Find all ordered pairs (x, y) of positive integers x, y such that $x+y$ divides 2014 and (simultaneously) $x^yy^x$ divides $(x+y)^{(x+y)}$ .
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.
The assertion that the product of your rationals can only be an integer if $x=y$ is correct, but needs proof.
I prefer to work directly with integers. We are told that $x^yy^x$ divides $(x+y)^{x+y}$. Let $d=\gcd(x,y)$, and let $x=da$ and $y=db$. Then $a$ and $b$ are relatively prime. Our divisibility condition says that $d^{x+y} a^yb^x$ divides $d^{x+y}(a+b)^{x+y}$. It follows that $a^yb^x$ divides $(a+b)^{x+y}$. Suppose that one of $a$ or $b$, say $a$, is greater than $1$. We have that $a^y$ divides $(a+b)^{x+y}$. This is impossible since $a$ and $a+b$ are relatively prime.
Thus $a=b=1$ and therefore $x=y$.