1.Introduction

I found some pretty identities such as 1= (4x^4 + 1)^4 - (4x^4)^4 - (4x^3)^4 - (2x)^4 - (3x^2)^4 - (2x^2)^4 + (x^2)^4.

1 is expressed in the sum/difference of seven integral 4th powers!

1 = (4x^5 - x)^4 - (4x^5 + x)^4 + (4x^4 + 1)^4 + (4x^4)^4 
  - (4x^3)^4 - (3x^2)^4 + (2x^2)^4 + (x^2)^4 - (2x)^4.

1 is expressed in the sum/difference of nine integral 4th powers!

I added it in the case of the square and the cubic, too.

Proof is as very easy.

2.Theorem
      
      x:integer
     
　 　1 = (4x^4 + 1)^4 - (4x^4)^4 - (4x^3)^4 - (2x)^4 - (3x^2)^4 - (2x^2)^4 + (x^2)^4.

     1 = (4x^4 - 1)^4 - (4x^4)^4 + (4x^3)^4 + (2x)^4 - (3x^2)^4 - (2x^2)^4 + (x^2)^4.

     1 = (4x^5 - x)^4 - (4x^5 + x)^4 + (4x^4 + 1)^4 + (4x^4)^4 
       - (4x^3)^4 - (3x^2)^4 + (2x^2)^4 + (x^2)^4 - (2x)^4.

     1 = (4x^5 - x)^4 - (4x^5 + x)^4 + (4x^4 - 1)^4 + (4x^4)^4
       + (4x^3)^4 - (3x^2)^4 + (2x^2)^4 + (x^2)^4 + (2x)^4.


     1 = (2x^2 + 1)^2 -(2x^2)^2 -(2x)^2

     1 = (2x^2 - 1)^2 -(2x^2)^2 +(2x)^2
   
     1 = (3x^3 + 1)^3 - (3x^3)^3 -(3x^2)^3 -(2x)^3 -(x)^3

     1 = (3x^3)^3 -(3x^3 - 1)^3  -(3x^2)^3 +(2x)^3 +(x)^3



     
Proof.

(ax^4+1)^4 - (ax^4)^4 = 4a^3x^12+6a^2x^8+4ax^4+1.............................(1)

Set a=4,then right hand side of (1) becomes to 256x^12+96x^8+16x^4+1.........(2)

Substitute 96=3^4+2^4-1 to (2) and simplify,we get

1 = (4x^4 + 1)^4 - (4x^4)^4 - (4x^3)^4 - (2x)^4 - (3x^2)^4 - (2x^2)^4 + (x^2)^4.

Similarly,we can get remaining identities.

 
Q.E.D.