From a silly (long division) puzzle comes an interesting number-theory "theorem" (quotes indicate some doubt). I've worked a BUNCH of this type of long division puzzle. 
EDIT (The problem represents LOELPE/MNTN where EP is the quotient and LEAC is the remainder, with LIONP representing E times MNTN, PPMC representing the subtraction LOELP-LIONP (with E "brought down" at the right end per algorithm, etc.) . A similar thread showed that not everyone is familiar with this structure.)

Modular arithmetic has proven important in quite a few. So I've seen more than a few instances of this pattern (which occurs in the above problem: EN = P mod 10 and PN = E mod 10):

Given that {N,I,K} is a subset of [0..9] it seems to always be the case that N+K=10. I have computer evidence suggesting further that {N,K} is a subset of [2..8]-{5} and I is in {4,9} (in fact, more specifically, the only way I can be 4 is if {N,K}={2,8}; otherwise, I=9 [but we can have (N,K,I) = (2,8,4)].
(In the problem above, it happens that P=2, E=8, N=9.)
This isn't obvious to me and I don't remember when it hit me, but the sound and light that I imagined occurring reminded me of the scene from Monty Python and The Holy Grail where The Lord appeared from a cloud to tell them of their Quest.
I have what might pass for a brute-force proof, but I won't submit it owing to cowardice. 
I just want to know if what I think I know is actually true and if, to a number theorist (me, an analysisisissst), it's at all obvious.
 A: I wanted something conceptual, but my proof devolves into case-based reasoning at the end, as you complained that your own does.
If 
$$ ni = k \mod 10$$
and
$$ ki = n \mod 10$$
then (add the equations and rearrange)
$$
(n + k)(i-1) = 0 \mod 10.
$$
The reason your result "wants to" be true is that in a ring which is an integral domain (which $\mathbb{Z}/10$ isn't, of course), this should either give $n + k = 0$ or $i = 1$, the latter being impossible from the constraint of the puzzle that $n \neq k$, and the former giving what you want because of the constraint of the puzzle that $n$ and $k$ be between $0$ and $9$.
But unfortunately we're not in an integral domain because $2 \cdot 5 = 0$, so we need to nitpick the possibility that one of these numbers is even and the other is divisible by five away.
To do this, consider another equation (subtract the equations and rearrange)
$$
(n - k)(i + 1) = 0 \mod 10.
$$
We know that $n \neq k$, so the second equation guarantees that $i + 1$ is divisible by $2$ or $5$. If it is divisible by both then $ i = -1$ mod $10$ which quickly gives the theorem. So we omit this case.
Assume (to try to get a contradiction) that $n + k$ is not divisible by $10$, so the first equation guarantees that $i - 1$ is a zero divisor. Thus $i \neq 4$. In particular, neither $i + 1$ nor $i - 1$ is divisible by 5. 
But this says both $n + k$ and $n - k$ are divisible by $5$; since $10 > n - k > 0$ (if $n < k$ swap the names of the two variables in the proof) we conclude $n - k = 5$, but then $n + k = 5 + 2k$ is only divisible by $5$ if $k = 5$; and this then gives a nonsense value for $n$.
A: [A]..From the 2 equations in P.E,N, none of P,E,N is 0 as they are 3 distinct digits, and for the same reason we know N is not 1.......    [B]..  For all case,substitute the value of P, in the 1st equation into the 2nd, and,separately,substitute the value for E, in the 2nd equation, into the 1st equation. From this we find that 10 divides P(N2 - 1) and divides E(N2 - 1) [where N2 is N squared]......    [C].. From this,we see that 5 divides N2 - 1 .Otherwise 5 divides both E and P, giving (E,P) = (0,5) or (5,0), but nether E nor P is 0.....Hence N is 4, 6,or 9, as,by [A] it can't be 1. ......   [D].. For N=4,or N=6, E and P are even by[B], and positive by [A], implying E and P lie among 2,4,6,8 . But from the original equations, if N=6 and E=2,4,or 8,then P=E. So N is not 6. The case N=4 yields 2 solutions (E,P) = (2,8) or (8,2). Finally, if N=9.adding N to both sides of the original equations shows that 10 divides E+P, yielding  6 more solutions (E,P) = (2,8),(3,7),(4,6),(6,4),(7,3), or (8,2).
