Divisibility by $33^{33}$ Let $P_n=(19+92)(19^2+92^2)\cdots(19^n+92^n)$ for each positive integer $n$. Determine , with proof the least positive integer $n$, if it exists , for which $P_n$ is divisible by $33^{33}$.
I have made no progress concrete enough to show . 

Modulo $3$, we have
$$19^k+92^k\equiv 1^k+(-1)^k\pmod 3 $$
and this is $0$ iff $k$ is odd. So for each odd $k$, the factor $19^k+92^k$ adds (at least) one factor $3$; and for even $k$, it doesn't. 
This alone gives us 
$3^m\mid P_{2m-1}$, or equivalently
$$3^{\lfloor\frac{n+1}2\rfloor}\mid P_n. $$
There may be higher powers of $3$ added by such a factor. Indeed, 
$$19^k+92^k\equiv 1+2^k\equiv 0\pmod 9\iff k\equiv 3\pmod 6.$$
This improves our estimate to 
$$3^{\lfloor\frac{n+1}2\rfloor+\lfloor \frac{n+3}6\rfloor}\mid P_n. $$
Hence for $3^{33}$, $n=49$ would be sufficient, but this is still not optimal (though we may suspect that we should rather do more work for the second part - divisibility by $11^{33}$).
So let's look at $11$:
We have $19\equiv -3\pmod{11}$ and $92\equiv 4\equiv (-3)\cdot 6\pmod{11}$, hence
$$19^k+92^k\equiv(-2)^k(1+6^k)\equiv 0\pmod{11} \iff k\equiv 5\pmod {10}.$$
This gives us
$$ 11^{\lfloor\frac{n+5}{10}\rfloor}\mid P_n.$$
Incidentally, we verify that $11^2\|19^5+92^5$ so that 
$$ 11^{2\lfloor\frac{n+5}{10}\rfloor}\mid P_n.$$
From this, we find that $n=165$ will certainly be sufficient.
For the exact result (which is readily found numerically: $n=155$), you will need to investigate the factors modulo $1331$ (and even $14641$), I am afraid.
