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Topic: ----------- math reference
Replies: 3   Last Post: Aug 10, 2011 6:29 PM

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 Ulrich D i e z Posts: 34 Registered: 1/2/10
Re: ----------- math reference
Posted: Aug 10, 2011 5:44 PM

Tim Little wrote:

> On 2011-08-08, Deep <deepkdeb@yahoo.com> wrote:
> > I would appreciate it very much if someone can refer me to an
> > appropriate reference where following statement is justified.
> >
> > Statement: Every integer N that is not 2mod4 can be expressed as the
> > difference of two squares.

>
> No reference necessary.
>
> For odd numbers: 2n+1 = (n+1)^2 - n^2.
> For numbers divisible by 4: 4n = (n+1)^2 - (n-1)^2.
>
> They may also have other representations, but these suffice to justify
> the statement.

How about justifying a slight variant of the above statement:

Statement: Every integer N that is 2 (mod 4) cannot be
expressed as the difference of two squares.

Ulrich

One has the identity: a^2 - b^2 = (a+b)(a-b) .

That means the difference N of the squares of two integral numbers
a and b can always be written as a product of the form (a+b)(a-b);
a,b in Z:

N := a^2 - b^2 = (a+b)(a-b) ; a,b \in Z .

If one has a factorization
N = Factor_1 * Factor_2 ; Factor_1, Factor_2 \in Z,
then

Factor_1 = (a+b);
Factor_2 = (a-b);

leads to a = (Factor_1 + Factor_2)/2
and b = (Factor_1 - Factor_2)/2.

Here a and b represent integral numbers only if Factor_1 and
Factor_2 are of the same parity. That circumstance in turn implies
that in that case N is constitutable as a product of two factors of
equal parity which is not possible if N is an odd multiple of 2 / if N is
2 (mod 4). [N = 2 (mod 4) -> N = 4k + 2 = 2*(2k+1) , 2k+1 is odd.]

---------------------------------------------------------------------------

a^2 is the sum of the first abs(a) immediately consecutive
odd natural numbers:

a^2 = \sum{i=0}^{|a|-1}{2i+1}

Thus the absolute value of the difference of two arbitrary different
squares can be constituted by a sum of immediately consecutive
odd natural numbers.

A sum of immediately consecutive odd natural numbers is of
the pattern:

\sum_{i=j}^{i=k}{2i+1} ; 0<= j <= k
= \sum_{i=0}^{k-j}{2j+2i+1}
= (k-j+1)*2j + \sum_{i=0}^{k-j}{2i+1}
= (k-j+1)*2j + (k-j+1)^2

If (k-j) is even, then this expression represents an odd number
and thus a number which is not 2 (mod 4) and where
multiplying by -1 does not yield a number which is 2 (mod 4).

If (k-j) is odd, then (k-j+1) is even, then (k-j+1)*2j is 0 (mod 4)
and (k-j+1)^2 is 0 (mod 4), thus in this case the entire expression
represents a number which is 0 (mod 4) / a number which is not
2 (mod 4) and where multiplying by -1 does not yield a number
which is 2 (mod 4).

In all the cases above, which are all cases, neither the absolute
value of the difference of two arbitrary different squares, nor the
product of that difference and (-1) is 2 (mod 4).

Ulrich

Date Subject Author
8/8/11 Deep Deb
8/8/11 Tim Little
8/10/11 Ulrich D i e z
8/10/11 Ulrich D i e z