Prove n^3-n is divisible by 3 for all n >0, n is an integer.
I have seen the proof of this problem with induction. So I thought of a different ordinary way. It is too simple logic.
Let's factorize the expression. n^3-n=n.(n-1).(n+1).Let's denote n with the help of 3 and analyze the nature of the expression.
only 2 cases here for n . n must be divisible by 3 or not.If not n should be in the form 3k+r where k,r >0 and 3>r>0.You can change this k and obtain any integer with the help of 3k+r. r is the the remainder. remainder can be 0,1,2,3.But for the occasion,r was taken as 1,2. If n is divisible by 3 ,then n.(n-1).(n+1) is divisible by 3. if n is not ,it should be 3k+1 or 3k+2.
case 1: n=3k+1
then n.(n-1).(n+1) =(3k+1).3k.(3k+2) which is obviously divisible by 3.
case 2: n=3k+2
then n.(n-1).(n+1) =(3k+2).(3k+1).(3k+3) => (3k+2).(3k+1).3(k+1) which is obviously divisible by 3.
so if n is a multiple of 3 or not ( all n can be a multiple of 3 or in the form of 3k+1, 3k+2 ) . n^3-n is divisible by 3 for all n >0.
I have seen the proof of this problem with induction. So I thought of a different ordinary way. It is too simple logic.
Let's factorize the expression. n^3-n=n.(n-1).(n+1).Let's denote n with the help of 3 and analyze the nature of the expression.
only 2 cases here for n . n must be divisible by 3 or not.If not n should be in the form 3k+r where k,r >0 and 3>r>0.You can change this k and obtain any integer with the help of 3k+r. r is the the remainder. remainder can be 0,1,2,3.But for the occasion,r was taken as 1,2. If n is divisible by 3 ,then n.(n-1).(n+1) is divisible by 3. if n is not ,it should be 3k+1 or 3k+2.
case 1: n=3k+1
then n.(n-1).(n+1) =(3k+1).3k.(3k+2) which is obviously divisible by 3.
case 2: n=3k+2
then n.(n-1).(n+1) =(3k+2).(3k+1).(3k+3) => (3k+2).(3k+1).3(k+1) which is obviously divisible by 3.
so if n is a multiple of 3 or not ( all n can be a multiple of 3 or in the form of 3k+1, 3k+2 ) . n^3-n is divisible by 3 for all n >0.
Last edited:
