Question

fun question binomial :)



Please note that a correction is there in the question-http://prntscr.com/9ibt7n

Answer 1

Prove that if \(p\) is a prime \(p>2\) then the difference- \(\ ([2+\sqrt{5}]^p-2^{p+1})\) is divisible by p.
Here [.] is the greatest integer function.

Answer 2

Qwerty... ^_^ I hope if I take the time to answer this "fun question" I get owl bucks :D

Answer 3

wait... o.o your a qualified helper? :O what the hell man... even you got accepted and I didn't ;-;

Answer 4

qwerty is that floor function ?

Answer 5

qwerty on the last 2... it seems like you put "P + 1" as to the power is that correct?

Answer 6

yeah its also known as floor function :)
http://mathworld.wolfram.com/FloorFunction.html

Answer 7

I got the answer :D

Answer 8

"p+1" is in the power there and the question is 100% clear
if you got the answer then share it :) no more spam

Answer 9

spam

Answer 10

what interest me the most is that
[2+sqrt5]=4 :O
and its not like that
[(2+sqrt5)^p] which would be much more not lovely way xD

Answer 11

oh damn its this-\(\ ([(2+\sqrt{5})^p]-2^{p+1})\)

Answer 12

see i thought that!!

Answer 13

thanks :)

Answer 14

It doesn't matter if there's sqrt(5) or sqrt(4) or sqrt(6) ...
It doesn't change anything. The number should be in [4,9)
anyhow, it should be simplified this way:

Why this fraction is always equal to a natural number:
\[\frac{2 ^{2P} - 2 ^{P+1}}{ P }\]

Answer 15

well @Yavar that was a typo only
the origin question is\( \ ([(2+\sqrt{5})^p]-2^{p+1})\)

Answer 16

factoring the numerator gives this:

\[\frac{2 ^{P}\times(2 ^{P}-2) }{ P }\]

Answer 17

OH, COME ON. i WAS THINKING THAT WAY.

Answer 18

@imqwerty does it have a solution? And you are sure there's a solution?

Answer 19

yeah :)

Answer 20

i can give some hints

Answer 21

So I'm not wasting my time for a fake question?

Answer 22

yeah you are not :)
i'll share the solution its related to binomial

Answer 23

there's a simple point which no one's considering.

Answer 24

yeah :) think of \(2+\sqrt{5}\) and a similar number to add or subtract so that you can get something helpful

Answer 25

:))))))))))))))))))
Aaaaaaaah, you make me crazy!!!!!!!!!!!!!!!!!!!!!!!!!!
@SolomonZelman could you join us?

Answer 26

@SolomonZelman no idea about what @imqwerty just said?

Answer 27

I am trying to see if I can get something worthy...

Answer 28

No, I don't really know, sorry -:(

Answer 29

I would however be curious to see the solution when it comes out, so I won't remove what I say, to get a notification....

Answer 30

neither do I....
I'm juz waiting for the answer!

Answer 31

okay i'll tell the answer :)

Answer 32

lets say
\(x=(2+\sqrt{5})^p\)
when we expand \(x\) we get this-
\(x= ~^pC_0(\sqrt{5})^p(2)^0 +~ ^pC_1(\sqrt{5}^{p-1}(2)^1+.....~^pC_p(\sqrt{5})^0(2)^p\)

now lets say \(N=(\sqrt{5}-2)^p\)
here we can clearly see that the inside part of \(N\) which is \((\sqrt{5}-2)\) is less than 1
so therefore \((\sqrt{5}-2)^p\) must also be less than 1

:) wanna try from here?

Answer 33

I swear I'll find a very FUN QUESTION for you later. :)))

Answer 34

:) thanks
should i continue?

Answer 35

I can't continue.
My brain needs some rest.
I will save your question and prefer to work on it later.
But I'll use your hint.

Answer 36

okay :)

Answer 37

Binomial ha!

Answer 38

so the question is \[4^p-2^{p+1}\] divisible by p?

Answer 39

or am i making wrong interpretation ?

Answer 40

sorry i made a typo there but it is this-
\(\ ([(2+\sqrt{5})^p]-2^{p+1})\)
it need not be \(4^p-2^{p+1}\)

Answer 41

oh i see the correction (2+root5) in the integer function

Answer 42

the whole power i mean

Answer 43

now it becomes interesting

Answer 44

yea:)

Answer 45

so by binomial you mean we expand (2+root5)^p

Answer 46

i just wrote binomial there because i was taught this in binomial class
well we should not really call it a binomial question but yeah it does have relation with expansion

Answer 47

Ok!

Answer 48

hey
what i see so far; \(2^p(\lfloor( \sqrt{\frac{5}{2}})^p+(\sqrt{\frac{5}{2}})^{p-1} p+...+1 \rfloor -2)\)

Answer 49

i don't see how can proceed so far

Answer 50

@imqwerty why using N=(root5-2)^p

Answer 51

@xapproachesinfinity how did you get that?

