# One Plus One

```
1 ---------------------- is.

1 + 1 ------------------ is "the next number after 1" or "the number 2".

1 + 1 + 1 -------------- is "the next number after 1+1" or "the next number after 2" or "is the number 3".

1 + 1 + ......... + 1 -- is "the number a".
a times

a + 1 ------------------ is "the next number after a".

a + 2 ------------------ is "the next number after a+1" or "the 2nd number after a" or "the number a plus 2".

a + b ------------------ is "the b-th number after a".

a + a ------------------ is "the a-th number after a" or "two times a" or "2*a" or "a added together two times".

(a + a) + a ------------ is "three times a" or "3*a" or "a added together three times".

a + a + ......... + a -- is "b*a" or "b times a" or "a added together b times".
b-times

a + a + ......... + a -- is "a*a"  or "a times a" or "a added together a times" or "a^2" or "the number a squared".
a-times

(a + a + ......... + a) + (a + a + ......... + a)  --------------------------------------- is "two times the number a squared" or "2*a^2".
a-times                   a-times

(a + a + ......... + a) + (a + a + ......... + a) + ......... + (a + a + ......... + a) -- is "a*a*b" or (a^2)*b or "b times, a squared".
a-times                   a-times                               a-times
b-times

(a + a + ......... + a) + (a + a + ......... + a) + ......... + (a + a + ......... + a) -- is "a*a*a" or "a^3" or "the number a cubed".
a-times                   a-times                               a-times
a-times

The line just above, it could be repeated "b-times" and each line added to the next.  The result would be b*(a^3) or (a^3) added to itself b times.

--- OR ---

The line just above, it could be repeated "a-times" and each line added to the next.  The result would be a*(a^3) or (a^3) added to itself a times, or a^4.

So one question to ask is: Can you find three numbers so that:
a + b = c
And the answer is yes you can.
The way you compute 'c' is to choose the number 'c' to be equal to the 'b-th' number after 'a'.
Another way to think about this is to think of 'a' as a long string of 1 + 1 + ... + 1 (a times) and in a sumular way
think of 'b' as a long string of '1's ('b' of them).
Just choose 'c' to be string of '1's you get when you concatenate the 'a's string of '1's and 'b's string of '1's.

Another question to ask is: Can you find three numbers so that:
a^2 + b^2 = c^2
And the answer is yes you can.
However, in this case we can not just define c^2 to be the result of a^2 + b^2.
One way to think of this is to visualize a^2 as follows:
1 + 1 + ... + 1 ('a' of these)
1 + 1 + ... + 1 ('a' of these)
.
. ('a' of these lines)
.
1 + 1 + ... + 1 ('a' of these)

In a simular manner, and with out loss of generality (assume b>a), set down the list for b^2.
Next ask the question under what conditions is is possible to take the set of '1's that represent b^2 and
arrange them in a fashion that extendes the square that is created from the set of '1's that represent a^2 and
get a result that is still a square.
Just any old three numbers will not do.  The smallest triple that will do is (3,4,5)
Check that "three squared plus four squared is five squared".
Or to say this differently Check that "9 plus 16 is 25"
Here is the number three squared plus four squared equals five squared represented as a bunch of dots (or 1+1+...+1)
If you think about it each set of dots is a 'square-of-dots'

...   ....   .....
... + .... = .....
...   ....   .....
....   .....

And here are the 'dots' for four squared converted to 'X'es'
and then wrapped around the 'dots' for three squared
and we see a five by five square.

...xx
...xx
...xx
xxxxx

The number of '1's in the 4^2 set fit perfectly around the square in '3^2's set.

So now the question arises.  What is special about 3^2 and 4^2 that allows the group from 4^2 to extend group for
3^2 perfectly so that the group for 5^2 results?

The answer is: Well it is 'tricky'

Another question to ask is: Can three numbers be found so that: a^3 + b^3 = c^3?
If you think about arrays of dots then the question becomes is it possible to extend a cube with bits of another
cube in such a manner as the resulting figure is again a cube?

The answer is no.

And in general is there a set of three numbers such that a^n + b^n = c^n (when n>=3)?

The answer again is no.

```