Prove that for all is composite

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- Jul 19th 2009, 09:26 PMnewtoinequalityProve:
Prove that for all is composite

- Jul 19th 2009, 11:23 PMTwistedOne151
I'm pretty sure that the numbers are all divisible by 11, and therefore composite (since they are all greater than 11). I'm not sure how to prove divisibility, though.

--Kevin C. - Jul 20th 2009, 05:00 AMOpalg
This is not true if n is a multiple of 5, because and therefore .

On the other hand, if n is not a multiple of 5 then and so has an inverse (mod 11). Therefore , so the result is true in that case.

But I don't see how to prove that is composite when n is a multiple of 5.