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TrojanPoem
 one year ago
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
 one year ago
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

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phi
 one year ago
Best ResponseYou've already chosen the best response.0I would first solve for the roots. what are they?

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.03 (+ or ) 2 sqrt(2)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\[x^26x+1=0\implies x^2=6x1\] Since \(z,y\) are roots, we have \[y^2 = 6y1\implies y^n=6y^{n1}y^{n2}\tag{1}\] \[z^2=6z1\implies z^n=6z^{n1}z^{n2}\tag{2}\] Add them both and conclude by induction

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0Hmm, I have to solve it using Combinations

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0Maybe 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
 one year ago
Best ResponseYou've already chosen the best response.0y = 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
 one year ago
Best ResponseYou've already chosen the best response.0Now, it asks to apply what I learnt in the question above.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Thats pretty clever!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Your 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
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, I understood from the explanation which is too clear. The problem is to solve "try yourself" problem

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Not getting any ideas yet but im still trying to see if the induction method that i suggested before really works

dan815
 one year ago
Best ResponseYou've already chosen the best response.3since u are mentioned permutations and combinations

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0I think you are right.

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0Y^n + z^n we will sum them , and maybe the result can help.

dan815
 one year ago
Best ResponseYou've already chosen the best response.3what will z^n+y^n be equal to from that binomial expansion

dan815
 one year ago
Best ResponseYou've already chosen the best response.3yeah play around with that, i think u can show its divisible by 5 fomr it

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0I failed , try this before.

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0((3+sqrt(2))^2 + (3sqrt(2))^2) = 2 Singles

phi
 one year ago
Best ResponseYou've already chosen the best response.0** (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
 one year ago
Best ResponseYou've already chosen the best response.1yeah it doesn even work for n=1

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0You think 5 is a typo from the book ?

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0so It's 2 not 5 as 2 will work for any n

phi
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0It's written as I have copied it.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1if psble take a screenshot of as much as possible and attach

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0The second example was intended to show how the writer of the book thinks and what he wants in answering my problem

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.3maybe its x^2 5 x + 1 = 0 then for every odd n z^n + y^n is divisible by 5

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0It'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
 one year ago
Best ResponseYou've already chosen the best response.3k 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

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0Lol me ? IDK I am even stuck with 5 more in this book.

dan815
 one year ago
Best ResponseYou've already chosen the best response.3oh this is kidn of silly lol, the sqrt 29 in the roots will cancel out for ever odd integer

dan815
 one year ago
Best ResponseYou've already chosen the best response.3okay well we need a different question

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0Find the coefficient of x^n in the expansion of (1 + x + 2x^2 + 3x^3 + ..... + nx^n)^2

dan815
 one year ago
Best ResponseYou've already chosen the best response.3u can write out on liek a dot product and it will be obvious

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0WOW, seems nice although it's out of my curriculum.

dan815
 one year ago
Best ResponseYou've already chosen the best response.3no no u know this stuff, u know summation of 1+2+3.... +n that formula right

dan815
 one year ago
Best ResponseYou've already chosen the best response.3im just using that to get an expression for the summation

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0But I have never seen this E or epslion.

dan815
 one year ago
Best ResponseYou've already chosen the best response.3k 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
 one year ago
Best ResponseYou've already chosen the best response.3u 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
 one year ago
Best ResponseYou've already chosen the best response.0Each time you are multiplying by x and increase 1 to the coefficeitn

dan815
 one year ago
Best ResponseYou've already chosen the best response.3maybe its easier to see if i write it out in polynomial form

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0At 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
 one year ago
Best ResponseYou've already chosen the best response.3u do the same thing with your bigger question ull see a pattern and u can see what they add upto

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0Oh , clarify a bit this multiplication

dan815
 one year ago
Best ResponseYou've already chosen the best response.3because all other multiplications will give different degree term

dan815
 one year ago
Best ResponseYou've already chosen the best response.3i dunno how else to say it

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0I got it to some extent , focusing with myself I will get full understanding of it.

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0Now there is more 4 questions , wanna try ?

dan815
 one year ago
Best ResponseYou've already chosen the best response.3okay try a similar problem, can u tell me waht the coefficient on the n1 th degree term is

dan815
 one year ago
Best ResponseYou've already chosen the best response.3are you in highschool btw?

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0yeah, but  not to lie I never took this until now.

asnaseer
 one year ago
Best ResponseYou've already chosen the best response.1if z,y are indeed the roots of: \(x^26x+1=0\) then this implies that: \(z+y=6\implies z=6y\) therefore: \(z^n+y^n=(6y)^n+y^n=6^nn6^{n1}y+...y^n+y^n\) Note that the last two terms cancel out leaving an expression that is always a multiple of 6

asnaseer
 one year ago
Best ResponseYou've already chosen the best response.1so it cannot be a multiple of 5 for all n  are you sure the question is stated correctly?

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0I wrote it as It was in the book but maybe typo of the book writer . But WOW you got it

asnaseer
 one year ago
Best ResponseYou've already chosen the best response.1looking at the other replies I believe @dan815 proposal is probably correct  i.e. the question must be a misprint of \(x^25x+1=0\). With this equation you would indeed get a multiple of 5.

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0I like this z+y = 6 from the equation really nice !

dan815
 one year ago
Best ResponseYou've already chosen the best response.3oh 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^25x+1
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