Unique integer solutions

[Changabooniggiwan]

Not exactly a puzzle, because I have no solution, but one of you maths whizzes might be able to help me...

The equation

n = x^2 + (2y - 1)x + (y^2 - 3y + 1)

where n is a positive odd integer, has a unique solution for x and y where both x and y are positive non-zero integers. Trust me on this. As you might expect, however, there are endless non-integer and negative solutions.

For example, if n = 99, x = 5 and y = 6.

99 = 5^2 + (2*6 - 1)*5 + (6^2 - 3*6 + 1)

Put another way, if you have positive, non-zero integer values for x and y, n will be an odd positive integer unique to that combination of x and y values. Again, trust me here.

Can the positive integer solutions of x and y be determined algebraically, given n?

Take painkillers before you try this.
Changa

[Changabooniggiwan]

Actually, you shouldn't just take my word for it...

Imagine the odd numbers arranged thus:
(apologies if this comes out misaligned)

________1
_______3 5
______7 9 11
___13 15 17 19
_21 23 25 27 29

You get the idea.

The formula n = x^2 + (2y - 1)x + (y^2 - 3y + 1) gives a value on the grid unique to each diagonal intersection. For example, 15 occurs in the 2nd diagonal from right to left (x = 2) and in the 3rd diagonal from left to right (y = 3).

________1
_______3 5
______7 9 11
___13 15 17 19
_21 23 25 27 29 x = 2

________1
_______3 5
______7 9 11
___13 15 17 19
_21 23 25 27 29 y = 3


Since each odd number only occurs once in the table, and since each x and y value is an integer, that's the reason for my certainty.

[esme]

x^2 + (2y - 1)x + (y^2 - 3y + 1) = x^2 + 2xy -x + y^2 -3 y +1= (x + y - 1/2)^2+ 3/4 -2 y which gives 4n = (2s -1)^2 + 3 -8 y

The formula is more or less given by taking the next-largest square above 4n with appropriate remainder of s modulo 4 and then calculating the rest.

If it is indeed true that the answer is unique, the formula goes something like this:

Take the ceiling (i.e. round up) of the square root of 4n, round up to the next odd integer, this is 2s -1, calculate 8y - 3 = (2s-1)^2 - 4n, this gives y, calculate x= s-y.

It is not difficult to write down the formulas, but here on this forum, it is no fun without mathjax or similar.
_________________
Mundus vult decipi, ergo decipiatur.

[DejMar]

Not the solution, but one might simplify the quadratic polynomial using the quadratic formula into the equations of the roots and provide the discriminant that one could use to in finding the real solutions and assist in determining possible integer solutions to the given equation.

discriminant: 8y - 3 + 4n
x1 = (1/2)*(1 - 2y + SQRT(8y - 3 + 4n)) and
x2 = (1/2)*(1 - 2y - SQRT(8y - 3 + 4n))

As can be seen in the equations, the square root of the discriminant must be odd (and an integer) in order for the roots to be possible integers. As had been stated, for a positive non-zero integer y, only one root [x1] can be positive.
[n=99; y=6; x1=5; x2=-16]

[Thok]

The geometric description makes the equation want to be in coordinates s=x+y and t=x-y, so x=(s+t)/2 and y=(s-t)/2

Then x^2 + (2y - 1)x + (y^2 - 3y + 1) = (x^2+2xy+y^2)-x-3y+1
=s^2-(s+t)/2-3(s-t)/2+1 = s^2-2s+t+1= (s-1)^2+t

Setting this equal to n, we have (s-1)^2+t=n. Solving for s, we get
s = 1+sqrt(n-t) (we get the positive square root, since s=x+y>=1). But there doesn't seem to be a good way to determine t except by trying a bunch of possibilities. It's not quite so bad, since s<sqrt(n)-1, and -s<t<s.

Edit: Yeah, take the square root of n, round to get s-1, set t=n-(s-1)^2, solve for x and y.