Prove:

**newtoinequality** · July 19th 2009, 08:26 PM

Prove that for all $n\ge1, 1+5^n+5^{2n}+5^{3n}+5^{4n}$ is composite

**TwistedOne151** · July 19th 2009, 10:23 PM

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.

**Opalg** · July 20th 2009, 04:00 AM

Originally Posted by TwistedOne151

I'm pretty sure that the numbers are all divisible by 11, and therefore composite (since they are all greater than 11).

This is not true if n is a multiple of 5, because $5^5\equiv1\!\!\!\pmod{11}$ and therefore $1+5^{5k}+5^{10k}+5^{15k}+5^{20k} \equiv1+1+1+1+1\not\equiv0\!\!\!\pmod{11}$ .

On the other hand, if n is not a multiple of 5 then $5^n\not\equiv1\!\!\!\pmod{11}$ and so $5^n-1$ has an inverse (mod 11). Therefore $1+5^{n}+5^{2n}+5^{3n}+5^{4n} \equiv (5^{5n}-1)(5^n-1)^{-1}\equiv0\!\!\!\pmod{11}$ , so the result is true in that case.

But I don't see how to prove that $1+5^{n}+5^{2n}+5^{3n}+5^{4n}$ is composite when n is a multiple of 5.

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