Question

If z, y were the roots of the equation :

x^2 -6 x + 1 = 0

Prove that : (y^n + z^n) is a Positive integer which can be divided by 5 for any integer n

Answer 1

I would first solve for the roots. what are they?

Answer 2

3 (+ or -) 2 sqrt(2)

Answer 3

\[x^2-6x+1=0\implies x^2=6x-1\]
Since \(z,y\) are roots, we have \[y^2 = 6y-1\implies y^n=6y^{n-1}-y^{n-2}\tag{1}\]
\[z^2=6z-1\implies z^n=6z^{n-1}-z^{n-2}\tag{2}\]

Add them both and conclude by induction

Answer 4

Hmm, I have to solve it using Combinations

Answer 5

And permutations

Answer 6

Maybe this can help , The book has this solved example y = x^4 + 6 x^3 + 11 x^2 + 6 x prove that it can be divided by 24 and was solved like this

Answer 7

y = x^4 + x^3 + 5x^3 + 5x^2 + 6 x^2 + 6x
y = x^3(x+1) + 5x^2(x+1) + 6x(x+1)
y = x(x+1)(x^2 + 5x +6 )
y = x(x+1)(x+2(x+3) = (x+3)P4
(x+3)P4 / 4! = x+3C4 always positive integer so it can be divided by (24 = 4!)

Answer 8

Now, it asks to apply what I learnt in the question above.

Answer 9

Thats pretty clever!

Answer 10

Your book is using the fact that \(\binom{n}{r}\) is an integer.

You could also view it like this :
product of any \(n\) consecutive integers is divisible by \(n!\)

Answer 11

Yeah, I understood from the explanation which is too clear. The problem is to solve "try yourself" problem

Answer 12

Any ideas ?

Answer 13

Not getting any ideas yet but im still trying to see if the induction method that i suggested before really works

Answer 14

Ok.

Answer 15

binomial theorem

Answer 16

since u are mentioned permutations and combinations

Answer 17

I think you are right.

Answer 18

look at
(z+y)^n = ....

Answer 19

Y^n + z^n we will sum them , and maybe the result can help.

Answer 20

what will z^n+y^n be equal to from that binomial expansion

Answer 21

yeah play around with that, i think u can show its divisible by 5 fomr it

Answer 22

I failed , try this before.

Answer 23

I tried*

Answer 24

((3+sqrt(2))^2 + (3-sqrt(2))^2) = 2 Singles

Answer 25

** (y^n + z^n) is a Positive integer which can be divided by 5 for any integer n ***

I am having a problem with the divide by 5.
if we do a simple numerical check with n=2
and y= 3+2 sqr(2), z= 3 - 2 sqr(2)
I get y^2 + z^2 = 34

Answer 26

yeah it doesn even work for n=1

Answer 27

You think 5 is a typo from the book ?

Answer 28

so It's 2 not 5 as 2 will work for any n

Answer 29

Any ideas ?

Answer 30

recheck the question

Answer 31

I would assume the problem would be closer to the example you posted, and you start with factors that are consecutive. And then you play the same trick.
But that is not the case here, so I have no clear idea of what is the purpose of this question is here.

Answer 32

It's written as I have copied it.

Answer 33

if psble take a screenshot of as much as possible and attach

Answer 34

The second example was intended to show how the writer of the book thinks and what he wants in answering my problem

Answer 35

I can't transfer my photos to the PC or laptop. But tell me what you mean by as much as possible and I will fetch it.

Answer 36

maybe its
x^2 -5 x + 1 = 0
then for every odd n
z^n + y^n is divisible by 5

Answer 37

It's written as x^2 -6 x + 1 = 0 in the book. But fine , all I want is to know it's idea solve yours

Answer 38

k lets work on this new problem then

x^2 -5 x + 1 = 0
then for every odd positive integer n
z^n + y^n is divisible by 5

Answer 39

prove

Answer 40

Lol me ? IDK I am even stuck with 5 more in this book.

Answer 41

oh this is kidn of silly lol, the sqrt 29 in the roots will cancel out for ever odd integer

Answer 42

okay well we need a different question

Answer 43

Post here ?

Answer 44

yeah

Answer 45

Find the coefficient of x^n in the expansion of

(1 + x + 2x^2 + 3x^3 + ..... + nx^n)^2

Answer 46

u can write out on liek a dot product and it will be obvious

Answer 47

Show me

Answer 48

uh wait no lemme see