1.Introduction

Ramanujan gave the identity (4x5-5x)4 + (6x4-3)4 + (4x4+1)4 - (4x5+x)4 - (2x4-1)4 = 81.
(http://mathworld.wolfram.com/DiophantineEquation4thPowers.html)

I found a similar identity as Ramanujan's one and several related identities.



2. Result
          
         I show some results under the condition c2 < 100.

         (4x5-5x)4 + (6x4-3)4 + (4x4+1)4 - (4x5+x)4 - (2x4-1)4 = 81 (Ramanujan)

         (4x5-x)4 + (6x4+3)4 + (4x4-1)4 - (4x5+5x)4 - (2x4+1)4 = 81
          
         (2x5-4x)4 + (3x4-3)4 + (2x4-1)4 - (2x5-x)4 - (x4+1)4 + (3x)4 = 81

         (2x5-2x)4 + (3x4-1)4 + (2x4+1)4 - (2x5+x)4 - (x4-1)4 - (2x)4 + x4 = 1
       
         (2x5)4 + (3x4+2)4 + (2x4)4 - (2x5+3x)4 - (1x4)4 - (2x)4 + x4 = 16



3. Method

   

   (2a1x5-b1x)4+(3a1x4+c2)4+(2a1x4+c3)4-(2a1x5+b2x)4-(a1x4-c3)4
   =(96a14-32b2a13-32b1a13)x16
   +(36c3a13+24b12a12-24b22a12+108c2a13)x12
   +(-8b13a1+18c32a12-8b23a1+54c22a12)x8
   +(b14+12c33a1-b24+12c23a1)x4
   +c24....................................................(1)

   f16:coefficient of x16.
   f12:coefficient of x12.
   f8 :coefficient of x8.
   f4 :coefficient of x4.


   Let solve the simultaneous equation of (f16=0,f12=0,f8=0).   
   We obtain the two nontrivial solutions. 
   First solution is  a1 = -1/2c2+1/2c3, b1 = -3/2c2+1/2c3, b2 = c3.  
   Second solution is a1 = 1/2c2-1/2c3, b2 = -1/2c3+3/2c2, b1 = -c3.


   Case 1: f4=0

   Set c2=-3c3, then f4=0. 

   First solution gives the Ramanujan's one (4x5-5x)4+(6x4-3)4+(4x4+1)4-(4x5+x)4-(2x4-1)4 = 81.
   
   Second solution gives the (4x5-x)4 + (6x4+3)4 + (4x4-1)4 - (4x5+5x)4 - (2x4+1)4 = 81.
  
 
   Case 2: f4 <> 0

   Using the first solution, we obtain an identity.

   ((-c2+c3)x5+(3/2c2-1/2c3)x)4 + ((-3/2c2+3/2c3)x4+c2)4 + ((-c2+c3)x4+c3)4
   - ((-c2+c3)x5+c3x)4 - ((-1/2c2+1/2c3)x4-c3)4 - ((3/2c3-1/2c2)x)4 + (c2x)4
   =c24

   This identity is a parameter solution of A14 + A24 + A34 - A44 - A54 - A64 + A74 = n4,too.
   At the case (c2,c3)=(-3,-1),we obtain 

   (2x5-4x)4 + (3x4-3)4 + (2x4-1)4 - (2x5-x)4 - (x4+1)4 + (3x)4 = 81

   At the case (c2,c3)=(-1,1),we obtain 

   (2x5-2x)4 + (3x4-1)4 + (2x4+1)4 - (2x5+x)4 - (x4-1)4 - (2x)4 + x4 = 1
 
   
  
4. Example

       (4x5-x)4 + (6x4+3)4 + (4x4-1)4 - (4x5+5x)4 - (2x4+1)4 = 81

       1<=x<=20

       x       
 
       1        34 +       94 +       34 -       94 -       34 = 81
       2      1264 +      994 +      634 -     1384 -      334 = 81
       3      9694 +     4894 +     3234 -     9874 -     1634 = 81
       4     40924 +    15394 +    10234 -    41164 -     5134 = 81
       5    124954 +    37534 +    24994 -   125254 -    12514 = 81
       6    310984 +    77794 +    51834 -   311344 -    25934 = 81
       7    672214 +   144094 +    96034 -   672634 -    48034 = 81
       8   1310644 +   245794 +   163834 -  1311124 -    81934 = 81
       9   2361874 +   393694 +   262434 -  2362414 -   131234 = 81
      10   3999904 +   600034 +   399994 -  4000504 -   200014 = 81
      11   6441934 +   878494 +   585634 -  6442594 -   292834 = 81
      12   9953164 +  1244194 +   829434 -  9953884 -   414734 = 81
      13  14851594 +  1713694 +  1142434 - 14852374 -   571234 = 81
      14  21512824 +  2304994 +  1536634 - 21513664 -   768334 = 81
      15  30374854 +  3037534 +  2024994 - 30375754 -  1012514 = 81
      16  41942884 +  3932194 +  2621434 - 41943844 -  1310734 = 81
      17  56794114 +  5011294 +  3340834 - 56795134 -  1670434 = 81
      18  75582544 +  6298594 +  4199034 - 75583624 -  2099534 = 81
      19  99043774 +  7819294 +  5212834 - 99044914 -  2606434 = 81
      20 127999804 +  9600034 +  6399994 -128001004 -  3200014 = 81