Prove that for all is composite

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- July 19th 2009, 08:26 PMnewtoinequalityProve:
Prove that for all is composite

- July 19th 2009, 10:23 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. - July 20th 2009, 04:00 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.