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This can by done by Induction
Show that for all integers n >1, n^3 - n can be divided by 3. (Note: n^3 stands for n*n*n)
( M5 m" [' B: {) W- Y/ x) Y7 V4 ]. }+ \9 k4 M
Proof: 7 [+ M4 P4 }2 h
Let n >1 be an integer
; D& i1 f" h' rBasis: (n=2)6 ]7 \! k: Q2 d! N: a" g' p
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
: B* x c4 D/ Z" ?# e( c! m" y" U! T6 m$ A6 G5 {5 A* d
Induction Hypothesis: Let K >=2 be integers, support that
* I& {8 Z0 w1 t7 ^& O& D K^3 – K can by divided by 3.' M% F! U% o# s3 {6 k4 X
, N+ @* ?: B$ A" O1 a, q, B2 p- v" g' D
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3- d4 t9 ~9 Q4 o! e
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem. I- ^. q# v* f* s
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
4 L4 L& ]3 B/ ]2 O" J$ o = K^3 + 3K^2 + 2K
# D3 }# _: q; ]: L; {1 F; S = ( K^3 – K) + ( 3K^2 + 3K)1 e( S( u6 Q Y- ^, H/ a" @
= ( K^3 – K) + 3 ( K^2 + K): d4 ^( Q. `9 `- Q6 T* W& @
by Induction Hypothesis, we know that k^3 – k can by divided by 3 which means that k^3 – k = 3 X for some integer X>06 W/ }. D/ z2 k4 `6 }$ i$ H7 G
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
+ L" E+ e/ @9 K: `" M = 3X + 3 ( K^2 + K)
5 e6 p8 |$ t3 W% k = 3(X+ K^2 + K) which can be divided by 37 k: O% p% S- R0 k
& F9 y, g7 N3 t1 b$ H& I. G' `8 O
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.3 Q, m, V1 s1 d3 |
5 w1 _+ {& t% N$ N4 z2 y' G[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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