Question

Let \( n \ge 2 \) and consider the polynomial of degree \( n - 1\)
\[
f(x) =\sum_{i=0}^{n-1} x^i
\]
Show that \( f(x) \) divides \( f \left(x^{2n} \right) -n \)

For the case n=8, see the proof on
http://openstudy.com/study#/updates/4fcdafe0e4b0c6963ad89d51

Answer 1

For n=2
\[ f(x) = 1 + x \\
f(x^4) -2= 1 + x^4 -2= x^4-1=(x+1)(x-1)(x^2+1)
\]

Answer 2

For n=3
\[
f(x) = 1 + x + x^2\\
f(x^6)-3 = 1 + x^6 + x^{12} -3 = (x-1) (x+1) \left(x^2-x+1\right)
\left(x^2+x+1\right)
\left(x^6+2\right)
\]

Answer 3

\[f(x^8) = 1 + x^8 + x^{16} + x^{24} =\frac{x^{32}-1}{x^8-1}\]
\[f(x^{2n})= \frac{x^{2n^2}-1}{x^{2n}-1} \]
Does it make sense?

Answer 4

I got it.

Answer 5

@experminetX @Davidc
\[
f(x^{2n})-n = x^{2n} + \cdots x^{2n(n-1)} -(n-1)\\
= (x^n)^2 + \cdots +(x^n)^{2(n-1)} -(n-1) = g(x^n)
\]
Where g is

\[
g(x) = x^2 + x^4 +\cdots + x^{2(n-1)} -(n-1)\\
g(1)=0\\
\]

so g is divisible by x -1

\[
g(x) = (x-1)h(x)\\
f(x^{2n})-n=g(x^n) =(x^n-1)h(x^n)=(x-1)( 1+ x + \cdots + x^{n-1}) h(x^n)\\
= (x-1) f(x) h(x^n)
\]

Answer 6

@experimentX @Davidc

Answer 7

\[\sum_{k=0}^{n-1}(x^{2n})^k=g_{k}(x)\sum_{k=0}^{n-1}x^k+n\]I don't know how to write g(x)....but it is easy to see the pattern

Answer 8

Did you understand my proof?

Answer 9

@eliassaab yes :)

Answer 10

but i see a pattern of the quotient. Anyone can help me yo find it?

Answer 11

BY the proof above,
\[
g(x) = (x-1) h(x^n)
\]