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Message: <blockquote data-quote="Persius" data-source="post: 19484656" data-attributes="member: 477838"><p><span style="font-size: 15px">Prove  n^3-n is divisible by 3 for all n >0, n is an integer.</span></p><p><span style="font-size: 15px"></span></p><p><span style="font-size: 15px">I have seen the proof of this problem with induction. So I thought of a different ordinary way. It is too simple logic.</span></p><p><span style="font-size: 15px"></span></p><p><span style="font-size: 15px">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.</span></p><p><span style="font-size: 15px"></span></p><p><span style="font-size: 15px">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.</span></p><p><span style="font-size: 15px"></span></p><p><span style="font-size: 15px">case 1: n=3k+1</span></p><p><span style="font-size: 15px"></span></p><p><span style="font-size: 15px">then n.(n-1).(n+1) =(3k+1).3k.(3k+2)  which is obviously divisible by 3.</span></p><p><span style="font-size: 15px"></span></p><p><span style="font-size: 15px">case 2: n=3k+2</span></p><p><span style="font-size: 15px"></span></p><p> <span style="font-size: 15px">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.</span></p><p><span style="font-size: 15px"></span></p><p><span style="font-size: 15px">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.</span></p></blockquote><p></p>

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