Hello,

Look for every possibility of n if 5 divides the product.

Try , then and so on

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

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- Mar 12th 2008, 12:28 AM #2

- Mar 12th 2008, 02:05 AM #3
If you multiply out the brackets then you get , which is true by Fermat's little theorem.

- Mar 12th 2008, 08:20 AM #4

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Hello, asw-88!

Prove that: .

must one of five possible forms: .

5k-2)([5k-2]^2-1)([5k-2]^2 + 1)" alt="[1]\;n \,= \,5k-2\!:\;\;N \:=\5k-2)([5k-2]^2-1)([5k-2]^2 + 1)" />

. _ . . . . . . . . . .

. _ . . . . . . . . . . . . . a multiple of 5

5k-1)([5k-1]^2-1)([5k-1]^2+1)" alt="[2]\;n \,=\,5k-1\!:\;\;N \:=\5k-1)([5k-1]^2-1)([5k-1]^2+1)" />

. . . . . . . . . . . .

. . . . . . . . . . . . . . . a multiple of 5

5k)([5k]^2 - 1)([5k]^2+1)" alt="[3]\;n \,=\,5k\!:\qquad N \:=\5k)([5k]^2 - 1)([5k]^2+1)" />

. . . . . . . . . . . . . . . a multiple of 5

. . . . . . . . . . . .

. . . . . . . . . . . . . . . a multiple of 5

. . . . . . . . . . . .

. . . . . . . . . . . . . . . a multiple of 5

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

- Mar 12th 2008, 11:48 AM #5

- Mar 13th 2008, 08:03 AM #6