TITLE: Why are Jucys-Murphy elements' eigenvalues whole numbers?
QUESTION [21 upvotes]: The Jucys-Murphy elements of the group algebra of a finite symmetric group (here's the definition in Wikipedia) are known to correspond to operators diagonal in the Young basis of an irreducible representaion of this group. As one can see from the Wikipedia entry, all of the elements of such a diagonal matrix (in other words the operator's eigenvalues) are integers.
I'm looking for a simple way of explaining this fact (that the eigenvalues are wholes). By simple I mean without going into more or less advanced representation theory of the symmetric group (tabloids, Specht modules etc.), so trying to prove the specific formula given in Wikipedia is not an option. (I'm considering the Young basis as the Gelfand-Tsetlin basis of the representation for the inductive chain of groups $S_1\subset S_2\subset \ldots\subset S_n$, which is uniquely defined thanks to this chain's spectrum's simplicity, not as a set of vectors in correspondence with the standard tableaux.)
In fact, I'm trying to prove the first statement ($a_i\in \mathbb{Z}$) of proposition 4.1 in this article.

REPLY [3 votes]: Awfully sophisticated proof for the fact :) 
Just to relate it with the question about Knizhnik-Zamolodchikov equation:
Find polynom p(z) with values in C[S_n] such that p'(z) = \sum_i (Id+(1i))/(z-i) p(z).  [Knizhnik-Zamolodchikov equation for S_n]
Consider the following KZ ODE:
$ p'(z) = \sum_{i=2...n} \frac{ Id + \pi( (1i) )}{z-z_i} p (z) $
As it is discussed in MO-question above it is known to have polynomial solution.
The reside at infinity is equal to  $Res=-\sum_{i=2...n} { Id + \pi( (1i) )}$.
Which is our beloved JM-element up to sign and n*Id.
Hence its eigenvalues must be non-positive integers (this is obvious since
at infinity the solution looks like $(1/z)^{Res}, so in order to be polynomial in z
they must be non-positive ints).
Hence we are done.
Moreover we got that eigs are greater or equal -n (as David Speyer proved directly above).