1.Introduction

I found two pretty identities such as 2 = (8x^5-2x)^4 + (8x^4+1)^4 + (8x^4-1)^4 - (8x^5+2x)^4 - (4x^2)^4.
2 is expressed in the sum/difference of five integral 4th powers!

2 = (32x^5-4x)^4+(32x^4+1)^4+(32x^4-1)^4-(32x^5+4x)^4 -(32x^4)^4+(8x^2)^4.  
2 is expressed in the sum/difference of six integral 4th powers!

Proof is as very easy.

2.Theorem
      
      x:integer
     
　　 2 = (8x^5-2x)^4 + (8x^4+1)^4 + (8x^4-1)^4 - (8x^5+2x)^4 - (4x^2)^4.

     2 = (32x^5-4x)^4 +(32x^4+1)^4 +(32x^4-1)^4 -(32x^5+4x)^4 -(32x^4)^4 +(8x^2)^4.
     
Proof.

(ax^5+bx)^4 + (ax^4+1)^4 + (ax^4-1)^4 - (ax^5-bx)^4 ..............................................(1)
=(2a^4+8ba^3)x^16+(12a^2+8b^3a)x^8+2

When (a,b)=(8,-2),2a^4+8ba^3=0 and 12a^2+8b^3a=4^4.

Substitute (a,b)=(8,-2) to (1),and obtain　

2 = (8x^5-2x)^4 + (8x^4+1)^4 + (8x^4-1)^4 - (8x^5+2x)^4 - (4x^2)^4.

When (a,b)=(32,-4),we obtain

2 = (32x^5-4x)^4+(32x^4+1)^4+(32x^4-1)^4-(32x^5+4x)^4 -(32x^4)^4+(8x^2)^4.



Here is one more similar identity,

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

(1+2x^4)^4-(1-2x^4)^4 = 16x^4+64x^12.
 
Substitute x=1 to above identity,then we get 64=3^4-2^4-1^4.

Again,substitute 64=3^4-2^4-1^4 to above identity,we get first identity. 

Q.E.D.