I am having trouble with this homework question and was hoping someone could help. Cheers

Prove that 5|n(n2-1)(n2=1)

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- Mar 11th 2008, 10:59 PM #1

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- Mar 11th 2008, 11:28 PM #2

- Mar 12th 2008, 01:05 AM #3
If you multiply out the brackets then you get $\displaystyle 5|n^5-n$, which is true by Fermat's little theorem.

- Mar 12th 2008, 07:20 AM #4

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Hello, asw-88!

Prove that: .$\displaystyle 5\,|\,n(n^2-1)(n^2+1)$

$\displaystyle n$ must one of five possible forms: .$\displaystyle 5k-2,\:5k-1,\:5k,\:5k+1,\:5k+2$

$\displaystyle [1]\;n \,= \,5k-2\!:\;\;N \:=\5k-2)([5k-2]^2-1)([5k-2]^2 + 1)$

. _ . . . . . . . . . . $\displaystyle =\;(5k+2)(25k^2-20x+3)(25k^2-20k + 5)$

. _ . . . . . . . . . . $\displaystyle = \;(5k+2))(25k^2+20k + 3){\color{red}5}(5k^2 - 4k + 1)$ . . . a multiple of 5

$\displaystyle [2]\;n \,=\,5k-1\!:\;\;N \:=\5k-1)([5k-1]^2-1)([5k-1]^2+1)$

. . . . . . . . . . . . $\displaystyle = \;(5k-1)(25k^2 - 10k)(25k^2-10k + 2)$

. . . . . . . . . . . . $\displaystyle = \;(5k-1){\color{red}5}k(5k- 2)(25k^2 - 10k+2)$ . . . a multiple of 5

$\displaystyle [3]\;n \,=\,5k\!:\qquad N \:=\5k)([5k]^2 - 1)([5k]^2+1)$

. . . . . . . . . . . . $\displaystyle = \;{\color{red}5}k(25k^2 - 1)(25k^2+1)$ . . . a multiple of 5

$\displaystyle [4]\;n\,=\,5k+1\!:\;\;N \;=\;(5k+1)([5k+1]^2-1)([5k+1]^2+1)$

. . . . . . . . . . . . $\displaystyle = \;(5k+1)(25k^2 + 10k)(25k^2+10k + 2)$

. . . . . . . . . . . . $\displaystyle = \;(5k+1){\color{red}5}k(5k + 2)(25k^2 + 10k + 2)$ . . . a multiple of 5

$\displaystyle [5]\;n\,=\,5k+2\!:\;\;N \;=\;(5k+2)([5k+2]^2-1)([5k+2]^2+1)$

. . . . . . . . . . . . $\displaystyle = \;(5k+2)(25k^2 + 20k + 3)(25k^2 + 20k + 5)$

. . . . . . . . . . . . $\displaystyle = \;(5k+2)(25k^2 + 20k+3){\color{red}5}(5k^2 + 4k + 1)$ . . . a multiple of 5

. . . . . . . . . . . . . . . . .*Q.E.D.*

- Mar 12th 2008, 10:48 AM #5

- Mar 13th 2008, 07:03 AM #6