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Question

# Use Euclid's Division Lemma, prove that for any positive integer n,n3−n is divisible by 6.

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Solution

## Any positive integer is of the form 6m,6m+1,6m+2,6m+3,6m+4,6m+5 for some positive integer n.When n=6m,n3−n=(6m)3−6m=216m3−6m=6m(36m2−1)=6q where q=m(36m2−1)n3−n is divisible by 6When n=6m+1n3−n=n(n2−1)=n(n−1)(n+1)=(6m+1)(6m)(6m+2)=6m(6m+1)(6m+2)6q where q=m(6m+1)(6m+2)When n=6m+2n3−n=n(n2−1)=n(n−1)(n+1)=(6m+1)(36m2+30m+6)=6m(36m2+30m+6)+1(36m2+30m+6)=6[m(36m2+30m+6)]+6(6m2+5m+1)=6p+6q, where p=m(36m2+30m+6) and q=6m2+5m+1n3−n is divisible by 6When n=6m+3n3−n=(6m+3)2−(6m+3)=(6m+3)[(6m+3)2−1]=6m((6m+3)2−1)+3((6m+3)2−1)=6[m((6m+3)2−1)]+3[36m2+36m+8]=6p+3q,where p=m((6m+3)2−1) and q=36m2+36m+8n3−n is divisible by 6When n=6m+4n3−n=(6m+4)3−(6m+4)=(6m+4)[(6m+4)2−1]=6m[(6m+4)2−1]+4[(6m+4)2−1]=6m[(6m+4)2−1]+4[36m2+48m+15]=6m[(6m+4)2−1]+12[12m2+16m+5]=6m[(6m+4)2−1]+6[24m2+32m+10]=6p+6q, where p=m[(6m+4)2−1] and q=24m2+32m+10n3−n is divisible by 6When n=6m+5n3−n=(6m+5)3−(6m+5)=6m[(6m+5)2−1]+5[(6m+5)2−1]=6m[(6m+5)2−1]+5[36m2+60m+24]=6p+30q=6(p+5q)=6(p+5q) where p=m[(6m+5)2−1] and q=6m2+10m+5n3−n is divisible by 6Hence, n3−n is divisible by 6 for any positive integer n

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