# Prove:

• Jul 19th 2009, 08:26 PM
newtoinequality
Prove:
Prove that for all $\displaystyle n\ge1, 1+5^n+5^{2n}+5^{3n}+5^{4n}$ is composite
• Jul 19th 2009, 10:23 PM
TwistedOne151
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, 04:00 AM
Opalg
Quote:

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 $\displaystyle 5^5\equiv1\!\!\!\pmod{11}$ and therefore $\displaystyle 1+5^{5k}+5^{10k}+5^{15k}+5^{20k} \equiv1+1+1+1+1\not\equiv0\!\!\!\pmod{11}$.

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

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