Contest Problem - Divisibility Find all ordered pairs (x, y) of positive integers x, y such that
$x+y$ divides 2014 and (simultaneously) $x^yy^x$ divides $(x+y)^{(x+y)}$
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This is a contest problem from U Tenn, FERMAT contest.
My try:
Let S = x+y
The prime factorization of 2014 is {2,19,53} 
S divides 2014 would mean S = 2,19,38,53,106,1007.
The condition that it should simultaneously hold the second condition would lead to 
$$x^{(S-x)}(S-x)^x *K = S^S$$
$$\frac{(x+y)^x}{y^x}\frac{(x+y)^y}{x^y}= k$$
$$\left(1+\frac{x}{y}\right)^x\left(1+\frac{y}{x}\right)^y = k$$
If x and y are positive integers the Left Hand side could never be a whole number and k will not be an integer.  Thus x = y.
Further x = y can only hold for even possibilities of S which again is 38 and 106.  
Therefore the ordered pair (x,y) that satisfy both divisibility rules are:
(19,19) and (53,53). 
Correct me if I am wrong.

Your reasoning is nice but you missed an easy one:  $x=y=1$.  That is what you get if you look at $S=2$.
