Assume $n$ is even. Prove that $323$ divides $20^n+16^n-3^n-1$. I'm unclear what is the best method to teach this with minimum math experience. 
 A: We have $n=2k$. Hence,
$$f(k) = 400^k + 256^k - 9^k - 1$$You can in fact prove that $646$ divides $f(k)$. 
Note that $646 = 2 \cdot 19 \cdot 17$. In general, to prove that $abc \mid n$, where $a$, $b$, and $c$ are mutually relatively prime, it suffices to prove that $a \mid n$, $b \mid n$ and $c \mid n$.
First it is easy to prove that $f(k)$ is even, since
$$f(k) = 400^k + 256^k - 9^k - 1 = \text{even} + \text{even} - \text{odd} - \text{odd} = \text{even}$$ Hence, $2 \mid f(k)$.
Now note that
$(400-9) \mid (400^k - 9^k)$, i.e., $391 \mid (400^k-9^k)$ and $17 \mid 391$.
Similarly,
$(256-1) \mid (256^k-1)$, i.e., $255 \mid (256^k-1)$ and $17 \mid 255$.
Hence, $17 \mid f(k)$.
Now note that
$(400-1) \mid (400^k - 1)$, i.e., $399 \mid (400^k-1)$ and $19 \mid 399$
Similarly,
$(256-9) \mid (256^k-9^k)$, i.e., $247 \mid (256^k-9^k)$ and $19 \mid 247 $
Hence, $19 \mid f(k)$.
Hence, $(2 \cdot 17 \cdot 19) \mid f(k)$
A: We have $323 = 17 \cdot 19$ with $\gcd(17, 19) = 1$. Also, $19 = 20-1 \mid 20^n - 1, 19 = 16 + 3 \mid 16^n - 3^n \Rightarrow 19 \mid 20^n+16^n-3^n-1$
and $17 = 20 - 3 \mid 20^n - 3^n$, $17 = 16+1 \mid 16^n +1 \Rightarrow 17 \mid 20^n+16^n-3^n-1$
$\gcd(17,19)=1 \Rightarrow 323 = 17 \cdot 19 \mid 20^n+16^n-3^n-1$
A: Hint $\ $ Below, put $\rm\, m=17,\ n=19,\ a=20,\ b=16,\ c=3,\ d = 1$
$\rm mod\ m\!:\ a\equiv c,\, b\equiv -d\:\Rightarrow\:a^n + b^n - c^n - d^n \equiv\, c^n + (-d)^n - c^n -d^n \equiv\, 0$
$\rm mod\ \,n\!:\ a\equiv d,\, b\equiv -c\:\Rightarrow\: a^n + b^n - c^n - d^n \equiv\, d^n + (-c)^n - c^n -d^n \equiv\, 0$
A: Since $n$ is even:
$20^n + 16^n - 3^n - 1 \equiv 1^n + (-3)^n - 3^n - 1 = 1 + 3^n - 3^n - 1 = 0 \bmod 19$
Also:
$20^n + 16^n - 3^n - 1 \equiv 3^n + (-1)^n - 3^n - 1 = 3^n + 1 - 3^n - 1 = 0 \bmod 17$
So the quantity is divisible by $17\times 19 = 323$
