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Difference of two Squares[edit]

Numbers pattern[edit]

Hi,
I recently came across a rather beautiful pattern. It's a little hard to explain (and I don't know how to go about writing a mathematical proof) so I'll be relying heavily on demonstration by examples.

Get any even number - in this example 10

Divided it by 2 → 5 - (this becomes your "middle number")

Times that number by itself → 5 x 5 = 25

Then get the next two numbers from "the middle" (4 and 6) and times them together → 4 x 6 = 24

Repeat again until you reach your original number and zero → 3 x 7 = 21, 2 x 8 = 16, 1 x 9 = 9, and 0 x 10 = 0.

OK lets examine the resulting products → 25, 24, 21, 16, 9, 0.

Now the difference between each number and the next is acending odd numbers! 1, 3, 5, 7, 9 (25 - 24 = 1), (24 - 21 = 3 etc) - intersting!

Then I noticed another pattern, if you get the difference between each consecutive number and the original you get ascending square numbers! 1, 4, 9, 16, 25 (25 - 24 = 1, 25 - 21 = 4, 25 - 16 = 9, 25 - 9 = 16, 25 - 0 = 25) - pretty nifty!

Which then lead onto the pattern that if you get any odd number and add the sum of itself and all previous prime numbers you get a square! For example 9 → 1+3+5+7+9=25; and what's more you can tell the square because it is the odd number divided by two + 0.5 squared (9/2=4.5; 4.5+0.5=5; 5^2=25)

Well that was all very nice, but then I decided to have a look at how this pattern applied to odd numbers.

OK for example I'll use 19

19/2= 9.5

+/- 0.5 gives us our "middle numbers" of 9 and 10

9x10=90, 8x11=88, 7x12=84, 6x13= 78, 5x14=70, 4x15=60, 3x16=48, 2x17=34, 1x18=18, 0x19=0

Right the differences give us 2,4,6,7,10,12,14,16,18 - sequence of ascending even numbers!

Now to the point of putting this on the reference desk - has this been noticed before and is there an explanation for it? Thanks, --Fir0002 10:50, 23 December 2006 (UTC)[reply]

Read difference of two squares and square number.--.7g. 10:56, 23 December 2006 (UTC)[reply]

Let 2n be the number you start with. Half that, the middle number, is n, the square is $n^{2}$ , the numbers either side of the middle are n-1, n+1. Their product is $(n-1)(n+1)=n^{2}-1$ , the next pair is $(n-2)(n+2)=n^{2}-4$ and the i-th pair is $(n-i)(n+i)=n^{2}-i^{2}$ , so you see the sequence of squares emerge. --Salix alba (talk) 11:09, 23 December 2006 (UTC)[reply]

Well, the difference of successive integer squares is the sequence of odd numbers: 0,1,4,9,16,25,36,49... the differences: 1,3,5,7,9,11,13... so having a sequence of odd numbers turning up in your algorithm there isn't really that surprising. I've seen this property used as a simple way to compute an integer square root (subtract odd numbers until you go negative, and the square root is how many times you subtracted). - Rainwarrior 22:17, 23 December 2006 (UTC)[reply]

Wow that's really good, thanks for your comments. Is there an explanation for the odd numbers sequence? --Fir0002 22:38, 23 December 2006 (UTC)[reply]

Take a look at Figurate number#Gnomon for a "proof without words". —Keenan Pepper 23:38, 23 December 2006 (UTC)[reply]

And if you'd prefer a proof with some words:

(n+1)^{2}-n^{2}=(n^{2}+2n+1)-n^{2}=2n+1

Or alternatively:

n^{2}-(n-1)^{2}=n^{2}-(n^{2}-2n+1)=2n-1

Where 2n - 1 is the nth odd number. A little bit of algebra can go a long way! -- Meni Rosenfeld (talk) 20:54, 24 December 2006 (UTC)[reply]

Can someone find this citation for me?[edit]

John von Neumann has a reference for the famous "Any one who considers arithmetical methods of producing random digits is, of course, in a state of sin." quote; it's cited as being from John von Neumann, "Various techniques used in connection with random digits," in A.S. Householder, G.E. Forsythe, and H.H. Germond, eds., Monte Carlo Method, National Bureau of Standards Applied Mathematics Series, 12 (Washington, D.C.: U.S. Government Printing Office, 1951): 36-38. As this was published by the GPO, it should be publically available (as far as I can tell, this is a work of the US federal government, and so public domain), but I can't find it on gpo.gov, and nist.gov only goes back to the mid-1990s. Where can I verify this? grendel|khan 20:36, 24 December 2006 (UTC)[reply]

Apparently this was reprinted in Abraham H. Taub, editor, John von Neumann: Collected Works, Volume V: Design of Computers, Theory of Automata and Numerical Analysis, pages 768–770, Pergamon Press, New York, 1963. That should be in any decent university library. --Lambiam ^Talk 22:25, 24 December 2006 (UTC)[reply]

Don Knuth, who is famously meticulous, begins volume 2 of The Art of Computer Programming with this quotation, except using "Anyone" instead of "Any one". I trust that he got the text and the attribution exactly right. The only caveat is to check the latest edition and the (online) errata for any corrections. Apparently, if you need to know where von Neumann said it, Knuth is silent. --KSmrq^T 08:24, 25 December 2006 (UTC)[reply]