Question

If n be any natural number then by which largest number n3-n is always divisible

Answer 1

n3 = (n-1)n(n+1)product of three consequent integers. At least one of them is even, so 2|(n3−n)2|(n3−n). They all have different residues modulo 33, so one of them is divisible by 33, so 3|(n3−n)3|(n3−n). So, 6|(n3−n)6|(n3−n) for every nn, so m≥6m≥6. Let's prove that m=6m=6. Let kk be a natural number, such that k|(n3−n)k|(n3−n) for every nn. If kk has a prime divisor p≠2,3p≠2,3 then k/|(p+2)3−(p+2)=(p+1)(p+2)(p+3)k⧸|(p+2)3−(p+2)=(p+1)(p+2)(p+3) since p<p+1,p+2,p+3<2pp<p+1,p+2,p+3<2p. So, kk is of the form 2a3b2a3b. aa can't be greater than 11because 4/|23−24⧸|23−2 and bb can't be greater than 11 because 9/|23−29⧸|23−2. So, k≤2131=6k≤2131=6 and m=6m=6.

Answer 2

Answer:

Step-by-step explanation: