Question

\[Let\ \ f(x)=x^7+x^6+x^5+x^4+x^3+x^2+x+1\]\[Find\ the\ remainder\ when\ f(x^{16})\ is\ divided\ by\ f(x).\]

Answer 1

\[ x^{7}+x^{6}+x^5+x^4+x^3+x^2+x+1 = \frac{x^8 - 1}{x - 1}\]
\[ f(x^{16})(x-1)/(x^8 - 1)\]

Answer 2

Wild Guess
\[ 8(x-1)\]
???

Answer 3

no

Answer 4

lol ...

Answer 5

The remainder is 8.

Answer 6

yeah! :)

Answer 7

Haha .. how??

Answer 8

One way is to do very long division.

Answer 9

It has an easy way to find the remainder.

Answer 10

Stay tuned. I am trying to find it.

Answer 11

http://www.wolframalpha.com/input/?i=PolynomialRemainder [(x^(7*16)%2Bx^(6*16)%2Bx^(5*16)%2Bx^(4*16)%2Bx^(3*16)%2Bx^(2*16)%2Bx^(16)%2B1)%2C+(x^8-1)%2C+x]

Answer 12

For some weird reason, the remainder from these two divisions are same
http://www.wolframalpha.com/input/?i=PolynomialRemainder [%28x^%287*16%29%2Bx^%286*16%29%2Bx^%285*16%29%2Bx^%284*16%29%2Bx^%283*16%29%2Bx^%282*16%29%2Bx^%2816%29%2B1%29%2C+x^7%2Bx^6%2Bx^5%2Bx^4%2Bx^3%2Bx^2%2Bx%2B1%2C+x]

Answer 13

oh ..

Answer 14

It seems that is a special case of this general case: For any integer \( n \ge 2 \) if we put
\[
f(x) =\sum_{i=0}^{n-1} x^i
\]
then \( f(x) \) divides \( f\left( x^{2n}\right)- n \)

Answer 15

I still do not know how to prove it. I will post it as a new problem and see if someone can come up with a clever solution.