$13$ divides $10^{2p}-10^p+1$ for any prime $p > 3$ In a curse of number theory I need show that  for any prime $p > 3$ prove that $13$ divides $10^{2p}-10^p+1$.
but I failed You can help me? Thanks!
 A: As $10\equiv-3\pmod{13},10^2\equiv(-3)^2\equiv3^2$ and $p$ is odd
we have $$10^{2p}-10^p+1\equiv(3^2)^p-(-3)^p+1\equiv(3^2)^p+(3)^p+1\pmod{13}$$
Now we can prove for integer $m>0$ 
$$x^{2m}+x^m+1$$ is divisible by $x^2+x+1$ if $3\nmid m$
Do you recognize $x$ here?
So, we don't need $p$ to be prime, $p\equiv\pm1\pmod6$ will serve our purpose
A: Any prime $p \neq 2, 3$ is of from $6k + 1$ or $6k + 5$ so now consider the following:
$$10 ^ 6 \equiv 1\; (\bmod\; 13)$$ 
$$ \implies 10 ^ {6k} \equiv 1^k\; (\bmod\; 13) $$
$$ \implies 10 ^ {6k} \equiv 1\; (\bmod\; 13) $$
$$ \implies 10 ^ {6k + 1} \equiv 1 \times 10 \; (\bmod\; 13) $$
$$ \implies 10 ^ {6k + 1} \equiv 10 \; (\bmod\; 13) $$
Now consider $10 ^ {2p} - 10 ^ p + 1$ where $p = 6k + 1$
$$10 ^ {2 (6k + 1)} - 10 ^ {6k + 1} + 1 (\bmod\; 13)$$
$$= (10 ^ {6k + 1})^2 - 10 + 1 (\bmod\; 13)$$
$$= 100 - 10 + 1 (\bmod\; 13)$$
$$= 91 (\bmod\; 13)$$
Thus as $13 | 91$ so $13 | 10^{2p} - 10^p + 1$ for $p = 6k + 1$. 
And similarly for the case when $p = 6k + 5$ and our proof is complete.$\blacksquare$
