# Thread: Prove:

1. ## Prove:

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

2. 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.

3. 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.