Relationship between subsequent squares of integers

Question

Today I realized a relationship between squares of integers that I had not noticed before.

If you start with 1, then the squares are 1, 4, 9, 16, 25, 36, 49, 64, 81... The differences between these squares are all odd numbers which keep increasing, i.e., add 3 to 1 and you get 4; add 5 to that and you get 9; add 7 to that and you get 16; add 9 to that and you get 25; add 11 - get 36, etc.

Namely, you have to add the next odd number to the square of an integer to get the square of the next integer.

How is relationship formed? Is is something that inherently arises from the square functionality?

Does this relationship continue to infinity?

score 2 · Accepted Answer · edited Jun 25 '14 at 07:46

2

That is simpler than you'd expect.

$(n+1)^2 - n^2= 2 n+1$ it gives the odd numbers series in $n$.

P.S: To tell you the truth, I myself was very much impressed by this fact when I realised it sometime ago.

edited Jun 25 '14 at 07:46

Andreas Caranti

68,873

answered Jun 25 '14 at 05:58

O_huck

500
5
16

Thank you for the answer. Now when you express it as the equation above, the relationship becomes truly obvious and logical :) – Harijs Deksnis Jun 25 '14 at 06:05
I have done some mild $\LaTeX$ editing to your answer. – Andreas Caranti Jun 25 '14 at 07:47
@AndreasCaranti : Thanks! – O_huck Jun 27 '14 at 13:45

score 2 · Answer 2 · edited Mar 09 '17 at 17:31

2

Proof without words. . . sorry I don't know how to crop white space from PDFs. . .

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edited Mar 09 '17 at 17:31

Community

1

answered Jun 25 '14 at 06:43

David

82,662

Relationship between subsequent squares of integers

2 Answers2