Question

Let \(r\) and \(s\) be two real numbers such that
\[ r+s =-1 \\ r s =-1\] and let \(k \ge 1\) be an integer. Show that
\[ r^k + s^k \] is a positive or negative integer.

Answer 1

\((r+s)^k=\sum_{j=0}^{k }\binom{k}{j} r^js^{k-j}=r^k+s^k+\sum_{j=1}^{k-1 }\binom{k}{j} r^js^{k-j} \\
r^k+s^k=(r+s)^k-\sum_{j=1}^{k-1 }\binom{k}{j} r^js^{k-j} \\ \)

Answer 2

\(
r^k+s^k=(-1)^k-\sum_{j=1}^{k-1 }\binom{k}{j} r^j \frac{s^k}{s^j}\\= (-1)^k-\sum_{j=1}^{k-1 }\binom{k}{j} \left (\frac{r}{s} \right )^j s^k \\ \)

Answer 3

rs=-1
r=-1/s

Answer 4

r/s=-1/s^2

Answer 5

`\(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ f(n)}\)` use this for inside the text sigma.

Answer 6

\(\large r^k+s^k= (-1)^k- \sum_{j=1}^{k-1 }\binom{k}{j} (-1)^j s^{k-2j} \)

Answer 7

hmm this is what i got so far xD

Answer 8

when k is even
1- ( negative + positive + negative + positive +.....+negative )
if s>1 , second term also >1
1-negative is positive (case one )
if s<1
not sure (case 2)

when k is odd
-1 -(positive +negative+.....+ positive )
when s>1 then all term is negative (case 3)
case 4 :- when s< 1 hmm

Answer 9

hehe case 2 and 4 idk what to do in them :(

Answer 10

Kinda wondered about induction when I saw this.

Answer 11

induction for signs :|

Answer 12

I don't think induction for signs work
just make few steps, we can see that it doesn't work
if k odd, the basic step =-1, assume it works on k step, next is stuck: (r+k)^(k+1) = 1, positive.:(

Answer 13

haha :D

Answer 14

Is there any outlet if we factor the term inside the sum as \(r^js^j (s^{k-2j})\) and then argue of j is odd or even?

Answer 15

When k is even, it's positive but that's all I know lol.

Answer 16

*if not of

Answer 17

lol thats what i got , but wat if s<1 kai ?

Answer 18

Negative numbers squared are always positive =P

Answer 19

lol imean
<1 like 1/2 :P

Answer 20

Oh I see. We are showing that any two real numbers is not only positive or negative but that it is also an integer?

Answer 21

its not necessary to be integer , but unlike integers real which <1 and >0
increase with decreasing power hmm

Answer 22

so we wont know which term is max, specially with binomial coefficient or i might dint see it yet

Answer 23

\[
(r+s)(r^k+s^k) = r^{k+1}+s^{k+1} +rs^{k}+sr^{k} = r^{k+1}+s^{k+1} -(r^{k-1}+s^{k-1})\\
\implies r^{k+1} +s^{k+1} = (r+s)(r^k+s^k)+(r^{k-1}+s^{k-1})
\]We assume that by indictive hypothesis that \(p(k-1)\land p(k)\).\[
r^{k+1} +s^{k+1} = -m+n \in\mathbb Z
\]Then we have to show the base case?

Answer 24

i see

Answer 25

inductive hypothesis

Answer 26

cool !

Answer 27

I usually am not this clever... I just guessed.

Answer 28

wait the trick is by being integer xD

Answer 29

wait lol all above is not needed :|

Answer 30

s+r=-1 and rs=-1 is solvable equation

Answer 31

we only need to find set on solution in R :(

Answer 32

so from wolfram
http://www.wolframalpha.com/input/?i=r%2Bs%3D-1+%2C+rs%3D-1

Answer 33

and remember this post
http://openstudy.com/users/marki#/updates/54987c7ae4b0e4319f8bf7ae

Answer 34

what i was thinking before :|

Answer 35

Here is an easy proof:
r and s are solutions of \[ x^2 + x-1=0\] hence
\[ r^2 = 1 -r\\
r^k = r^{k-2} - r^{k-1}\\

s^2 = 1 -r\\
s^k = s^{k-2} - s^{k-1}\\

r^k+ s^k = r^{k-2}+ s^{k-2} - (r^{k-1}+s^{k-1})\\
\]
And use induction. Noticing that
\[
r+s=-1\\
r^2+s^2=(r+s)^2 - 2 r s=1+2=3
\]

Answer 36

@marki @ganeshie8

Answer 37

This problem was inspired by
http://openstudy.com/users/marki#/updates/54987c7ae4b0e4319f8bf7ae
where it was asked that
\[
r^6+ s^6 =18
\]
This is a kind of generalization of it.
@campbell_st

Answer 38

\[ s^2 = 1- s\]
in my solution above.

Answer 39

nice !

Answer 40

Clever xD
I think this problem is equivalent to proving \((-1+\sqrt{5})^n + (-1-\sqrt{5})^n\) is divisible by \(2^n\) for all integers \(n\ge 1\)

Answer 41

Yes it is @ganeshie8