Why do all primes $n>3$ satisfy $\,309\mid 20^n-13^n-7^n$ Solve the following... $309|(20^n-13^n-7^n)$ in $\mathbb{Z}^+$.
I invested lotof time to it and finally went to WolframAlpha for help by typing... 
Solve $309k=20^n-13^n-7^n$ over the integers. It returned the following...  Note that all the primes are generated here. Can someone please explain why? Thanks! EDIT: the formulas suggested in the answerare too complicated, but this is so simple!

$20^6 \equiv 13^6 \equiv 7^6 \equiv 229 \bmod 309$, so $20^n \equiv 13^n + 7^n \mod 309$ if and only if $20^{n+6} \equiv 13^{n+6} + 7^{n+6} \bmod 309$.  Since it does work for $n=1$ and $n=5$, but not for $0$, $2$, $3$ or $4$, we find that $20^n \equiv 13^n + 7^n \bmod 309$ if and only if $n \equiv 1$ or $5 \bmod 6$.  
All primes except $2$ and $3$ are congruent to $1$ or $5 \bmod 6$.  $2$ or $3$ can't divide a number congruent to $1$ or $5 \bmod 6$, so it's not easy for a small number of this form to be composite. The first few composites are $25 = 5^2$, $35 = 5 \cdot 7$, $49 = 7 \cdot 7$, $55 = 5 \cdot 11$, ...
