<s>[INST] Show that $1+5^n + 5^{2n} + 5^{3n} + 5^{4n}$ is never prime for n > 0 integer Show that $1+5^n + 5^{2n} + 5^{3n} + 5^{4n}$ is never prime for n > 0 integer
Trying cases on a computer I can see that the equation is divisible by 11 when $n \neq 5^k$. That's easy to prove if you write the sum as a fraction using the fact that the series is geometric.
Is there a more general way to solve the problem, one which includes a solution for $n=5k$ and doesn't require ansatz?
edit: $ n = 5k$ not $ n = 5^k $
 [/INST] Case $1:n=2k$
Then our sum is $\frac{5^{10k}-1}{5^{2k}-1}=\frac{(5^{5k}-1)(5^{5k}+1)}{5^{2k}-1}$. Since the denominator is smaller than each of the factors we conclude the number is not prime.
Case $2: n=2k+1$
Let $a=5^{2k+1}$ and notice $1+a+a^2+a^3+a^4=(a^2+3a+1)^2-5a(a+1)^2$.
Since $a=5^{2k+1}$ this is equal to: $(a^2+3a+1)^2-5^{2k+2}(a+1)^2=(a^2+3a+1)^2-(5^{k+1}(a+1))^2$
We recognize this as a difference of squares and obtain:
$((a^2+3a+1)-5^{k+1}(a+1))(a^2+3a+1)+5^{k+1}(a+1))$.
It is very easy to see the first factor is greater than $1$.
</s>