prove for n>2,
n^3 - n always divisible by 12 using euclid division lemma
Answers
(n^3-n)/12
hence n > 2
hope u can understand
Answer:
Step-by-step explanation:
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,
n
3
−n=(6m)
3
−6m=216m
3
−6m
=6m(36m
2
−1)=6q where q=m(36m
2
−1)
n
3
−n is divisible by 6
When n=6m+1
n
3
−n=n(n
2
−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+2
n
3
−n=n(n
2
−1)=n(n−1)(n+1)
=(6m+1)(36m
2
+30m+6)
=6m(36m
2
+30m+6)+1(36m
2
+30m+6)
=6[m(36m
2
+30m+6)]+6(6m
2
+5m+1)
=6p+6q, where p=m(36m
2
+30m+6) and q=6m
2
+5m+1
n
3
−n is divisible by 6
When n=6m+3
n
3
−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[36m
2
+36m+8]
=6p+3q,
where p=m((6m+3)
2
−1) and q=36m
2
+36m+8
n
3
−n is divisible by 6
When n=6m+4
n
3
−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[36m
2
+48m+15]
=6m[(6m+4)
2
−1]+12[12m
2
+16m+5]
=6m[(6m+4)
2
−1]+6[24m
2
+32m+10]
=6p+6q, where p=m[(6m+4)
2
−1] and q=24m
2
+32m+10
n
3
−n is divisible by 6
When n=6m+5
n
3
−n=(6m+5)
3
−(6m+5)
=6m[(6m+5)
2
−1]+5[(6m+5)
2
−1]
=6m[(6m+5)
2
−1]+5[36m
2
+60m+24]
=6p+30q
=6(p+5q)
=6(p+5q) where p=m[(6m+5)
2
−1] and q=6m
2
+10m+5
n
3
−n is divisible by 6
Hence, n
3
−n is divisible by 6 for any positive integer n