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#1 2005-07-30 02:00:29

juki
Member
Registered: 2005-07-30
Posts: 3

what u think of this problem

1. x and y are two natural numbers such that 3x^2 + x = 4y^2 +y . Prove that x - y is the square of a whole nnumber rate this problem from 0 to 10 and tell ur soloution

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#2 2005-07-30 14:38:50

kylekatarn
Member
Registered: 2005-07-24
Posts: 445

Re: what u think of this problem

is this what we need to prove?
(x,y)∈N and 3x² +x=4y² +y => x-y=n² and n∈Z

//Mathisfun - Thanks for adding the symbol '∈'! smile

Last edited by kylekatarn (2005-07-30 23:38:18)

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#3 2005-07-30 17:19:34

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,535

Re: what u think of this problem

Thanks for the puzzle, juki.

Oh, and kyle, I just added the "∈" up top, so you can use that if you want smile

(BTW kylekatarn is confirming: "is x and y a Natural Number {1,2,3,4,...} and is n an integer {... -3, -2, -1, 0, 1, 2, 3, ...} ?")

Let's plug in some numbers just to get this started:

x=1: 3x^2 + x = 4y^2 +y  becomes 5 =  4y^2 +y

4y^2 +y - 5 = 0 has the solutions 1 and -1.25

Now, -1.25 is not included because y should be a Natural Number, so we are left with

x=1, y=1, and x-y=0, so I suppose 0 is the square of 0, so that is a good start.


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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#4 2005-07-30 23:41:52

kylekatarn
Member
Registered: 2005-07-24
Posts: 445

Re: what u think of this problem

Another question - has this been already proved or did you toss a random problem?
If its random maybe we should try to disprove first.

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#5 2005-07-31 18:21:00

ganesh
Moderator
Registered: 2005-06-28
Posts: 14,381

Re: what u think of this problem

juki wrote:

x and y are two natural numbers such that 3x^2 + x = 4y^2 +y

Put x=2,
we get 4y² + y - 14 = 0, Solving,
we get y = [-1 ± √(1 + 224)] / 8
y = -2 or 1.75
Neither of them are Natural Numbers!
x - y is NOT always a square of a natural number roll

Last edited by ganesh (2005-07-31 21:02:11)


Character is who you are when no one is looking.

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#6 2005-07-31 18:36:24

ganesh
Moderator
Registered: 2005-06-28
Posts: 14,381

Re: what u think of this problem

MathsIsFun wrote:

Let's plug in some numbers just to get this started:

x=1: 3x^2 + x = 4y^2 +y  becomes 5 =  4y^2 +y

When we put x=1,
3x² + x = 3(1² ) + 1 = 4
Therefore,
4y² + y - 4 = 0
y = [-1 ± √ (1 + 64)]/8 = [-1 ± √ 65 ]/8
which is an irational number.
We see that here too x-y is not the square of a natural number smile


Character is who you are when no one is looking.

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#7 2005-07-31 20:54:52

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: what u think of this problem

ganesh wrote:

Put x=2,
we get 4y² + y - 14 = 0, Solving,
we get y = [-1 ± √(1 + 224)] / 2

y=[-1 ± √(1 + 224)]/8, actually. That means that y=-2 or 1.75 and the original proof wanted x and y to both be natural, meaning that these values are disregarded. The same applies to your second post where x=1. Sorry!


Why did the vector cross the road?
It wanted to be normal.

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#8 2005-07-31 21:03:13

ganesh
Moderator
Registered: 2005-06-28
Posts: 14,381

Re: what u think of this problem

Corrected the post; thanks, Mathsyperson. smile


Character is who you are when no one is looking.

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