Answer 52

hmm i think i made errors nevermind

Answer 53

that shoould be root(5)/2 all to the power p
i expanded and factored

Answer 54

how did you get that \(2^p\) out of the Greatest integer function?
consider this case-> \([2 \times 1.5] \ne 2[1.5]\)
here [.] is the greatest integer function

Answer 55

hmm true! that's another mistake

Answer 56

think about the expansion of \(N\) and \(x\) :)
and remember that \(x=[x]+\left\{ x \right\} \)
here [x] is G.I.F
and {x} represents the fractional part of x

Answer 57

i'm not used to fractional part but it seems not that frightening
considering its definition

Answer 58

just an example-
if you have x=3.14
then
\(x=[x]+\left\{ x \right\}\)
here
\(x=3.14\)
\([x]=3\)
\(\left\{ x \right\}=0.14\)

Answer 59

i see

Answer 60

i gave up hahaha

Answer 61

u tryin to use [x]=x-{x}
when u said think about expansion of N?

Answer 62

i wanted to do some useful mathematical operation with N and x
and i wrote x=[x]+{x}
so that i can replace x from that expression with [x]+{x} and get a meaning full result :)

Answer 63

wait i havn't gave up yet :O

Answer 64

:)

Answer 65

:)

Answer 66

haha :)

Answer 67

ShadowLegendX You've already chosen the best response. Medals 0
spam


LOL

Answer 68

Waiting for reply.

Answer 69

\(x=(\sqrt{5}+2)^p\)
and \(N=(\sqrt{5}-2)^p\)

expansion of \(x\) is-> \(x=~ ^pC_0(\sqrt{5})^p(2)^0 +~^pC_1(\sqrt{5})^{p-1}(2)^1~+.....+~^pC_p(\sqrt{5})^{p-p}(2)^p\)

expansion of \(N\) is-> \(x=~ ^pC_0(\sqrt{5})^p(-2)^0 +~^pC_1(\sqrt{5})^{p-1}(-2)^1~+.....+~^pC_p(\sqrt{5})^{p-p}(-2)^p\)

now
\(x-N=~2\underbrace{(~^pC_1(\sqrt{5}^{p-1}(2)^1+~^pC_3(\sqrt{5})^{p-3}(2)^3+.....)} \\\hspace{118pt}{It~is~an~Integer~say~I}\\\\\)

replacing \(x\) with \([x]+\left\{ x \right\}\)

\([x]+\left\{ x \right\}-N=~2I\)

here on LHS \([x]\) is integer so \(\left\{ x \right\}-N\) must be an integer
now we know that \(0\le \left\{ x \right\} \le 1 \)
and \(0 < N < 1\)
therefore \(-1 < \left\{ x \right\}-N<1\)
now only integer value between -1 and 1 is 0 so \(\left\{ x \right\}-N=0\)

so now we have this-

\([x]=2(~^pC_1(\sqrt{5}^{p-1}(2)^1+~^pC_3(\sqrt{5}^{p-3}(2)^3+.....)\)
\([x]=~^pC_1(\sqrt{5})^{p-1}(2)^2+~^pC_3(\sqrt{5})^{p-3}(2)^4+......~^pC_p(\sqrt{5})^{p-p}(2)^{p+1}\)
\( [x]-2^{p+1}=~^pC_1(\sqrt{5})^{p-1}(2)^2+~^pC_3(\sqrt{5})^{p-3}(2)^4+......\)


clearly all the right hand side terms have a factor of p in them so RHS is divisible by p
Thus the LHS that is - \([x]-2^{p+1}\) is also divisible by p :)

Answer 70

@ganeshie8 remember us doing this? :D

Answer 71

Man, I saw your question just two seconds after you posted your answer. Could've seen it earlier -_-

Answer 72

Here it is
http://openstudy.com/users/parthkohli#/updates/5666f691e4b00a591e8e2dff

Answer 73

You have caused a dilemma on who should I give the medal to.

Answer 74

oh damn, I had the feeling that i needed to take the difference x-N
but couldn't figure out what would that be used for

Answer 75

how do you show that what you called I is an integer?
something does not feel write here?

Answer 76

the whole proof hinges on I being an integer

Answer 77

p is a prime greater than 2
and its clear that \(^pC_1 , ~^pC_3,~^pC_5,~.....\) will be integer
and also \(2^1,~2^2,~2^3....\) will be integer
now if the root5 terms are like -> \(\sqrt{5}^{p-(2n-1)}\) here the 2n-1 part is an odd integer less than p in each term and also p is odd because primes >2 are all odd
so we have- \(\sqrt{5}^{odd-odd}\)
odd-odd is always even thus all powers of \(\sqrt{5}\) are even or we can say that they are multiple of 2
thus we can say that this part will also be integral
so all part are integral therefore that whole thing is an integer

Answer 78

oh i see thanks for clarification

Answer 79

np :)

Answer 80

Um, shouldn't this be closed? o-o

Answer 81

no