ELMO SL 2014 A7
by Wolstenholme, Aug 1, 2014, 9:49 PM
Find all positive integers 
with 
such that the polynomial
![\[ P(a_1, a_2, ..., a_n) = a_1^n+a_2^n + ... + a_n^n - n a_1 a_2 ... a_n \]](//latex.artofproblemsolving.com/a/8/d/a8de9f925985591d6d69b099266a49dd1c5c7b9f.png)
in the variables 
, 
, 
, 
is irreducible over the real numbers, i.e. it cannot be factored as the product of two nonconstant polynomials with real coefficients.
Solution:
Note that
and that 
so the relevant polynomial is factorable for 
and 
Now I shall prove that for
this polynomial is irreducible. Treating it as a polynomial in 
it suffices to show that its constant term, the polynomial 
, is irreducible. Treating this new polynomial as a polynomial in 
and then in 
and so on it suffices to show that 
is irreducible for all .
Note that![$ a_{1}^n + a_{2}^n + a_{3}^n = \prod_{k = 1}^n\left(a_1 + \omega_{k}\sqrt [n] {a_{2}^n + a_{3}^n}\right) $](//latex.artofproblemsolving.com/0/e/5/0e57df7f7d1eaa91b640cb5fbbb2d63242e79312.png)
where 
for all 
.
Looking at this as a polynomial in
, if it was reducible then one of its factors must have constant term of the form 
for some 
-th root of unity 
and some integer 
. But this itself must be some polynomial 
.
Therefore
. Now, take 
. It now suffices to show that there is no polynomial 
such that 
. But note that the roots of the RHS all have multiplicity a multiple of and since the roots of the LHS are precisely the 
-th roots of unity that are not -th roots of unity, all roots of the LHS have multiplicity 
so we have the desired result.
Now, I would like to include some motivation. Dealing with irreducibility of multivariable polynomials is much more difficult than with normal polynomials... certain criterion like that of Eisenstein still work but usually the first thing one should try is to reduce the problem to a single variable polynomial problem. Moreover, when looking for irreducibility over the reals as opposed to the rationals, looking at roots (bounding them, multiplicities, etc...) is usually the way to go. The combination of these ideas motivates the above solution (at least, that's how I was motivated).
with 
such that the polynomial![\[ P(a_1, a_2, ..., a_n) = a_1^n+a_2^n + ... + a_n^n - n a_1 a_2 ... a_n \]](http://latex.artofproblemsolving.com/a/8/d/a8de9f925985591d6d69b099266a49dd1c5c7b9f.png)
in the
variables 
, 
, 
, 
is irreducible over the real numbers, i.e. it cannot be factored as the product of two nonconstant polynomials with real coefficients.Solution:
Note that
and that 
so the relevant polynomial is factorable for 
and 
Now I shall prove that for
this polynomial is irreducible. Treating it as a polynomial in 
it suffices to show that its constant term, the polynomial 
, is irreducible. Treating this new polynomial as a polynomial in 
and then in 
and so on it suffices to show that 
is irreducible for all .Note that
![$ a_{1}^n + a_{2}^n + a_{3}^n = \prod_{k = 1}^n\left(a_1 + \omega_{k}\sqrt [n] {a_{2}^n + a_{3}^n}\right) $](http://latex.artofproblemsolving.com/0/e/5/0e57df7f7d1eaa91b640cb5fbbb2d63242e79312.png)
where 
for all 
.Looking at this as a polynomial in
, if it was reducible then one of its factors must have constant term of the form 
for some 
-th root of unity 
and some integer 
. But this itself must be some polynomial 
.Therefore
. Now, take 
. It now suffices to show that there is no polynomial 
such that 
. But note that the roots of the RHS all have multiplicity a multiple of and since the roots of the LHS are precisely the 
-th roots of unity that are not -th roots of unity, all roots of the LHS have multiplicity 
so we have the desired result.Now, I would like to include some motivation. Dealing with irreducibility of multivariable polynomials is much more difficult than with normal polynomials... certain criterion like that of Eisenstein still work but usually the first thing one should try is to reduce the problem to a single variable polynomial problem. Moreover, when looking for irreducibility over the reals as opposed to the rationals, looking at roots (bounding them, multiplicities, etc...) is usually the way to go. The combination of these ideas motivates the above solution (at least, that's how I was motivated).
August 2016
June 2016