Talk:Square number/Archive 1

Note on layout
note on layout: don't use FONT tags, especially not to specify "windings" as the font - not all users will have it. -- Tarquin 22:11, 21 April 2003 (UTC)


 * I have seen now how it was supposed to look; earlier my browser was set "ignore fonts". - Patrick 23:27, 21 April 2003 (UTC)


 * Okay -- Sorry about that, I didn't realize. Jacquerie27 (talk) 07:46, 22 April 2003 (UTC)


 * Fredrik, couldn't you also leave the other method there? It might be useful for less experienced people who cannot understand the method that you have written. - 80.229.152.246 18:13, 15 February 2005 (GMT)


 * Can we add another formula? This one is for finding the next square number in the sequence from any square number.  I have written it and await approval/disapproval.  It can be found here: User:Whiteheadj/Square Number.  Enjoy! --Whiteheadj 20:46, 16 November 2005 (UTC)

Negative squared isn't positive
The fact that any negative number squared becomes positive is actually false. If you ever see the expression -3^2 evaluated as 9, that's incorrect. The exponentiation is always done before the negation unless there are parentheses there to indicate otherwise.

However, there are some contexts in which it _looks_ like texts are saying that -3^2 = 9, but a closer inspection will either reveal a subtle interpretation or a misunderstanding. For instance, what is the difference between the following statements:

"If I take negative three and square it, I get nine." "If I square negative three, I get nine." "If I evaluate negative three squared, I get negative nine." "If I take the opposite of three squared, I get negative nine."

All of the above statements are correct. The reason some of them end up with 9 as the answer and some end up with -9 is that some of the statements have groupings implied in their phrasing. The first two statements translate into algebraic notation as (-3)^2 = 9, the third statement translates to -3^2 = -9, and the fourth statement translates to -(3^2) = -9. This has been taken from the webite http://mathforum.org/library/drmath/view/55709.html — Preceding unsigned comment added by 24.17.78.220 (talk) 02:21, 24 March 2006 (UTC)

Chen's theorem

 * ''Chen Jingrun showed in 1975 that there always exists a number P that $$\Omega (P)\le 2 $$ between n2and (n+1)2. See also Legendre's conjecture and Big Omega function.

This is dubiously relevant to the article. We cannot include every theorem that ever dealt with squares, so why this one?

However, my real objection is to the wording. User:WATARU may be proud that he knows what the Big Omega function is, but that is no reason to require our readers to do so. To say that there is either a prime or a product of two primes between two consecutive squares is both shorter and more comprehensible. Septentrionalis 19:31, 13 September 2006 (UTC)

Showing numbers are not perfect squares
I want to see if I can come up with some criteria that the digits of a number must follow every piece of for it to be a perfect square. Look at Talk:241 (number) and it will show you what piece of criteria can be used to rule it out as a perfect square. In turn, however, there still are some numbers that meet all of the criteria that are not perfect squares, the smallest of these is 721. Anyone know how to rule out 721?? Georgia guy 22:49, 24 February 2006 (UTC)


 * The simplest reason why 721 cannot be square is that it is divisible by 7 but not 49. As you point out, divisibility by seven can be determined by the digits; a similar test can be worked out for forty-nine. But a general test, in this manner, would probably be infinitely long, since it would requite a test for every prime. Septentrionalis 19:35, 13 September 2006 (UTC)


 * I couldn't follow the discussion at 241, but perhaps what he is getting at is that the decimal representation of a square must end in 0, 1, 4, 5, 6, or 9. This test doesn't rule out 241 or 721.  But one can take it further:   the decimal representation of a square must end in 00, 01, 04, 09, 16, 21, 24, 25, 29, 36, 41, 44, 49, 56, 61, 64, 69, 76, 81, 84, 89, or 96; this rules out 78% of all integers as possible squares---although still not 241 or 721.  But taking it one step further, one finds that the decimal representation must end in one of the 159 patterns 000, 001, ..., 489, ..., 976, 984, or 996, thus ruling out 84.1% of all integers as possible squares, and including 241 and 721.  -- Dominus 03:21, 14 September 2006 (UTC)
 * The discussion at 241 is a demonstration that 241 = 7n + 3, for some n. Squares must be 49n, 7n+1, 7n +2,, or 7n+4 (proved by enumeration in base 7) Septentrionalis 01:23, 15 September 2006 (UTC)


