BdMO Online Forum

Problem 7:
$f(x)=x^6+x^5+\cdots +x+1$Find the remainder when dividing $f(x^7)$ by $f(x)$.

I am quite sure that the solution is the following:
f(x)= 1+x+x^2+x^3+.....+x^6
Now if we write it like this:
f(x)= 1+x+x^2+x^3+.....infinite, then
f(x)= (1-x)^(-1)
f(x^7)=(1-x^7)^(-1)
so f(x^7)/f(x) = (1-x)/(1-x^7)
according to remainder theorem, the remainder is = 0
so Answer: 0.
(Osman, a friend of mine solved this)

Actually that is applicable when x is less than 1 but this was not given in the question, and I'm looking for a solution of this one long time so I think some of the seniors could help

Hint: solution

nice and clean:D

nafistiham wrote: here, $x^7-1=f(x)\cdot(x-1)$
so, \[f(x)|(x^{42}-1),(x^{35}-1),(x^{28}-1),(x^{21}-1),(x^{14}-1),(x^7-1)\]
so, the remainder will be
\[7\]

How does the green line lead to the red line?

sakib.creza wrote:

nafistiham wrote: here, $x^7-1=f(x)\cdot(x-1)$
so, \[f(x)|(x^{42}-1),(x^{35}-1),(x^{28}-1),(x^{21}-1),(x^{14}-1),(x^7-1)\]
so, the remainder will be
\[7\]

How does the green line lead to the red line?

actually, what I didn't think much important to mention was this
\[f(x)|x^{7}-1|x^{7n}-1\]
It should be clear enough now