TrojanPoem
  • TrojanPoem
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
Mathematics
schrodinger
  • schrodinger
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phi
  • phi
I would first solve for the roots. what are they?
TrojanPoem
  • TrojanPoem
3 (+ or -) 2 sqrt(2)
ganeshie8
  • 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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TrojanPoem
  • TrojanPoem
Hmm, I have to solve it using Combinations
TrojanPoem
  • TrojanPoem
And permutations
TrojanPoem
  • 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
  • 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
  • TrojanPoem
Now, it asks to apply what I learnt in the question above.
ganeshie8
  • ganeshie8
Thats pretty clever!
ganeshie8
  • 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
  • TrojanPoem
Yeah, I understood from the explanation which is too clear. The problem is to solve "try yourself" problem
TrojanPoem
  • TrojanPoem
Any ideas ?
ganeshie8
  • ganeshie8
Not getting any ideas yet but im still trying to see if the induction method that i suggested before really works
TrojanPoem
  • TrojanPoem
Ok.
dan815
  • dan815
binomial theorem
dan815
  • dan815
since u are mentioned permutations and combinations
TrojanPoem
  • TrojanPoem
I think you are right.
dan815
  • dan815
look at (z+y)^n = ....
TrojanPoem
  • TrojanPoem
Y^n + z^n we will sum them , and maybe the result can help.
dan815
  • dan815
what will z^n+y^n be equal to from that binomial expansion
dan815
  • dan815
yeah play around with that, i think u can show its divisible by 5 fomr it
TrojanPoem
  • TrojanPoem
I failed , try this before.
TrojanPoem
  • TrojanPoem
I tried*
TrojanPoem
  • TrojanPoem
((3+sqrt(2))^2 + (3-sqrt(2))^2) = 2 Singles
phi
  • 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
  • ganeshie8
yeah it doesn even work for n=1
TrojanPoem
  • TrojanPoem
You think 5 is a typo from the book ?
TrojanPoem
  • TrojanPoem
so It's 2 not 5 as 2 will work for any n
TrojanPoem
  • TrojanPoem
Any ideas ?
dan815
  • dan815
recheck the question
phi
  • 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
  • TrojanPoem
It's written as I have copied it.
ganeshie8
  • ganeshie8
if psble take a screenshot of as much as possible and attach
TrojanPoem
  • TrojanPoem
The second example was intended to show how the writer of the book thinks and what he wants in answering my problem
TrojanPoem
  • 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
  • dan815
maybe its x^2 -5 x + 1 = 0 then for every odd n z^n + y^n is divisible by 5
TrojanPoem
  • 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
  • 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
  • dan815
prove
TrojanPoem
  • TrojanPoem
Lol me ? IDK I am even stuck with 5 more in this book.
dan815
  • dan815
oh this is kidn of silly lol, the sqrt 29 in the roots will cancel out for ever odd integer
dan815
  • dan815
okay well we need a different question
TrojanPoem
  • TrojanPoem
Post here ?
dan815
  • dan815
yeah
TrojanPoem
  • TrojanPoem
Find the coefficient of x^n in the expansion of (1 + x + 2x^2 + 3x^3 + ..... + nx^n)^2
dan815
  • dan815
u can write out on liek a dot product and it will be obvious
TrojanPoem
  • TrojanPoem
Show me
dan815
  • dan815
uh wait no lemme see
dan815
  • dan815
|dw:1434805428709:dw|
dan815
  • dan815
so that series
dan815
  • dan815
|dw:1434805474806:dw|
dan815
  • dan815
|dw:1434805575985:dw|
dan815
  • dan815
n+(n-1)*(n+1)*(n)/2
TrojanPoem
  • TrojanPoem
WOW, seems nice although it's out of my curriculum.
dan815
  • dan815
no no u know this stuff, u know summation of 1+2+3.... +n that formula right
dan815
  • dan815
im just using that to get an expression for the summation
TrojanPoem
  • TrojanPoem
arithmetic ?
dan815
  • dan815
yeah
TrojanPoem
  • TrojanPoem
But I have never seen this E or epslion.
dan815
  • 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
  • dan815
|dw:1434805819564:dw|
dan815
  • 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
  • TrojanPoem
Each time you are multiplying by x and increase 1 to the coefficeitn
dan815
  • dan815
maybe its easier to see if i write it out in polynomial form
dan815
  • dan815
|dw:1434806006840:dw|
TrojanPoem
  • 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
  • dan815
|dw:1434806044605:dw|
dan815
  • dan815
u do the same thing with your bigger question ull see a pattern and u can see what they add upto
TrojanPoem
  • TrojanPoem
Oh , clarify a bit this multiplication
dan815
  • dan815
because all other multiplications will give different degree term
dan815
  • dan815
hm
dan815
  • dan815
i dunno how else to say it
TrojanPoem
  • TrojanPoem
I got it to some extent , focusing with myself I will get full understanding of it.
TrojanPoem
  • TrojanPoem
Now there is more 4 questions , wanna try ?
dan815
  • dan815
okay try a similar problem, can u tell me waht the coefficient on the n-1 th degree term is
dan815
  • dan815
and sure
TrojanPoem
  • TrojanPoem
let me think
dan815
  • dan815
are you in highschool btw?
TrojanPoem
  • TrojanPoem
yeah, but - not to lie- I never took this until now.
dan815
  • dan815
is this AP algebra?
asnaseer
  • 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
  • asnaseer
so it cannot be a multiple of 5 for all n - are you sure the question is stated correctly?
TrojanPoem
  • TrojanPoem
I wrote it as It was in the book but maybe typo of the book writer . But WOW you got it
asnaseer
  • 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
  • TrojanPoem
I like this z+y = 6 from the equation really nice !
asnaseer
  • asnaseer
thx :)
dan815
  • 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
  • asnaseer
good spot!

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