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 . 
 A: 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.
A: $19\equiv 1 (mod 3)$
$92\equiv -1(mod 3)$
Checking $P_1$
$19+92\equiv 1-1 (mod 3)$
Hence $P_1$ is a multiple of $3^1$,
Similarly you can find values of $n$ for which $3^k|19^n+92^n$
Then add all $k$ until sum of $k=33$
Can you repeat a similar argument for $11$
A: COMMENT:
$19^n+92^n ≡ 8^n+4^n \ mod (11)$
⇒ $4^n(2^n+1)=11 k$
One solution of this equation is $n=5$, $k=3 \times 1024$, we have:
$4^5(2^5+1)=1024\times 33$
Therefor for $n=5$ and all powers such as $5(2t+1)$ , $19^n+92^n ≡ 0 \mod (33)$ due to geometric progression:
$(19^5)^{2t+1}+(92^5)^{2t+1}=(19^5+92^5)[(19^5)^{2t}-(19^5)^{2t-1}(92^5) + . . . + (92^5)^{2t}]$
We can see that $19^5+92^5$ is divisible by $11^2$ :
$19^5+92^5= 121\times 54490011 $
so we need to have $33/2=16.5$ factors of such kind; we may guess  $n=33\times 5=165$. But for this we have 165 times 3 and 17 times power of 5(from 5 to 165) so the product contains a factor like:
$P=3^{165 }\times 11^{2\times 17\times 5=170}$
So it seems $n=165$ is sufficient. The result from brute force , i.e $n=155=31\times 5$, indicates that there must be two hidden power for 33 such that  we can get equal powers for 3 and 11.That is  two of 5th powers must be divisible by $11^3$.If so power of 11 will be:
$(17-2=15)\times 2+1)\times 5=155$ 
