Prove that n^3-n is divisible by 6 for every integer n. Is it induction to be used here?...
You're going at this backwards assuming 6 divides n^3-n then showing 2 divides it?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.
oh that was even more easier! nice...that n^3-n is always even?...n3-n= n(n2-1)= n(n-1)(n+1)= (n-1)(n)(n+1), three consecutive integers. What does that tell you?
...now that was my problem...how can I show that 3 divides n^3-n?...I need to show that the sum of the digits is divisible by 3...but the number is encoded in a formula!...so what can I do?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?
That's correct. n successive integers starting at m look like m, m+1, ..., m+n-1. Use the division algorithm to write m=n*q+r, where 0<r<=n (note I've changed r slightly from the usual form). Then m+n-r is divisible by n and 0<=n-r<n so m+n-r is one of m, m+1, ..., m+n-1symplectic_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?
Because it worked with the sequence I had already typed and seemed easier than going back to modify it. If you believe the division algorithm will give you an r with 0<=r<n you should be able to wrangle this to a different r with 0<r<=n, there's little difference.symplectic_manifold said:OK, but why did you change r?