Even, odd, and square integers (AoT, nr.0)

Notation and terminology:

Different fonts represent different mathematical objects. In particular the bold font  Z  always represents the set of all (rational) integers, see below, while z z z z Z Z Z (can your eye see the optical difference between different fonts? -- don't worry! ☺) they represent some other objects; furthermore any of these other letters may represent different object on different occasions as it would be stated explicitly each time.

Z  =  { ... -2  -2  0  1  2 .., } -- the set of all (rational) integers;

N  :=  { n \in Z : n > 0 }  =  { 1  2  3  ... } -- the set of all natural numbers (positive integers).

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Introduction
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Let's add the first six positive (natural) integers, and we get:

         1 + 3 + 5 + 7 + 9 + 11  =   6^2

This holds for any number of the first odd numbers. Add  12 of them and you'll get 144 = 12^2, etc.

Equation

              1 + 3 + ... + (2*n-1)  =   n^2

is among the most elementary and elegant formulas.

Even, odd, and square integers hide more secrets like this. Mathematics is a mine of elegant formulations. It takes some effort to discover them -- so much better.

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Even and odd integers
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The set  Z  of all integers,

         Z := { ... -2  -1  0  1  2 ...}

splits into even and odd:

        Ev = { ... -4  -2  0  2  4 ... }        &        Odd = { ... -3  -1  1  3 ... }

Formally:

        Ev := {2*n : n \in Z}        &        Odd := { 2*n-1 : n \in Z} 

Of course:       Odd = { 2*n-1 : n \in Z} = { 2*n+1 : n \in Z}.

We see that sets  Ev and Odd  are disjoint (i.e. they do not have a common element; in other words, an integer cannot be both even and odd).

The following three statement about any integer  A  are equivalent:

1.  integer  A is odd (i.e. A \in Odd);
2.  (A-1)/2  is an integer;
3. (A+1)/2  is an integer.

Observe that the last two integers are consecutive,

                    (A+1)/2  =  (A-1)/2 + 1

i.e.

                    (A+1)/2  -  (A-1)/2  =  1

Also

                    (A+1)/2  +  (A-1)/2  =  A

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Integer squares and positive integer squares
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The set  Sq  of all integer squares is

                    Sq = { 0   1   4   9   16   25   36 ...} = { 0^2   1^2   2^2   3^2   4^2   5^2   6^2   ... }

while positive squares  PSq  miss  0:

                   PSq  :=  Sq \ {0}  =  { 1   4   9   16   25   36 ...}

or explicitly:

                   

                  Sq  :=  { n^2 :  n\in Z }           &               PSq := { s \in Sq : s > 0 }

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Terminology:   the set of all natural integers (or natural numbers)  N  is defined as follows:

                                    N  :=  { n \in Z :  n > 0 }

We see that

                                    N  =  { 1 2 3 ... }

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Now we may write:

                     PSq  =  { b^2 : n \in N }

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A mathematician cannot exist without the following elementary formulas:

                   (a+b)^2   =   a^2  +  2*a*b  +  b^2

                   (a-b)^2   =   a^2  -  2*a*b  +  b^2

                       a^2 - b^2  =  (a-b)*(a+b)

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Integers as the difference of two squares
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Above, we have already encountered formula

                               (A+1)/2  +  (A-1)/2  =  A

When A is an odd integer then the above three terms  (A+1)/2  and  (A-1)/2  and  A  are all integers. Now we can involve squares:

                              ((A+1)/2)^2  -  ((A-1)/2)^2  =  A

Neat!  We see that every odd  A  is the difference of the squares of two consecutive integers (A+1)/2  and  (A-1)/2.

Are even integers differences of two squares? -- you can convince yourself that  2  is NOT while 8 is:

                                        8  =  3^2 - 1^2

Also, going back to odd numbers, we see a multiple representation example:

                                   15   =   8^2 - 7^2   =   4^2 - 1^2

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The parity of an integer  a  and of it's square  a^2  is the same, either both are even or both are odd. The same for integers  b  and. b^2. Thus,

    a^2-b^2  is odd whenever integers  a and b  are of different parities.

In general,  a^2-b^2 = (a-b)*(a+b);  when  a and b  are integers then  a-b  and  a+b  are of the same parity (both the difference  a-b and the sum  a+b are integers of the same parity). Thus,

    a^2-b^2  is divisible by 4   whenever integers  a and b  are of the same parities

It follows that  a^2-b^2, when  a and b. are integers, is either oddd or divisible by 4 (but never

divisible by 2 without being divisible by 4).

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In the opposite direction:

Question:  Let integer  C  be odd or divisible by 4.  Do there exist integers  a b  such that  S = a^2-b^2 ?

Let's answer it step by step., well, in two steps.

(i) Assume that integer  C  is divisible by 4. Then  C-1  is even. We obtain a decomposition of  C  into a difference  of two integer squares:

                             C  =  ((C+1)/2)^2 - ((C-1)/2)^2

(ii) Assume that integer  C  is divisible by  4.  Then  (C + 4)/4  and  (C - 4)/4  are integers such that:

                             C  =  ((C + 4)/4)^2 - ((C - 4)/4)^2

 We proved:

Theorem:  Every integer C that is odd or divisible by 4  admits a decomposition into a difference

                                 C = a^2-b^2

where. a b are integers.

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Now we can see justify the equation

       1 + 3 + 5 + 7 + 9 + 11  =   6^2

from the Introduction:

       1 + 3 + 5 + 7 + 9 + 11   =   11 + 9 + 7 + 5 + 3 + 1

   =  (6^2-5^2)  +  (5^2-4^2)   +  (4^2-3^2)  +  (3^2-2^2)  +  (2^2-1^2)   +  1^2

                       =  6^2

People like to call  expressions like the middle line above a telescope.