# An odd integer series formula?

A few months ago I posted a simple equation that shows an interesting nexus between the difference between the squares of successive integers and the sums of their roots, viz:

Where y = x+1 then (x + y) = (y2 - x2)

Recently I expanded this relationship as follows:

Where n is any integer and y = (x + n), then n(x+y) = (y2 - x2)

Starting with x = 1 and y = 2 and increasing the x and y values by 1 at each iteration, this seems to produce an odd integers sequence as follows:

(1 + 2) = 3 = (4 - 1)
(2 + 3) = 5 = (9 - 4)
(3 + 4) = 7 = (16 - 9)
(4 + 5) = 9 = (25 - 16)
(5 + 6) = 11 = (36 - 25)....

Using the y = (x + n) form with the x value at 1 and increasing y by (x + n) gives a similar result:

1(1 + 2) = 3 = (4 - 1)
2(1 + 3) = 8 = (9 - 1)
3(1 + 4) = 15 = (16 - 1)
4(1 + 5) = 24 = (25 - 1)
5(1 + 6) = 35 = (36 -14)...

The difference between the successive results values being a sequence of odd integers.

Using the y = (x + n) form with the x value at 2 gives a similar result:

1(2 + 3) = 5 = (9 - 4)
2(2 + 4) = 12 = (16 - 4)
3(2 + 5) = 21 = (25 - 4)
4(2 + 6) = 32 = (36 - 4)
5(2 + 7) = 45 = (49 - 4)...

The difference between the successive results values again being a sequence of odd integers.

Not being very sophisticated mathematically I looked through Wikipedia’s integer series page (https://en.wikipedia.org/wiki/Integer_sequence) and found nothing like this series. Fibonacci numbers seem similar. I am curious to learn if this relationship is already known and whether it has any relationship to other known mathematical relationships.

To be frank here, I'm not quite sure what the interesting thing about the relationships that you found is - so if I missed it, please do point it out.

Starting with x = 1 and y = 2 and increasing the x and y values by 1 at each iteration, this seems to produce an odd integers sequence as follows:

(1 + 2) = 3 = (4 - 1)
(2 + 3) = 5 = (9 - 4)
(3 + 4) = 7 = (16 - 9)
(4 + 5) = 9 = (25 - 16)
(5 + 6) = 11 = (36 - 25)....
Well, if you increased x and y by 1 each, the sum (x+y) increases by 2 for each step, so if u started with an odd value, you will naturally generate a sequence of odd integers.

The difference between the successive results values being a sequence of odd integers.
Well the difference between successive steps is
##[(y+1)^2 - x^2] - [y^2 - x^2] = 2y + 1##
which naturally forms a sequence of successive odd integers since you're increasing ##y## by 1 each step.

Dennis Plews
TeethWhitener
Gold Member
Difference of squares: ##y^2-x^2=(y+x)(y-x)##. Substitute ##y=x+n## to get ##y^2-x^2=(y+x)n##.

Dennis Plews
SammyS
Staff Emeritus
Homework Helper
Gold Member
A few months ago I posted a simple equation that shows an interesting nexus between the difference between the squares of successive integers and the sums of their roots, viz:

Where y = x+1 then (x + y) = (y2 - x2)

Recently I expanded this relationship as follows:

Where n is any integer and y = (x + n), then n(x+y) = (y2 - x2)
...
Yes, a little over nine months ago you started such a thread.

In that thread you discussed the equation x + y = y2 - x2 .

I assume you intend the 2 to be used as an exponent here also.

The equation ##\ y = x+n \ ## is equivalent to the equation ##\ y - x=n \ ##.

Multiplying that ##\ y+x\ ## gives ##\ (y - x)(y+x) =n(y+x) \ ##.

This equation has a set of solutions in addition to those for the initial equation. These are the solutions to ##\ y=x\ ##.

Of course, ##\ (y - x)(y+x) =n(y+x) \ ## is the same as ##\ y^2 - x^2 =n(y+x) \ ##.

Dennis Plews
SammyS
Staff Emeritus
Homework Helper
Gold Member
Let f(x) = x2 .

What you have is variations on f(x+1) - f(x), for integer values of x.

For the case with n, it's essentially a similar difference with a cubic function.

I appreciate all of your comments. I have learned something from each and am encouraged to further my math skills.