<s>[INST] 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 . 
 [/INST] 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:
$(17-2=15)\times 2+1)\times 5=155$ 
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