Help me with a proof...

Syth

Can every square be written as the difference between 2 squares?

Ie for all x>2 in N, does there exist a, b in N st x² = a² - b² ?

It's true when x is odd. Just write all the natural numbers down, their squares under them, and the difference between any 2 consequetive squares under then eg:

1   2   3   4    5    6    7    8 
1   4   9  16  25  36  49  64
   3   5  7   9    11  13  15

As you can see the odd numbers are the difference between squares, so if you pick an odd number it can be written as the difference between 2 squares, but if you square an odd number, you still have an odd number, and so it can be written as the difference between 2 squares. I had figured out the algebraic formula for an odd number written as 2 squares, but I lost it.

I haven't been able to prove it for the even numbers yet. I programmed my computer to search loads of numbers to find one can't be written as 2 squares, but no luck.

Anyone got an ideas?

dccarm

They can, if you use the zero. Add a line below the example you gave showing the difference between every SECOND square. You get 8,12,16,20,24,... As all the even square numbers are divisible by 4 (2^2), they too must be part of this sequence of numbers. I haven't a proof, but I'd imagine you could do it by induction (it's years since I proved anything by induction)

dccarm

It shouldn't be too difficult. You know it works for odd numbers, and every even number is just two times an odd number. So if x is even and equal to 2y where y is odd, then x^2 must equal 4y^2. And you know y^2 can be expressed as the difference between two squares.

It's not a proof, but it's in the right direction of one. I'm sure someone with a more recent training can help you.

Syth

The plan was to do it without zero, It's be too easy with zero.

every even number is just two times an odd number

Not all. Numbers that can be expressed as powers of 2, ie 8 can't be written as twice an odd number. But 8² = 10² - 6².

Yes every even square can be written as 4 times something, but it must be (2n)² = 4n².
When you subtract 2 squares, that're only 2 appart you get:
(n+2)² - n² = n²+4n+4-n² = 4n+4 = 4(n+1) =/= 4n².

If you subtract the second squares, the first number you get is 8 and 8 isn't a square, so 8 isn't in the sequence of even squares, so I don't think you're idea would work.

Sev

Ok, I just thought this through now.. so there could be some gross logical errors...

Well...

     x²	= a² - b²

  (2x)²	= 4x²
	= 4a² - 4b² 
	= (2a)² - (2b)²

Therefore, if any number squared can be written as the difference of two squares, then the square of twice that number can be written as the difference of two squares.

For example.. we can show the case is true for the number 4

	4² = 5² - 3²
	16 = 25 - 9          [True]

Therefore by my proof the case is true for 2x4 = 8

=>	[2(4)]²	= [2(5)]² - [2(3)]²
=>	8²	= 10² - 6² = 64

	64	= 100 - 36        [True]

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So.. you say that you can prove that the case is true for every odd number.

Well then I say even numbers come in two flavours.

1. The ones that can be represented as (2x2x.....x2)x(odd number) ie. (odd number) x 2^n

or

2. The ones that are written plainly written as powers of two. ie. 2^n

Well the first case is covered because by your proof the equation holds for odds, and by my proof the equation holds for twice odds, also by my proof, the equation also holds for twice twice odds, twice twice twice odds and so on.. and you could construct an inductive proof like so.

As for the second case, we can forget about the case for 2, because thats not covered by the proposition. I have already shown that the case is true for '4' and by my above proof, the case is true for twice 4, then twice twice 4, then twice twice twice 4.. etc. to infinity. just like the previous.

ecksor

Any chance you could EDIT instead of DELETE/REPOST so that all the Mods don't get 20 mails for each of your posts?

Sev

Sorry... a bit of the old obsessive compulsive disorder. I delete and repost to turn ^2's into ²'s among other stupid and pointless, obsessive amendments.

I didnt know you get emailed each time somebody posts on this forum.

Syth

Right, I've re-worked out the odd case, and it's easy to see that
(2n+1)² = (2n²+2n+1)² - (2n²+2n)²

I also worked out that if you have an even number that has at least one odd factor (excluding 1), then it can be written as (2^p)*y where y is an odd number and p is an integer >= 1, and y² = a² - b². Then:
((2^p)*y)² = 2^(2p)*y² = 2^(2p)*(a² - b²) = 2^(2p)*a² - 2^(2p)*b² = ((2^p)*a)² - ((2^p)*b)²
Thus the square of any even number with at least one odd factor can be written as the difference between 2 squares.

As for the 2^n cases, your proof seems to do it. I didn't think that you could just figure out for on case, say n=2 => 2^n = 4 and then just double it each time.

That's cool thanks guys.

Hmmm, I wonder where else this could be taken... Could you write a square as the sum of 2 sqaures? 3 squares? Some squares can be written as the difference of more than 2 squares, ie 8² = 10² - 6², and 8² = 17² - 15². What's the most different ways you can write a square? What about cubes? Man, the posibilities are endless...

dccarm

Yes every even square can be written as 4 times something, but it must be (2n)² = 4n².
When you subtract 2 squares, that're only 2 appart you get:
(n+2)² - n² = n²+4n+4-n² = 4n+4 = 4(n+1) =/= 4n².

If you subtract the second squares, the first number you get is 8 and 8 isn't a square, so 8 isn't in the sequence of even squares, so I don't think you're idea would work. [/B]

This is what I was trying to get at and you skipped right past it.

If n is even, then by its very nature n^2 MUST be divisible by 4. (n^2 = 4y^2 where y is an integer = n/2)

By your own workings, the difference between the square of n and (n+2) is 4(n+1) which gives the sequence 4,8,12,16,... all divisible by 4

For all even n, there are two integers [(n^2)/4]-1 and [(n^2)/4]-3 which when squared have a difference of n^2. If you multiply it out you'll see it's correct.

Syth

This is what I was trying to get at and you skipped right past it.

Yeah looking back at it, I see that my logic wasn't good/logical

But when I tried to sqaure out those 2 numbers you gave me, it didn't work out. Could you go through it?

Although the theorem is already proved, it's interesting to prove it in many ways.

dccarm

Sorry - I really should think before I type! The two integers should be

[(n^2)/4 +1] and [(n^2)/4 - 1]

When you square them out, you get

[(n^4)/16 + (2n^2)/4 +1] - [(n^4)/16 - (2n^2)/4 +1]

the n^4 and the 1's cancel out, leaving

(2n^2)/4 + (2n^2)/4 = n^2 as required.

And this holds for all n^2 divisable by 4 ie all even n.