TITLE: Find the greatest integer $k$ for which $1991^k$ divides $1990^{{1991}^{1992}}+1992^{{1991}^{1990}}$
QUESTION [6 upvotes]: Find the greatest integer $k$ for which $1991^k$ divides $$1990^{{1991}^{1992}}+1992^{{1991}^{1990}}$$
It is easy to see that $k \geq 1$ as $1990 \equiv -1$ and $1992 \equiv 1 \pmod{1991}$
Also, I thought that perhaps as $1991$ is the product of two distinct primes, it would be worth looking at small values of $(pq)^k||(pq-1)^{{pq}^{pq+1}}+(pq+1)^{{pq}^{pq-1}}$ for primes $p$ and $q$. Any help would be really appreciated.

REPLY [3 votes]: Let $a=1991$, and let $$b=(a-1)^{ a^{a+1} } + (a+1)^{ a^{a-1} }$$
The goal is to compute $\nu_{a}(b)$. To do so, we will evaluate $b$ modulo $a^{a+1}$.
Applying binomial expansions, $$b\equiv (-1)^{a^{a+1}} + 1 + a^{a-1} a\mod{a^{a+1}}$$ (because all the higher order terms vanish).
Since $a^{a+1}$ is odd, this simplifies to $$b \equiv a^a\mod{a^{a+1}}$$
Therefore, $\nu_{a}(b)=a$. To rephrase in the terminology of your question, $k=1991$.

REPLY [3 votes]: I shall prove a more general result: Let $n>1$ be an odd positive integer. Then $n^n \|[(n-1)^{n^{n+1}}+(n+1)^{n^{n-1}}]$.
Proof: 
\begin{align}
(n-1)^{n^2}+(n+1)=\sum_{i=0}^{n^2}{\binom{n^2}{i}(-1)^{n^2-i}n^i}+(n+1) & \equiv \binom{n^2}{1}n-1+(n+1) \pmod{n^2}\\
& \equiv n \pmod{n^2}
\end{align}
Now applying Lifting the Exponent Lemma on each prime factor of $n$, we have $n^n \|[(n-1)^{n^{n+1}}+(n+1)^{n^{n-1}}]$.
For your special case, $k=1991$.