 * When it comes to 721, I used the rule about adding 5 times the last digit to the rest of the number (72 + 5 = 77; 7 + 35 = 42) and this proves it is not a perfect square. However, even with all this criteria, there still are some numbers that meet the criteria but that are not perfect squares, the smallest of which is 1009. Georgia guy 00:55, 19 September 2006 (UTC)
 * And, in turn, I expanded the criteria to include divisibility by 11 (which the remainder must be 0, 1, 3, 4, 5, or 9) and it ruled out 1009, but didn't rule out 1969. Georgia guy 16:30, 19 September 2006 (UTC)
 * Now, I expanded to include divisibility by 13 and the remainder must be 0, 1, 3, 4, 9, 10, or 12. This does rule out 1969, but even then, there is a larger number, 5769, that still meets all the criteria that is not a perfect square. Georgia guy 16:28, 11 October 2006 (UTC)

Figuring out squares
Look at this! 0^2 is 0, 1^2 is 1, 2^2 is 4, 3^2 is 9, 4^2 is 16, etc. Well, i figured out that if you take the answer to a square and add an odd number that is next in the sequence starting at 1, you get the next square. Example: 0(first square) + 1(first odd number) = 1(second square), 1 + 3(second odd number) = 4(third square)! This continues on. my dad did the mathimatical proof. y=(x+1)^2 -x^2, where x = the number before the number you want to square and the y = the odd number to add. So if x = 10 (this means we are trying to get 11), then y=(10+1)^2 -100, or y=(11)^2 -100. The answer is 21. 100 + 21 = 121 = 11^2! If you factor this out, it becomes y = x^2 + 2x + 1 - x^2, or     y = 2x +1. Plug in 10 for x, and you get y = 20+1, or y = 21. This is an easy way to find hard square numbers. Isn't that amazing? Dogmanice 03:56, 7 December 2006 (UTC)


 * Unfortunately, that's a well known property, it already appears in the article, and it doesn't really assist in finding squares except for rather specific cases. Calculating something like 7253482 takes less than a minute with long multiplication but is virtually impossible to calculate with this method. -- Meni Rosenfeld (talk) 10:32, 8 December 2006 (UTC)

Something else
a^2 - b^2 = a + b  if  b+1=a

— Preceding unsigned comment added by AnarchyElmo (talk • contribs) 05:21, 17 February 2007 (UTC)

0 zeroth square or first square??
0 is the square of 0, but is the first, not the zeroth, number in the list of square numbers in this article. Georgia guy (talk) 23:57, 1 March 2008 (UTC)


 * In I have defined 0 as the zeroth square number. I think this use of zeroth is accepted and makes the article consistent. PrimeHunter (talk) 02:34, 2 March 2008 (UTC)

Square theory
I made this equation back in 7th grade, and I couldn't think of any thing that it applies to in the real world. Could you guys help me.

y^2=x^2+(2x+1)+(2x+3)+(2x+5)+...+(2y-1)

This equation will transfer x^2 into y^2 if x is a positive whole number and y is a positive whole number larger than x.

By the way, I'm still in 7th grade. —Preceding unsigned comment added by SuperCockroach (talk • contribs) 19:13, 13 September 2008 (UTC)

Sum of consecutive odds
Perfect squares are also the sum of a number of consecutive odd integers starting at 1. The number of odd integers to be summed is the value of their root. For example, 1=1, 1+3=4, 1+3+5=9, 1+3+5+7=16, 1+3+5+7+9=25, etc. Shouldn't this be in the article somewhere? --214.13.130.100 (talk) 13:48, 12 January 2009 (UTC)


 * It's already mentioned clearly in Square number. PrimeHunter (talk) 15:03, 12 January 2009 (UTC)