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.

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 (2005-08-01 19:02:11)

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!