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- Mar 1st 2010, 05:38 PMmathh18proofs
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- Mar 1st 2010, 07:05 PMDrexel28
What kind of math do you know?

Easy way, merely note that if $\displaystyle n\equiv 0\text{ mod }3$ this is trivial since $\displaystyle n^3-n^\equiv 0-0=0\text{ mod }3\implies n^3\equiv n\text{ mod }3$. Otherwise, $\displaystyle (n,3)=1\implies n^3\equiv n\text{ mod }3$ since $\displaystyle \phi(3)=2$ - Mar 2nd 2010, 03:54 AMArchie Meade
Hi mathh18,

to use proof by induction for the 1st one,

**F(k)**

$\displaystyle k^3-k$

is divisible by 3 ?

**F(k+1)**

$\displaystyle (k+1)^3-(k+1)$

is divisible by 3 ?

$\displaystyle (k+1)^3-(k+1)=(k+1)[(k+1)^2-1]=(k+1)(k^2+2k)$

$\displaystyle =k^3+2k^2+k^2+2k$

Express this using F(k)

$\displaystyle k^3+2k^2+k^2+2k=\left(k^3-k\right)+3k^2+3k=\left(k^3-k\right)+3\left(k^2+k\right)$

If F(k) is true, then F(k+1) is certainly true as the 2nd term is divisible by 3.

Therefore F(1) true causes F(2) to be true, causing F(3) to be true, causing ......

Hence you now only need test n=1.

$\displaystyle 1^3-1=0,\ 0(3)=0\ \Rightarrow\ 0=\frac{0}{3}$

true