Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

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

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

Offline

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

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 '∈'!

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

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,559

Thanks for the puzzle, juki.

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

(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

Offline

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

Another question - has this been already proved or did you toss a random problem?

If its random maybe we should try to disprove first.

Offline

**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 16,071

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

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

Character is who you are when no one is looking.

Offline

**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 16,071

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

Character is who you are when no one is looking.

Offline

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

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.

Offline

**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 16,071

Corrected the post; thanks, Mathsyperson.

Character is who you are when no one is looking.

Offline

Pages: **1**