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

- TrojanPoem

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- phi

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

- TrojanPoem

3 (+ or -) 2 sqrt(2)

- ganeshie8

\[x^2-6x+1=0\implies x^2=6x-1\]
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

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## More answers

- TrojanPoem

Hmm, I have to solve it using Combinations

- TrojanPoem

And permutations

- TrojanPoem

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

- TrojanPoem

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!)

- TrojanPoem

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

- ganeshie8

Thats pretty clever!

- ganeshie8

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 \(n!\)

- TrojanPoem

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

- TrojanPoem

Any ideas ?

- ganeshie8

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

- TrojanPoem

Ok.

- dan815

binomial theorem

- dan815

since u are mentioned permutations and combinations

- TrojanPoem

I think you are right.

- dan815

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

- TrojanPoem

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

- dan815

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

- dan815

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

- TrojanPoem

I failed , try this before.

- TrojanPoem

I tried*

- TrojanPoem

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

- phi

** (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

- ganeshie8

yeah it doesn even work for n=1

- TrojanPoem

You think 5 is a typo from the book ?

- TrojanPoem

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

- TrojanPoem

Any ideas ?

- dan815

recheck the question

- phi

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.

- TrojanPoem

It's written as I have copied it.

- ganeshie8

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

- TrojanPoem

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

- TrojanPoem

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.

- dan815

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

- TrojanPoem

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

- dan815

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

- dan815

prove

- TrojanPoem

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

- dan815

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

- dan815

okay well we need a different question

- TrojanPoem

Post here ?

- dan815

yeah

- TrojanPoem

Find the coefficient of x^n in the expansion of
(1 + x + 2x^2 + 3x^3 + ..... + nx^n)^2

- dan815

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

- TrojanPoem

Show me

- dan815

uh wait no lemme see

- dan815

|dw:1434805428709:dw|

- dan815

so that series

- dan815

|dw:1434805474806:dw|

- dan815

|dw:1434805575985:dw|

- dan815

n+(n-1)*(n+1)*(n)/2

- TrojanPoem

WOW, seems nice although it's out of my curriculum.

- dan815

no no u know this stuff, u know summation of 1+2+3.... +n
that formula right

- dan815

im just using that to get an expression for the summation

- TrojanPoem

arithmetic ?

- dan815

yeah

- TrojanPoem

But I have never seen this E or epslion.

- dan815

k ignore that look at the pattern that is coming up, here is a simple example
look at a simple example
(1+2x+3x^2)^2

- dan815

|dw:1434805819564:dw|

- dan815

u mulitply the first and last term
then the 2nd and 2nd last term
3rd and 3rd last term
and u add them all up to get the constant on the nth degree

- TrojanPoem

Each time you are multiplying by x and increase 1 to the coefficeitn

- dan815

maybe its easier to see if i write it out in polynomial form

- dan815

|dw:1434806006840:dw|

- TrojanPoem

At first I though of solving it as the sum of geometrical like this ( 1 + x+ x^2 + x^3 + x^n + x^2 + x^3 +...+ x^n + x^3 +... +x^n until x^n is alone ) but failed ( too long)

- dan815

|dw:1434806044605:dw|

- dan815

u do the same thing with your bigger question ull see a pattern and u can see what they add upto

- TrojanPoem

Oh , clarify a bit this multiplication

- dan815

because all other multiplications will give different degree term

- dan815

hm

- dan815

i dunno how else to say it

- TrojanPoem

I got it to some extent , focusing with myself I will get full understanding of it.

- TrojanPoem

Now there is more 4 questions , wanna try ?

- dan815

okay try a similar problem, can u tell me waht the coefficient on the n-1 th degree term is

- dan815

and sure

- TrojanPoem

let me think

- dan815

are you in highschool btw?

- TrojanPoem

yeah, but - not to lie- I never took this until now.

- dan815

is this AP algebra?

- asnaseer

if z,y are indeed the roots of: \(x^2-6x+1=0\)
then this implies that: \(z+y=6\implies z=6-y\)
therefore: \(z^n+y^n=(6-y)^n+y^n=6^n-n6^{n-1}y+...-y^n+y^n\)
Note that the last two terms cancel out leaving an expression that is always a multiple of 6

- asnaseer

so it cannot be a multiple of 5 for all n - are you sure the question is stated correctly?

- TrojanPoem

I wrote it as It was in the book but maybe typo of the book writer . But WOW you got it

- asnaseer

looking at the other replies I believe @dan815 proposal is probably correct - i.e. the question must be a misprint of \(x^2-5x+1=0\). With this equation you would indeed get a multiple of 5.

- TrojanPoem

I like this z+y = 6 from the equation really nice !

- asnaseer

thx :)

- dan815

oh allsoo dont forgget to not that equation z+y = 6
z^n + y^n is div by 6 only when n is ODD* positive int same thing with the x^2-5x+1

- asnaseer

good spot!

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