Question

number theory questionq
http://prntscr.com/58jn7n

Answer 1

@ganeshie8 @KamiBug @Miracrown @Kainui

Answer 2

http://prntscr.com/58jowo i dont understand this part, gimme an example

Answer 3

what is a non-constant polynomial mean

Answer 4

one that has terms other than ax^0

Answer 5

means
f(x)=g(x)I(x)
such that
\(g(x)=\sum a_ix^i\)
\(I(x)=\sum b_jx^j\)
and for all b's and a's they are 1 or 2 ( depend on binomial theorem )

Answer 6

basically you want to find the ordered pairs such that the polynomial is not prime

Answer 7

and yes i and j are not zero

Answer 8

f(x) = g(x) * h(x)

the degrees of g and h need to be atleast 1

Answer 9

why such that the polynomial is not prime

Answer 10

this is what it asked for :D

Answer 11

ok i see

Answer 12

why didnt they just write that out

Answer 13

takes too many words if u use degree and other stuff to state the problem

Answer 14

hmm lets carry on

Answer 15

interesting problem

Answer 16

man us 3 are the only ones that stick it out with these problems lol

Answer 17

so it is in this form\[f(x) = x^n + x^{n-1} + \cdots + 2(x^m+x^{m-1}+\cdots+1)\]

Answer 18

yeah

Answer 19

0<=M<=N<=25

Answer 20

yeah m<n is implied in above split up

Answer 21

0<=m<n<=25

Answer 22

m=n-k

Answer 23

ya okay bascally in that form

Answer 24

oh i see

Answer 25

okay i see where that combinatorics is gonna come out

Answer 26

for exampl,e of
f(x)=g(x) * h(x)
for g(x)=x
then u are gonna have that same coefficients , and degree just decreased by 1

Answer 27

\[\begin{align}f(x) &= x^n + x^{n-1} + \cdots + 2(x^m+x^{m-1}+\cdots+1)\\~\\

&=(x^n + x^{n-1} + \cdots +x+ 1)+ (x^m+x^{m-1}+\cdots+x+1)\\~\\
&=\dfrac{x^{n+1}-1}{x-1}+\dfrac{x^{m+1}-1}{x-1}\\~\\

\end{align}\]

Answer 28

of f(x)=x^n+x^n-1..+2x^m+x^m-1..+2
for g(x)=x,x^2,x^3,....
h(x) for g(x)=x
h(X)=x^n-1+x^m+2*x^m-1...+2/x <--- problem with this one

Answer 29

Recall the formula
\[x^{ab}-1 = \left(x^a\right)^b-1 = (x^a-1)(x^{a(b-1)} + x^{a(b-2)}+\cdots +x + 1) \]

Answer 30

where did that formula come from

Answer 31

we use the same idea as here :
http://math.stackexchange.com/questions/966747/show-that-if-a-positive-integer-n-is-composite-then-rn-frac10n-1/966765#966765

Answer 32

that follows from \(\large x^n-1 = (x-1)(x^{n-1} + x^{n-2}+ \cdots +x+1)\)

plugin \(n=ab\)