1)= (n-1)(n)(n+1), three consecutive integers. What does that tell you?
You've managed to show 2 divides n^3-n by considering n even or odd, but to show 6 divides n^3-n you need to also show 3 divides n^3-n. Consider Hall's factorization, can you show 3 divides (n-1)n(n+1) for any n?
symplectic_manifold said:Great!...Can I generalise it by saying that among any n successive integers there is exactly one divisible by n?...What is a formal proof for this?
symplectic_manifold said:OK, but why did you change r?