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Re:  math reference
Posted:
Aug 10, 2011 5:44 PM


Tim Little wrote:
> On 20110808, 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  (n1)^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)(ab) .
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)(ab); a,b in Z:
N := a^2  b^2 = (a+b)(ab) ; 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 = (ab);
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}^{a1}{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}^{kj}{2j+2i+1} = (kj+1)*2j + \sum_{i=0}^{kj}{2i+1} = (kj+1)*2j + (kj+1)^2
If (kj) 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 (kj) is odd, then (kj+1) is even, then (kj+1)*2j is 0 (mod 4) and (kj+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



