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be the positive number satisfied 
Find the minimum of 
, we have 
When does the equality hold?
. If 
, then prove the following: 
, and 
are all polynomials such that ![\[ P(x^5)+xQ(x^5)+x^2R(x^5)=(x^4+x^3+x^2+x+1)S(x),\]](http://latex.artofproblemsolving.com/1/6/a/16a71abc110ece558d427a07d7be2d1f72148024.png)
prove that 
is a factor of 
.
be a primitive 
th root of unity, and observe that for 
, 
. Hence 
, and 
are roots of the quadratic equation 
. A quadratic null at 3 different places must be null everywhere, implying 
, so we're done.
![$T\colon\mathbb R[x]\to\mathbb R^5$](http://latex.artofproblemsolving.com/6/8/8/68855b81fbc161a6f3a1e961dad6f2bb9c1bc2b6.png)
be the function given by![\[ \sum a_kx^k\mapsto\left(\sum_{5\mid k}a_k,\sum_{5\mid k-1}a_k,\ldots,\sum_{5\mid k-4}a_k\right). \]](http://latex.artofproblemsolving.com/5/6/1/561dc24cbb1a22a0740d2ca5cbdea919e9b211a7.png)
It is clear that![\[ T(P(x^5)+xQ(x^5)+x^2R(x^5))=(P(1),Q(1),R(1),0,0). \]](http://latex.artofproblemsolving.com/d/1/9/d191251d5402e6fd85d892a7ece504357b262949.png)
It is also easy to see that![\[ T((x^4+x^3+x^2+x+1)S(x))=(S(1),S(1),S(1),S(1),S(1)). \]](http://latex.artofproblemsolving.com/5/6/8/56868c5e37649b8b4a7b689600853baa00032acc.png)
Since 
, these two must be equal, giving![\[ P(1)=Q(1)=R(1)=S(1)=0. \]](http://latex.artofproblemsolving.com/0/d/3/0d3799bfd4e92541002a6055adb144050d877405.png)
Hence, 
. 
divides 
from ,of course assuming that is not 
.
divides from ,of course assuming that is not .
![\[x^4+x^3+x^2+x+1 \mid x^5-1 \mid \left[P(x^5)+xQ(x^5)+x^2R(x^5)\right]-\left[P(1)+xQ(1)+x^2R(1)\right]\]](http://latex.artofproblemsolving.com/8/1/0/8103bd9135d49b300d5572b67ca69686c97a6da6.png)
Then![\[x^4+x^3+x^2+x+1 \mid P(1)+xQ(1)+x^2R(1)\]](http://latex.artofproblemsolving.com/4/f/d/4fdf2a5d8b5aacdb20c2f15f5f86cb639bc60622.png)
But 
then![\[P(1)+xQ(1)+x^2R(1) \equiv 0 \ \forall \ x\]](http://latex.artofproblemsolving.com/5/8/5/585729b07c902a35838dc3194a7b5b2140aa944d.png)
Hence, 
.
is the inverse of function 
and thus, they must intersect that 
axis (as its clearly not self-inverse function). Thus: 
is the only solution.
be a fifth root of unity not equal to 
.
and 
one at a time for 
and add up the five resulting equations.
and that 
,
while the right hand side is 
,
,
.![\[ x P(x^5) + x^2 Q(x^5) + x^3 R(x^5) = x(x^4 + x^3 + x^2 + x + 1) S(x)\]](http://latex.artofproblemsolving.com/4/b/1/4b10de4e307ba9b817eaa08286dc863ab67c0fa2.png)
for the five roots 
to get![\[ 0 = P(1)\sum_{k = 0}^{4}{\omega^k} + Q(1)\sum_{k = 0}^{4}{\omega^{2k}} + R(1)\sum_{k = 0}^{4}{\omega^{3k}} = 5S(1) = 5P(1).\]](http://latex.artofproblemsolving.com/2/6/1/26187bacf0e34bc169d1f85f2e6a800873613e0c.png)
a primitive fifth root of unity, then the five equation that result from plugging in 
for 
for 
are
.
yields 
.
then from the first two equations we have
.
,
so 
which means 
is a factor of 
as desired.
![\[\begin{vmatrix}1&\omega&\omega^2\\1&\omega^2&\omega^4\\1&\omega^3&\omega\end{vmatrix}=\omega^3+2-2\omega^2-\omega^4\neq0,\]](http://latex.artofproblemsolving.com/c/f/d/cfdae26343895428bcb86efb7bca01910870dddc.png)
one can immediately deduce that 
.
is a root of 
,
to the original equation, we get 
.
gives 
.
denote the given assertion, and let 
Adding up all of these yields 
.
which is more than enough. Solving the system, we easily get 
and the fact that if x-1 is a factor of P(x) suggests P(1)=0 gives suggestions to find ways with powers of 5 and 1. With experience, you will find motivation to consider the roots of unity. Now summing over 
with x=a fifth root of unity not equal to 1, and 
,
(in some order, 1, x, …, x^4 each appear once in each root of unity) 
,
,
,
